Блог Анкара

In This Chapter

^ Settling down with the concept of chemical equilibrium

^ Using Le Chatelier to find your easy chair

^ Measuring reactions at rest with equilibrium constants

Equilibrium . . . it’s a pleasant-sounding word. What does it mean? In everyday language, equilibrium connotes calmness and composure, or some state of balance. The chemical definition of equilibrium most closely mirrors the last of these meanings: balance. In this chapter, we describe how the idea of balance applies to chemical reactions.

Shifting Matter: An OverView of Chemical Equilibrium

On the microscopic scale, most reactions can occur in both the "forward" and "reverse" directions. When we say that a reaction is "going forward," what we mean is that the rate of the forward reaction is larger than the rate of the reverse reaction (see Figure 15-1); in other words, at any given moment more reactant mass converts into product than there is product mass converting back into reactant. Remember, chemical reactions simply convert a given amount of mass between states.

Given enough time, a closed chemical system (one protected from outside influences) reaches a state in which the rate of the forward reaction equals the rate of the reverse reaction, as shown in Figure 15-2. On the microscopic scale, chemistry occurs constantly in both directions, but on the larger, "macroscopic" scale, the reaction appears to be "at rest." We call this state of chemical balance Dynamic equilibrium.

Figure 15-1:

Equilibrium occurs when forward and backward rates are equal.

Reaction Going TO rward":

Reaction Going "Backward":

Reaction in Dynamic Equilibrium:

R ( > P R < ‘-P

Figure 15-2:

Chemical reactions achieve equilibrium over time.

Time

Once a chemical system achieves equilibrium, the concentrations of reactants and products no longer change (macroscopically, at least). The resulting combination of concentrations is called an Equilibrium mixture, And can be used to characterize the reaction, as will be shown in a later section. Whether the equilibrium mixture contains more products or more reactants (and to what extent) depends on differences in energy between reactants and products. The details of this energy-equilibrium relationship are described in detail in Chapter 21.

Predicting Shifts with Le Chatelier’s Principle

After a chemical system has reached equilibrium, that equilibrium can be disrupted, or Perturbed. Think of systems at equilibrium as people who have finally found their easy chair at the end of a long day. You may rouse them to take out the trash, but they’ll return to the easy chair at the first opportunity. This concept is more or less the idea behind Le Chatelier’s Principle: The equilibrium of a perturbed system shifts in the direction that opposes the perturbation. Perturbations include changes in the following:

Concentration: If a system is at equilibrium, adding or removing reactant or product disrupts the equilibrium. Roused from its easy chair, the equilibrium reasserts itself in response.

• If reactant is added or product is removed, reactant converts into product.

• If product is added or reactant is removed, product converts into reactant.

Either way, chemistry occurs until concentrations are those of the equilibrium mixture once more. In other words, the equilibrium shifts to oppose the perturbation, as shown in Figure 15-3.

Pressure: Reactions that include gases as reactants and/or products are particularly sensitive to pressure perturbation. If pressure is suddenly increased, equilibrium shifts toward the side of the reaction that contains fewer moles of gas, thereby decreasing pressure. If pressure is suddenly decreased, equilibrium shifts toward the side of the reaction that contains more moles of gas, thereby increasing pressure. Consider the following reaction:

N2(g) + 3H2(g) <rҐ 2NH3(g)

A given amount of mass on the reactant side of the equation (as N2 and H2) corresponds to double the moles of gas as the same mass on the product side (as NH3). Imagine that the system is at equilibrium at a low pressure. Now imagine that the pressure suddenly increases, perturbing that equilibrium. Reactant (N2 and H2) converts to product (NH3) so the total moles of gas decrease, thereby lowering the pressure.

If the system suddenly shifts to lower pressure, NH3 converts to N2 and H2 so the total moles of gas increase, thereby raising the pressure. The equilibrium shifts to oppose the perturbation. These responses are summarized in Figure 15-4.

Temperature: Reactions that absorb or give off heat (that is, most reactions) can be perturbed from equilibrium by changes in temperature. The easiest way to understand this behavior is to explicitly include heat as a reactant or product in the reaction equation:

A + B C + heat

Imagine that this reaction is at equilibrium. Now imagine that the temperature suddenly increases. Product C absorbs heat, converting to reactants A and B. Because the heat "product" has been decreased, the temperature of the system decreases.

If the temperature suddenly shifts down from equilibrium, reactants A and B convert to product C, releasing heat. The released heat increases the temperature of the system. The equilibrium shifts to oppose the perturbation. These responses of the equilibrium to changes in temperature are reviewed in Figure 15-5.

Figure 15-3:

An equilibrium shifts to oppose changes in concentration of reac-tants or products.

Adding reactants

Reactants

Removing reactants

Removing products

Products

Adding products

Figure 15-4:

An equilibrium that includes gases shifts to oppose changes in pressure.

Increasing pressure

N2(g + 3H2(g)

22

2NH3(g)

Decreasing pressure

Figure 15-5:

An equilibrium that includes heat as a reactant or (as shown) as a product shifts to oppose changes in temperature.

Decreasing temperature

A + B

C + heat

Increasing temperature

What is the effect of adding a catalyst on an equilibrium mixture? Precisely nothing. Catalysts may help speed the journey to the equilibrium mixture, but they have no effect on its composition.

What if the reaction starts on one "side" of the equilibrium mixture as opposed to the other side? In other words, what if the same reaction starts from an "all reactants" state versus an "all products" state? Given time, each reaction will find its way to the same equilibrium mixture. In science, that kind of beautiful symmetry is known as a Satisfying Result.

So, by using Le Chatelier’s Principle, you can predict the direction in which an equilibrium will shift in response to a perturbation. But how far will the equilibrium shift? And where exactly was the equilibrium "resting" in the first place? The answers to these quantitative questions involve a new quantity: the equilibrium constant. You can find more information on the equilibrium constant in the following section.

Measuring Equilibrium: The Equilibrium Constant

In a previous section, we mentioned that the composition of an equilibrium mixture (that is, the concentrations of the reactants and products when the system is at equilibrium) could be used to characterize the reaction. In this section, we discuss the connection between concentrations and equilibrium in great detail.

Understanding the equilibrium constant, Keq

In essence, the greater the equilibrium concentration of products relative to reactants, the more spontaneous is the reaction. Spontaneous (or "favorable") reactions favor products. Nonspontaneous (or "unfavorable") reactions favor reactants. To quantitate the tendency of a reaction to proceed toward products, we use the Equilibrium constant, Keq. For the following reaction

BEll

AA + bB cC + dD the equilibrium constant is calculated as follows:

[ C ]C [ D ]d

In general, concentrations of the products are multiplied together in the numerator. Concentrations of the reactants are multiplied together in the denominator. Stoichiometric coefficients show up as exponents on their corresponding reactants or products. Once actual values (that is, concentrations from an equilibrium mixture) are entered for each concentration and coefficient, the whole expression reduces to a single number, one ranging from zero to infinity. Real-life values always occupy the middle ground between these two extremes. Whatever it is, that final number has meaning.

Very favorable reactions produce a lot of product, so they have Keq values much larger than 1. Very unfavorable reactions convert very little reactant into product, so they have Keq values between 0 and 1. In a reaction with Keq = 1, the amount of product equals the amount of reac-tant at equilibrium.

Note that you can only calculate Keq by using concentrations measured at equilibrium. Concentrations measured before a reaction reaches equilibrium can be used to calculate a Reaction quotient, Q:

[ c ic [ d r , ,

Q = ■=—~iarz, b ( Nonequilibrium) [ A1 [ B ]

If Q < Keq, the reaction will progress "to the right," making more product. If Q > Keq, the reaction will shift "to the left," converting product into reactant. If Q = Keq, the reaction is at equilibrium. Because equilibria are temperature dependent, the Keq at one temperature may be the Q At another temperature.

You may have noticed that the expression given for calculating the Keq lists each reactant or product within brackets, as in [X]. If those brackets look unfamiliar, you’ll want to review Chapter 11, where you’ll learn that they indicate the molar concentration of a solute, M = mol L1.

But what if you’re dealing with a gas phase reaction, like the one pictured in Figure 15-4? Well, in that case you’ll work with partial pressures of the reactant and product gases (see Chapter 9), and the expression for Keq is:

=

* eq

Fine, you say, but what if my reaction includes some combination of liquids, gases, and liquid or gas solutions? And what about solids? Relax, there’s an answer—but it comes at the price of some new vocabulary:

Homogenous equilbria Contain reactants and products that are all of the same phase. These are the easiest cases because you can simply apply the Keq Expression directly, as already described.

Heterogeneous equilibria Contain reactants and products of different phases. You can still calculate KEq for such reactions, but you’ve got to follow a few simple rules:

Pure liquids and pure solids are omitted from the Keq expression of a heterogeneous equilibrium; these concentrations simply equal 1. This includes the solvent (such as H2O in the case of aqueous solutions), even if solvent molecules are directly involved in the chemistry of the reaction. The solvent is present in such high concentration that its involvement as a reactant or product doesn’t change its concentration.

I Partial pressures (P) Of reactant or product gases and molar concentrations (M) Of reactant or product solutes can appear together in the KEq expression, because their concentrations may change significantly during the course of a reaction.

Using Keq to solve problems

Some equilibrium problems don’t simply list final, equilibrium concentrations, but instead give you a combination of initial concentrations, final concentrations, and possibly the KEq itself. At this point, you may have developed that sinking feeling in your stomach that suggests that all this can get pretty complicated. Not to worry; you just need to be organized. Follow these steps for dealing with equilibrium problems that confuse you:

1. Start with a balanced equation for the reaction, including the phases of the reactants and products.

For example

NH4Cl(s) NH3(g) + HCl(g)

2. Check the phases to see if you’re dealing with a homogenous equilibrium or a heterogeneous equilibrium. If you’ve got a heterogeneous equilibrium on your hands, cross out any pure (unmixed) liquid or solid substances to ensure that you omit them from the final KEq expression. Continuing with the example introduced in step 1:

NH4Cl(s) NH3(g) + HCl(g)

3. Set up the Keq Expression, including the actual Keq value, if known (0.036 in this case):

Keq = (P NH3)(P HCl) = 0.036

4. Underneath each remaining reactant or product, list the following quantities, in order: initial concentration/partial pressure, change in concentration/partial pressure and equilibrium concentration/partial pressure. Leave unknown quantities blank.

The resulting table might look like this:

NH4Cl(s) NH3(g) + HCl(g)

Initial: 1 0atm 0.072atm

Change: 1 +xatm +xatm

Equilibrium: 1 xatm 0.072 + xatm

This kind of table is known as an ICEbox, drawing on the initials of the quantities in the three rows: Initial, Change, and Equilibrium. How exactly you fill in the ICEbox depends on which values are known and which are unknown. In the given example, the partial pressures of both product gases were known. By examining the balanced equation, it is clear that any changes in NH3 and HCl products occur in a one-to-one ratio, because each product has a stoichiomet-ric coefficient of 1. So, the unknown change in partial pressure (x) experienced by the gases en route to equilibrium is the same for each gas. Having filled in the blanks for Initial and Change, the Equilibrium values are calculated as the sum of the two. Just substitute the "Equilibrium" values into the KEq expression and solve for the unknown:

KEq = (x)(0.072 + x) = 0.036

X2 + 0.072x – 0.036 = 0 (a quadratic equation) x = 0.16 OR x = -0.23

Solving the quadratic equation gives two possible values for x. We can exclude the -0.23 solution because it results in negative values for the partial pressures, which is a physical impossibility. So, x = 0.16atm is the solution. Therefore, the equilibrium mixture of the reaction contains 0.16atm NH3 and 0.072 + 0.16 = 0.23atm of HCl.

Here are a few additional tips for working with equilibrium constants:

I Equilibrium constants for the "forward" and "reverse" directions of the same reaction are simply inverses of each other:

If. . . Keq for Reactants Products is Kforward

And . . . Keq for Products Reactants is Kreverse

Then . . . Kforward = (Kreverse)-1

IU Multiple equilibria can be connected or "coupled" when a product from one equilibrium is the reactant for another. The equilibrium constant for the overall, coupled reaction is the product of the equilibrium constants for the component reactions:

If. . . Keq for A <rҐ B is KAB

And. . . Keq for B C is KBC

Then . . . Keq for A C is KAC = (KAB)(KBC)

Specializing Equilibrium: Constants for Different Reaction Types

In previous sections of this chapter, we discussed how to calculate the KEq for gas phase reactions by using the partial pressures of the gases in an equilibrium mixture. That’s great stuff, as far as it goes. But consider this: In a constant-volume system, increasing the total concentration of a gas also increases its partial pressure. So what? The upshot of this fact is that calculating the equilibrium constant in terms of pressures can sometimes give a value of Keq different from the one you’d get if you calculated in terms of molar concentrations. To account for this discrepancy, chemists sometimes use the quantities KP and KC to make clear whether a gas phase equilibrium constant was calculated using pressure or molar concentration, repectively.

Is there any reliable relationship between KP and KC? Yes:

KP = KC <rҐ (RT)An

In this equation, the exponent (n Corresponds to the difference in the moles of gas between the product and reactant sides of the reaction equation. For example, in the reaction

2NH3(g) N2(g) + 3H2(g)

The exponent An = nproduct – nreactant = 4 – 2 = 2.

Other kinds of specialized equilibrium constants you’ll encounter are Acid and base dissociation constants, KA and KB, and the Solubility product constant, KSp.

KA and KB are discussed at length in Chapter 17. Here, you need only understand that they are the equilibrium constants that correspond to acid dissociation and base dissociation reactions.

I Strong acids possess large values of KA. I Strong bases possess large values of KB.

KSp is the equilibrium constant for the dissociation reaction of an ionic solid as it dissolves. This constant is discussed in the context of solutions and solute concentrations in Chapter 11. Here, we simply show how KSp fits under the umbrella of "equilibrium constant" and discuss how KSp relates to a phenomenon called the common ion effect.

When an ionic solid (such as silver bomide) dissolves, its cation and anion components separate:

AgBr(s) + H2O(l) Ag+(aq) + Br (aq)

Because AgBr is a pure solid and H2O is the solvent, those two compounds are assigned concentrations of 1, and effectively fall out of the equilibrium constant expression, such that:

Ksp = [Ag+][Br-]

This equation shows that

Iu The Ksp is the Product Of the concentrations of the dissolved ions. Iu Very soluble compounds possess large Ksp values. Iu Very insoluble compounds possess small Ksp values.

The solubility product governs the Common ion effect, A shift in the solubility equilibrium of a dissolved ionic solid upon the addition of a second compound, one that contains an ion in common with the first compound. For example, if sodium bromide, NaBr, is added to a solution of silver bromide, AgBr, a certain amount of dissolved silver bromide shifts back into solid form and precipitates from solution.

Why? Adding sodium bromide results in the addition of Br – to solution. This addition perturbs the silver bromide equilibrium, effectively adding Br- "product." To oppose this perturbation, the equilibrium shifts mass toward the AgBr "reactant" side, as shown in Figure 15-6. How far does the equilibrium shift? How much AgBr precipitates? Answer: AgBr precipitates until the concentrations of Ag+ and Br – in solution once again match the Ksp for silver bromide.

Ion effect. N Addition of

Common ion

In This Chapter

^ Making sure you remember it all ^ Getting in some practice ^ Checking your answers

M Hapter 19 was charged with must-remember lists and tidbits. Oxidation-reduction reaction problems are certain to be a part of your AP exam experience, and they are the favorite villains of many chemistry students. Arm yourself for what you know is coming, and be sure to master the highlights of Chapter 19, listed here, and try your hand at the practice problems.

Remembering the Rules of Engagement

Conquering redox problems involves two major skills. First, you must be able to identify oxidizing and reducing agents so you can follow the flow of charge in a redox reaction. Second, you need to be able to balance redox reaction equations. The following two sections review the key points and processes of these skills.

Following the flow of charge in redox

During a redox reaction, the oxidizing Agent Becomes reduced and the reducing Agent Becomes oxidized.

Oxidation may or may not involve bonding with oxygen, breaking bonds with hydrogen or losing electrons — but oxidation always means an increase in oxidation number.

Reduction may or may not involve bonding with hydrogen, breaking bonds with oxygen or gaining electrons — but reduction always means a decrease in oxidation number.

Rules for figuring out oxidation numbers

• Atoms in elemental form have an oxidation number of zero.

• Single-atom (monatomic) ions have an oxidation number equal to their charge.

• In a neutral compound, oxidation numbers add up to zero. In a charged compound, oxidation numbers add up to the compound’s charge.

• In compounds, oxygen usually has an oxidation number of -2, in which its oxidation number is -1.

• In compounds, hydrogen has an oxidation number of +1 when it bonds to non-metals, and an oxidation number of -1 when it bonds to metals.

• In compounds

Group IA atoms (alkali metals) have oxidation number +1.

Group IIA atoms (alkaline earth metals) have oxidation number +2.

Group IIIA atoms have oxidation number +3.

Group VIIA atoms (halogens) usually have oxidation number -1.

E Oxidation occurs at an anode, which releases electrons that flow to a cathode, where reduction occurs.

E°cell = E°red(cathode) – E°red(anode) and E° = E°red (reduction half-reaction) – E°red (oxidation half-reaction)

E AG = – nFE and the related equation

Log K =7n§°^ Q 0.0592

E AG = A°G + RT ln Q

E Nernst equation

E = E °- RT Ln Q nF

Balancing redox reaction equations

To balance a redox reaction equation for a reaction under Acidic conditions

1. Separate the reaction equation into the oxidation half-reaction and the reduction half-reaction. Use oxidation numbers to identify these component half-reactions.

2. Balance the half-reactions separately, temporarily ignoring O and H atoms.

3. Balance the half-reactions separately, using H2O to add O atoms and using H+ to add H atoms.

4. Balance the half-reactions separately for charge by adding electrons (e-).

5. Balance the charge of the half-reactions with respect to each other by multiplying the reactions such that the total number of electrons is the same in each half-reaction.

6. Reunite the half-reactions into a complete redox reaction equation.

7. Simplify the equation by canceling items that appear on both sides of the arrow.

To balance a redox reaction equation for a reaction under Basic conditions, Perform steps 1 through 7 above, and then add the following steps:

1. Observe where H+ is present in the resulting equation. Add an identical amount of OH-to both sides of the equation such that all the H+ is "neutralized," becoming water.

2. Cancel any amounts of H2O that appear on both sides of the equation.

Testing Your Knowledge

Don’t let your hard work in reading Chapter 19 go to waste — seal redox into your brain by attempting these questions.

Questions 1 through 4 refer to the following reaction: Cl2(g) + Sb(s) — Cl-(aq) + SbO+(aq)

1. Which of the following are the correct oxidation numbers for, in order, Cl in Cl2, Sb, Cl – and the components of SbO+ (ions of Sb and O)?

(A) -1, 0, -1, 0, +1

(B) 0, 0, -1, +1, 0

(C) 0, 0, -1, +3, -2

(D) -1, 0, -1, +3, -2

(E) 0, +3, -1, +3, -2

2. How many electrons are transferred in the reaction?

(A) 8

(B) 6

(C) 4

(D) 2

(E) 0

3. What is the balanced reaction equation under acidic conditions?

(A) 2Sb + 3Cl2 + 2H2O — 2SbO+ + 6Cl – + 4OH-

(B) 2Sb + 3Cl2 + O2 — 2SbO+ + 6Cl – + 4H+

(C) Sb + 3Cl2 + 2H2O — SbO+ + 6Cl – + 4H+

(D) 2Sb + 3Cl2 + 2H2O — 2SbO+ + 6Cl – + 4H+ (C) 3Sb + 4Cl2 + 3H2O — 3SbO+ + 9Cl – + 6H+

4. What is the balanced reaction equation under basic conditions?

(A) 2Sb + 3Cl2 + 4OH – — 2SbO+ + 6Cl – + 2H2O

(B) 2Sb + 3Cl2 + 2H2O — 2SbO+ + 6Cl – + 4H+

(C) 2Sb + 3Cl2 — 2SbO+ + 6Cl-

(D) 2Sb + 3Cl2 + 2H2O — 2SbO+ + 6Cl – + 4OH-

(E) Sb + Cl2 + 2OH – — SbO+ + 2Cl – + 2H2O

5. Solid iron (II) sulfide reacts with atmospheric oxygen to form iron (II) oxide and sulfur dioxide. Which of the following statements are true about the reaction?

I. Sulfur is the reducing agent, oxygen is the oxidizing agent.

II. Sulfur is reduced, oxygen is oxidized.

III. Sulfur transfers electrons to iron and oxygen.

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) II and III only

Questions 6 and 7 refer to the following galvanic cell (Figure 20-1):

Nickel

6. Which of the following statements are true about the spontaneously running electrochemical cell shown? Use the following information and assume standard conditions:

Ag+(aq) + e" —Ag(s) Ј°red = +0.80 V

Ni2+(aq) + 2e" — Ni(s) Ј°red = -0.28 V

I. Nickel is the cathode.

II. Silver is the anode.

III. Electrons flow from nickel to silver.

(A) I only

(B) II only

(C) III only

(D) None are true.

(E) All are true.

7. What is the standard cell potential of the electrochemical cell shown? Use the following information:

(A) +0.52 V

(B) -0.52 V

(C) +1.36 V

(D) -1.08 V

(E) +1.08 V

Questions 8 through 12 refer to the following list of choices:

8. Occurs when electrodes have a potential difference

9. Mole of electron charges

10. Site of reduction

11. Provides relationship between EMF and nonequilibrium concentrations

12. Site of oxidation

13. Which of the following is true of a nonspontaneous reaction?

Questions 14 through 17 refer to the following reaction that takes place within an electrochemical cell:

Solid copper reacts with oxygen gas under acidic conditions to produce copper (II) cation and water.

Cu2+(a<7) + 2e – — Cu(s) E°red = +0.34 V

O2(g) + 4H+(a<7) + 4e – — 2H2O(l) E°red = +1.23 V

14. What is the balanced reaction equation?

(A) 2Cu + O2 — 2Cu2+ + H2O

(B) Cu + O2 + H+ — Cu2+ + H2O

(C) 2Cu + O2 + 4H+ — 2Cu2+ + 2H2O

(D) 2Cu + O2 + 2H+ — 2Cu2+ + H2O

(E) Cu + O + 2H+ — Cu2+ + H2O

15. What is the standard cell potential for the reaction?

(A) -0.89 V

(B) +0.89 V

(C) -1.57 V

(D) +1.57 V

(E) +1.23 V

16. What is the standard free energy change for the reaction? (Note: 1 J = 1 CV, so 1 F = 96,500 J V^mol"1)

(A) +610 kJmol"1

(B) -610 kJmol"1

(C) +150 kJmol"1

(D) -340 kJmol"1

(E) +340 kJmol"1

17. What effect would increasing the acid concentration beyond 1M have on the otherwise standard reaction?

I. It would make the reaction less favorable.

II. It would increase the cell voltage.

III. It would decrease the value of the reaction quotient, Q.

(A) I only

(B) II only

(C) III only

(D) II and III only

(E) I, II and III

Checking Your Work

You made it. Have a snack. Check your answers.

1. (C.) Both Cl2 and Sb are in elemental form, so their atoms have an oxidation number of zero. The Cl – oxidation number is the same as its charge. SbO+ is a bit tricky to unravel, but remember that oxygen has a -2 oxidation number. So, in that compound, Sb must have a +3 oxidation number.

2. (B.) Find out that six electrons are transferred by balancing the reaction. Doing the first six steps of balancing gives you:

Step 1 Sb — SbO+ (oxidation)

Cl2 — Cl- (reduction) Step 2 Sb — SbO+

Cl2 — 2Cl-Step 3 Sb + H2O — SbO+ + 2H+

Cl2 — 2Cl-Step 4 Sb + H2O — SbO+ + 2H+ + 3e-

Cl2 + 2e – — 2Cl-Step 5 2Sb + 2H2O — 2SbO+ + 4H+ + 6e-

3Cl2 + 6e – — 6Cl-Step 6 2Sb + 3Cl2 + 2H2O + 6e- — 2SbO+ + 6Cl – + 4H+ + 6e-

So, six electrons are transferred from Sb to Cl2.

3. (D.) Completing the final step of balancing under acidic conditions gives you

Step 7 2Sb + 3Cl2 + 2H2O — 2SbO+ + 6Cl – + 4H+

4. (A.) Adding steps 8 and 9 to account for basic conditions gives you

Step 8 2Sb + 3Cl2 + 2H2O + 4OH – — 2SbO+ + 6Cl – + 4H2O Step 9 2Sb + 3Cl2 + 4OH – — 2SbO+ + 6Cl- + 2H2O

5. (A.) To figure out this one, just assign oxidation numbers to all the reactants and products:

In iron (II) sulfide, FeS, iron has oxidation number +2, sulfur has oxidation number -2. In O2, which is in elemental form, oxygen has oxidation number 0. In iron (II) oxide, FeO, iron has oxidation number +2, oxygen has oxidation number -2. In sulfur dioxide, SO2, sulfur has oxidation number +4, oxygen has oxidation number -2.

So, the oxidation number of oxygen decreased from reactants to products, while the oxida" tion number of sulfur increased. This means that oxygen was reduced, and therefore is the oxidizing agent. Sulfur was oxidized, and therefore is the reducing agent. So, I is correct and II is not correct. Choice III is wrong because, although sulfur transfers electrons to oxygen, it does not do so to iron.

6. (C.) Electrons flow from the anode to the cathode in an electrochemical cell. This flow of electrons occurs spontaneously from the stronger reducing agent to the weaker reducing agent. Based on the standard reduction potentials given for nickel and silver, it is clear that nickel is the stronger reducing agent — reducing Ni2+ is unfavorable whereas reducing Ag+ is favorable. So, nickel is the anode and silver is the cathode, with electrons flowing from nickel to silver.

7. (E.) Calculate the standard cell potential by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode:

E°cell = E°red(cathode) – E°red(anode)

E°cell = +0.80 V – (-0.28 V) = +1.08 V

8. (C.) Electromotive force, or EMF, is the potential difference across the anode and cathode of an electrochemical cell. Electrons flow from high potential to low potential, just as a boul" der rolls down a hill from a position of high gravitational potential to one of low potential.

9. (D.) One Faraday (F) is the charge associated with one mole (Avogadro’s number, N) of electron charges (e):

F = NAe = (6.022 x 1023 mol-1)(1.602 x 10-19 C) = 9.649 x 104 C mol1

This constant is key to relating the number of electrons transferred during a redox reaction to the free energy change for that reaction.

10. (B.) Reduction occurs at cathodes. Think "red cat."

11. (E.) The Nernst equation combines Faraday’s constant, F, with the reaction quotient, Q, to produce a relationship between the EMF of an electrochemical cell and the nonstandard concentrations of the reactants and products within that cell.

12. (A.) Oxidation occurs at anodes. Think "an ox."

13. (E.) Nonspontaneous reactions do not occur on their own, Zbut require an input of energy. Spontaneous reactions occur on their own, releasing energy. Remember that the free energy change varies with conditions, so all conditions must be known for you to compute the sign

Of AG. Chemical species (that is, reactants and products) move spontaneously from high energy states to low energy states. So, in a spontaneous reaction, the final free energy is lower than the initial free energy. Because AG = Gfinal – Ginitial, AG > 0 for a nonspontaneous reaction, in which the final free energy would be greater than the initial free energy. The equilibrium constant, Keq, For a reaction is the quotient of the product concentration(s) divided by the reactant concentration(s). Nonspontaneous reactions favor reactants, and therefore have Keq < 1. Electromotive force is different from AG in that favorable redox reactions generate a positive electromotive force; this force is what produces current across the cell. So, a nonspontaneous reaction is associated with a negative electromotive force, where E< 0.

14. (C.) Balance the reaction as follows:

Step 1 Cu — Cu2+ (oxidation)

O2 — H2O (reduction) Step 2 (already done) Step 3 Cu — Cu2+

O2 + 4H+ — 2H2O Step 4 Cu — Cu2+ + 2e-

O2 + 4H+ + 4e – — 2H2O Step 5 2Cu — 2Cu2+ + 4e-

O2 + 4H+ + 4e – — 2H2O Step 6 2Cu + O2 + 4H+ + 4e – — 2Cu2+ + 2H2O + 4e-Step 7 2Cu + O2 + 4H+ — 2Cu2+ + 2H2O

15. (B.) First, figure out which is the reduction half-reaction and which is the oxidation half-reaction. This has already been done in Steps 1 through 4 of the answer to the previous question:

Cu — Cu2+ + 2e – (oxidation)

O2 + 4H+ + 4e – — 2H2O (reduction)

Next, recall the formula for calculating the standard, EMF, E°, for a redox reaction from the standard reduction potentials of the half-reactions:

E°= E°red(reduction half-reaction) – E°red(oxidation half-reaction)

E° = +1.23 V – (+0.34 V) = +0.89 V

16. (D.) The relationship between standard free energy change and standard electromotive force is given by

AG° = -nFE°

We know from the balanced reaction equation that four electrons are transferred during the reaction, so N = 4. Faraday’s constant is 96,500 Cmol1 (or J V^mol"1). The standard electromotive force or cell potential (calculated in question 15) is +0.89 V. So,

AG° = -4 ( (96,500 J V-1mol-1) x (0.89 V) = -340,000 Jmol-1 = -340 kJmol-1

17. (D.) The concentration of acid corresponds to the concentration of H+, a species that sits on the reactants side of the reaction equation. Chemical systems respond to an increase in reactant concentrations by shifting mass toward the products; in other words, increasing the concentration of H+ makes the reaction more favorable. Increasing H+ decreases the reaction quotient, Q, Which is calculated as the concentration of products divided by the concentration of reactants. If Q Decreases, then the electromotive force increases. This is made clear by the Nernst equation.

E = E °- RT Ln Q nF

If Q Decreases, then the entire term (RTlnQ) / (nF) Decreases. This in turn means that a smaller quantity is subtracted from E° To compute E. The electromotive force, E, Is therefore larger.

In This Chapter

^ Keeping the important points in mind

^ Trying your hand at some practice questions

^ Understanding the answers with explanations

Chapter 15 explained the teeterings and totterings of chemical equilibrium. This topic is a conceptual octopus, wrapping its long tentacles around a variety of other topics. In other words, knowing equilibrium well helps you to understand other things well. So, embrace the octopus, and give these practice questions a try.

Leveling Out the Facts on Equilibrium

Understanding equilibrium helps you to predict how a reaction at rest will respond to a change in conditions. Making these predictions requires you to understand some things about the reactants and products, such as what phase they are in, and whether heat acts as a reactant or product in the reaction. Different categories of reactions sometimes receive the high honor of an equilibrium constant with its own special name. The following points summarize the most important concepts and details about equilibrium.

Defining equilibrium

Like other concepts in chemistry, equilibrium is an important idea that is often expressed by using equations. These points describe equilibrium both verbally and mathematically.

Chemical equilibrium is "dynamic equilibrium," in which forward and reverse reactions occur at the same rate.

Equilibrium constants for forward and reverse reactions are inverses of one another.

The equilibrium constant for a coupled reaction is equal to the product of the equilibrium constants for each individual reaction.

The equilibrium constant Keq Quantifies the tendency of a reaction to favor products. Keq is calculated by taking the concentrations of products raised to the power of their coefficients divided by the concentrations of reactants raised to their coefficients. Thus a reaction of the form aA + bB cC + dD will have a Keq = [A]a[B]b4-([C]c[D]d).

If you are dealing with a reaction between gases, then you will use the expression

( Pr ) C ( Pd ) d

If a Keq value is significantly greater than 1, that is an indication that the reaction favors the product heavily. Keq values between 0 and 1 indicate that reactants are favored. A Keq value equal to 1 indicates that at equilibrium, the product of the concentrations of reactants (raised to the appropriate powers) is equal to the product of concentrations of products (raised to the appropriate powers). Note the individual concentrations themselves do not have to be equal for this to be true!

Reactions in which all of the products and reactants are in the same phase are subject to homogeneous equilibrium and Keq is calculated normally. If, however, the products and reactants are in more than one phase, you will have to apply the rules of heterogeneous equilibrium, which are as follows:

• Pure liquids (including solvents) and solids are omitted from the Keq expression.

• Partial pressures of gases take the place of concentrations in the Keq expression and both concentrations and partial pressures may appear in the same expression.

Predicting the shifting

Equilibria are all about rest. If you rouse them, they oppose you, shifting toward reactants or products — whatever it takes — to get back to their bean bag chairs. Here’s how they do it:

Le Chatelier’s Principle describes the tendency of reactions to act to reverse perturbations:

• Concentration: If reactant is added or product is removed, the forward reaction will be favored. If product is added or reactant is removed, the reverse reaction will be favored.

• Pressure: If pressure is increased, the reaction will favor the side with fewer moles of gas. If the pressure is decreased, the reaction will favor the side with more moles of gas.

• Temperature: A decrease in temperature will cause the reaction to favor the side that releases more heat. An increase in temperature will cause the reaction to favor the side that does not release heat.

• Catalysts: Catalysts speed reactions but do not affect the concentration of products or reactants at equilibrium.

If you are given a mixture of initial and final concentrations, use the following five steps to avoid confusion in solving the problem:

• Step 1: Start with a balanced equation for the reaction, including the phases of the reactants and products.

• Step 2: Check the phases to see if you’re dealing with a homogenous equilibrium or a heterogeneous equilibrium. If you’ve got a heterogeneous equilibrium on your hands, cross out any pure (unmixed) liquid or solid substances to ensure that you omit them from the final Keq expression.

• Step 3: Set up the Keq expression, including the actual Keq value, if known.

• Step 4: Make an ICEbox. Underneath each remaining reactant or product, list the following quantities, in order: initial concentration/partial pressure, change in concentration/partial pressure, and equilibrium concentration/partial pressure. For the reaction A + B C,

[A] [B] [C]

Initial:

Change:

Equilibrium:

Use known values to fill in the empty slots in the ICEbox.

• Step 5: Substitute the "Equilibrium" values into the Keq expression and solve for the unknown.

Special constants

Several different quantities are all calculated just like Keq, but go by other names. In most cases, these quantities are in fact Keq values, putting on airs because they think they’re special enough to deserve their own titles. In another case (the reaction quotient, Q), the special title occurs because the reaction quotient is not an equilibrium quantity at all.

The expressions KP And Kc Are used to distinguish between gaseous phase equilibrium constants in which partial pressures were used (KP) or molar concentrations (Kc). The relationship between the two is KP = KC x (RT)An, Where An corresponds to the difference in the moles of gas between the product and reactant sides of the reaction equation.

Other specialized equilibrium constants are discussed in Chapter 11 (Ksp) and 17 (Ka and KB). KSp is simply the equilibrium constant for a dissolution reaction. KA and KB are simply the equilibrium constants for acid or base dissociation.

Keq is only valid for a reaction that has reached equilibrium. A similar quantity called the reaction quotient Q Is used when the reaction is not at equilibrium. If Q < Keq, the reaction will progress "to the right," making more product. If Q > Keq, the reaction will shift "to the left," converting product into reactant. If Q = Keq, the reaction is at equilibrium.

Testing Your Knowledge

Shifting back and forth with equilibria is dizzying. Steady yourself for a moment and give these practice questions a shot.

Multiple choice

Questions 1 through 4 refer to the following choices.

(A) KA

(B) KSp

(C) KEq

(D) Q

(E) KP

1. This is used for a reaction that has not yet reached equilibrium.

2. This equilibrium constant is specific to gases.

3. This equilibrium constant is specific to dissociated solutions of ionic solids.

4. This equilibrium constant can be used to calculate pH.

CO2(g) + H2(g) CO(g) + H2O(Q

5. According to the balanced equation above, which of the following will favor the reverse reaction?

(I) Decreasing the pressure

(II) Adding CO to the reaction mixture

(III) Removing CO2 from the reaction mixture

(A) I only

(B) II and III

(C) I and III

(D) I, II, and III

(E) None of the above

SnO2(s) + 2CO(g) Sn(s) + 2CO2(g)

6. Which of the following is a correct equilibrium constant expression for the above reaction? (A)

K

(B)

(C)

K

(D)

[co ]

(E) r Sn ]

KEq = rSno]

[2

7. If the Keq of a forward reaction is 8.3x10-7, what is the Keq for the reverse reaction?

(A) 8.3 x 10-7

(B) 8.3 x 107

(C) 1.2 x 106

(D) 1.2 x 10-6

(E) Cannot be determined from the given information

_

_

8. What is the value of the equilibrium constant for the reaction H2(g) + I2(g) 2HI if the equilibrium mixture contains 0.092M H2, 0.023M I2, and 0.056M HI?

9. If the Ksp for the dissociation of nickel(II) hydroxide is 6.0x10 16, what is the concentration of nickel ions in solution?

(A) 5.3 x 10-6

(B) 1.1 x 10-5

(C) 3.0 x 10-16

(D) 6.0 x 10-16

(E) Cannot be determined from the given information

2NO2Cl(g) 2NO2(g) + Cl2(g) 10. If the KC of the above reaction is 8.90 at 350(C, what is its KP?

Free response

11. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Ka = 1.8x10-5

(a) Write the equilibrium constant expression KA for the reaction above.

(b) If the initial concentration of acetic acid is 0.25M, what is the equilibrium concentration of CH3COO-?

Checking Your Work

It feels good to be done with the practice questions, doesn't it? It feels even better to know that you answered them correctly. Make yourself feel good — check your work.

1. (D). The reaction quotient Q Is used in place of KEq when a reaction has not yet reached equilibrium.

2. (E). KP is used for gases.

3. (B). The solubility product constant KSp is used for the dissociation of ionic solids.

4. (A). KA can be used to calculate pH in a solution of an acid or a base.

5. (D). Decreasing the pressure will favor the side with the greater number of moles of gas (the reactants here). Adding products and removing reactants will also act to favor the production of reactants (the reverse reaction) so all three are valid.

6. (D). You eliminate all pure solids and liquids from heterogeneous equilibrium constant expressions, so D is the correct answer.

7. (C). The KEq of a reverse reaction can be calculated by taking the inverse of the forward reaction. 1 4 (8.3 x 10-7) = 1.2 x 106.

8. (C). The equilibrium constant expression is [ HI ]___0.0562

[ H 2112]

0.092 x 0.023

1.5.

9. (A). The Ksp expression for the dissociation of Ni(OH)2 is Ksp = [Ni2+][OH-]2. You also know that for every one nickel ion that dissolves in solution, there must be 2 OH – ions so [Ni2+] = 0.50[OH-]. Call this concentration X and plug it into the Ksp expression and you have 6.0 x 10-16 = 0.50X x X2. Solve this expression for X and you should get 1.06 x 10-5 = X, but as stated before, [Ni2+] is half of that value, which is 5.3 x 10-6.

10. (E). You will need to use the expression KP = KC x (RT)An. The only tricky parts to applying this equation are to remember to convert your temperature to K (623K) and to correctly solve for An. Do this by subtracting the number of moles of gaseous Reactants From the number of moles of gaseous Products, Giving you 3 – 2 = 1. Your expression then becomes

KP = 8.90 x (0.0821 x 623)1 = 455.

11. (a) Neglecting the pure liquid H2O, the equilibrium constant expression becomes

[cH 3COO -][ H 3O +]

K =–f-—-t-~

A

[cH3cooH]

(b) 2.1 x 10-3.You need to make an ICEbox for this

Initial

Change

Equilibrium

CH3COOH(aq)

0.25M

-xM

(0.25 – x)M

— H3O+(aq) +

0M +xM xM

CH3COO-(aq)

0M

+xM

XM

Then, use the equilibrium constant expression you developed in (a) to solve for X Through the following equation

K

[ X ][ X ]

[ 0.25 - X ]

1.8 x 10-

In such problems, we always make the approximation that XIs very, very small relative to 0.25. So, we can neglect the -x In the denominator, turning the expression into the much simpler

K

[ 0.25 ]"

1.8×10-

Solving the expression for X Gives you x2 = 4.5 x 10 6. Taking the square root of both sides gives you X = 2.1 x 10-3.

X

In This Chapter

^ Reviewing the seven common reaction types ^ Mastering the chemical balancing act

^ Familiarizing yourself with the mole and how to use it in conversions

^ Using net ionic equations

^ Counting what counts with limiting reagents

^ Quantifying success with percent yield

^^toichiometry. Such a complicated word for such a simple idea. The Greek roots of the

Word mean "measuring elements," which doesn’t sound nearly as intimidating. Moreover, the ancient Greeks couldn’t tell an ionic bond from an ionic column, so just how technical and scary could stoichiometry really be? Simply stated, stoichiometry (stoh-ick-eee-AH-muh-tree) is the quantitative relationship between atoms in chemical substances.

We will begin this chapter by walking you through each of the types of chemical reaction you will encounter on the AP chemistry exam. Several of them will be discussed in greater detail in later chapters. Then, we will show you how to use your understanding of chemical equations to do AP-type Chemistry problems, and Chapter 14 will allow you to practice your newfound skills.

The first half of the chapter describes the basics of stoichiometry, then delves into the seven types of reaction that you’d do well to recognize (notice how their names tell you what happens in each reaction). By recognizing the patterns of these seven types of reaction, you can often predict reaction products given only a set of reactants. There are no perfect guidelines, and predicting reaction products can take what is called "chemical intuition," a sense of what reaction is likely to occur based on knowing the outcomes of similar reactions. Still, if you are given both reactants and products, you should be able to tell what kind of reaction connects them, and if you are given reactants and the type of reaction, you should be able to predict likely products. Figuring out the formulas of products often requires you to apply knowledge about how ionic and molecular compounds are put together. To review these concepts, see Chapter 5.

The remainder of the chapter covers how to use your knowledge of these reaction types to calculate the quantities that allow you to quantify the success of a reaction.

Stoichiometry Lingo: How to Decipher Chemical Reactions

Chemists have a nice, ordered set of rules for describing chemical reactions. Reactants are always on the lefthand side of the equation, and products on the right, for example. Chemists would rather stick their hands in a vat of concentrated hydrochloric acid than reverse this order. There are also a nice, ordered set of chemical symbols that chemists use to describe reactions, and each symbol has one precise meaning. These fundamentals of stoichiometry are described in this section and, since they form the basis of all of stoichiometry, which is a huge point-gainer on the AP exam, it would behoove you to memorize them before moving on.

In compound formulas and reaction equations, you express stoichiometry by using subscripted numbers on atoms and coefficients in front of groups of atoms. Just like an accountant can’t afford to make a mistake when tracking the dollars and cents in your parents’ bank account, you can’t afford to lose track of any atoms on the AP exam. The road to poor scores in AP Chemistry is littered with students who couldn’t master stoichiometry!

In general, all chemical equations are written in the basic form:

Reactants — Products

Where the arrow in the middle means Yields Or "turns into." The basic idea is that the reactants react, and the reaction produces products. By Reacting, We simply mean that bonds within the reactants are broken, to be replaced by new and different bonds within the products.

Chemists fill chemical equations with symbols because they think it looks cool and, more important, because the symbols help pack a lot of meaning into a small space. Table 13-1 summarizes the most important symbols you’ll find in chemical equations.

After you understand how to interpret chemical symbols, compound names (see Chapter 5), and the symbols in Table 13-1, there’s not a lot you can’t understand. You’re equipped, for example, to decode a chemical equation into an English sentence describing a reaction. Conversely, you can translate an English sentence into the chemical equation it describes. When you’re fluent in this language, you regrettably won’t be able to talk to the animals; you will, however, be able to describe their metabolism in great detail.

Building and Breaking: Synthesis and Decomposition Reactions

The two simplest types of chemical reactions involve the joining of two compounds to form one compound or the breaking apart of one compound to form two. We describe both of these simple reactions in the sections that follow.

Synthesis

In synthesis (or combination) reactions, two or more reactants combine to form a single product, following the general pattern

A + B — C

For example,

2Na(s) + Cl2(g) — 2NaCl(s)

The combining of elements to form compounds (like NaCl) is a particularly common kind of combination reaction. Here is another such example:

2Ca(s) + O2(g) — 2CaO(s)

Compounds can also combine to form new compounds, such as in the combination of sodium oxide with water to form sodium hydroxide:

Na2O(s) + H2O(l) — 2NaOH(ag)

Decomposition

In a decomposition reaction, a single reactant breaks down (decomposes) into two or more products, following the general pattern

A —B + C

For example,

2H2O(l) — 2H2(g) + O2(g)

Notice that combination and decomposition reactions are the same reaction in opposite directions.

Many decomposition reactions produce gaseous products, such as in the decomposition of carbonic acid into water and carbon dioxide:

H2CO3(ag) — H2O(l) + CO2(g)

Swapping Spots: Displacement Reactions

Many common reactions involve the swapping of one or more atoms or polyatomic ions in a compound. You can check these reactions out in the following sections.

Single replacement

In a single replacement reaction, a single, more-reactive element or group replaces a less-reactive element or group, following the general pattern

A + BC — AC + B

For example,

Zn(s) + CuSO4(a<7) — ZnSO4(ag) + Cu(s)

Single replacement reactions in which metals replace other metals are especially common. You can determine which metals are likely to replace which others by using the Metal activity series, A ranked list of metals in which ones higher on the list tend to replace ones lower on the list. Table 13-2 presents the metal activity series.

Double replacement

Double replacement is a special form of Metathesis reaction (that is, a reaction in which two reacting species exchange bonds). Double replacement reactions tend to occur between ionic compounds in solution. In these reactions, cations (atoms or groups with positive

Charge) from each reactant swap places to form ionic compounds with the opposing anions (atoms or groups with negative charge), following the general pattern

AB + CD — AD + CB

For example,

KCl(a<7) + AgNO3(ag) — AgCl(s) + KNO3(ag)

Of course, ions dissolved in solution move about freely, not as part of cation-anion complexes. So, to allow double replacement reactions to progress, one of several things must occur:

One of the product compounds must be insoluble, so it precipitates (forms an insoluble solid) out of solution after it forms.

One of the products must be a gas that bubbles out of solution after it forms.

One of the products must be a solvent molecule, such as H2O, that separates from the ionic compounds after it forms.

Burning Up: Combustion Reactions

Oxygen is always a reactant in combustion reactions, which often release heat and light as they occur. Combustion reactions frequently involve hydrocarbon reactants (like propane, C3H8(g), the gas used to fire up backyard grills), and yield carbon dioxide and water as products. For example,

C3H8(g) + 5O2(g) — 3CO2(g) + 4H2O(l)

Combustion reactions also include combination reactions between elements and oxygen, such as:

S(s) + O2(g) — SO2(g)

So, if the reactants include oxygen (O2) and a hydrocarbon or an element, you’re probably dealing with a combustion reaction. If the products are carbon dioxide and water, you’re almost certainly dealing with a combustion reaction.

Getting Sour: Acid-Base Reactions

Acids and bases are complicated enough to deserve their own chapter, so flip forward to Chapter 17 for the nitty-gritty details of this reaction type. In the meantime, and so you have a comprehensive list of reaction types which the AP chemistry exam may quiz you on in one place, we’ll provide a summary here.

Neutralization reactions, which occur when an acid and a base are mixed together to form a salt and water, are the bread and butter of acid-base reactions. The best way to spot a reaction of this type is to look for water among the products of a reaction. If the water came from a double replacement reaction in which a hydrogen from one reactant (the acid) joined up with a hydroxide from the other (the base), then you are definitely dealing with an acid-base reaction.

Not all acid-base reactions are so simple, however. Bases do not necessarily contain hydroxide, and the products are not always water and a salt. See Chapter 17 for the details of the myriad of acid-base reactions.

Getting Charged: Oxidation-Reduction Reactions

Oxidation-reduction (or redox) reactions, like acid-base reactions, have an entire chapter (Chapter 19) devoted to them. Redox reactions are concerned with the transfer of electrons between reactants. Because they involve the transfer of charged particles, ions are a staple of redox reactions and are the things to look out for when trying to spot one. The species in a redox reaction that loses electrons to the other reactant is Oxidized In the reaction, while the species that gains the electrons is Reduced. There are two easy mnemonics commonly used to remember this, so take your pick or make up your own:

LEO And GER: Lose electrons, oxidation. Gain electrons, reduction. OIL RIG: Oxidation is lost, reduction is gained.

Oxidation and reduction reactions always occur together, but may be written separately as half reactions in a chemistry problem. In these half reactions, electrons being transferred are written explicitly, so if you see them, take it as a dead giveaway that you are dealing with a redox reaction.

The Stoichiometry Underground: Moles

Chemical equations are all well and good, but what do they actually mean? For translating chemical equations into useful chemical data, you can find no better or more fundamental tool than the mole. As an AP chemistry student, you’re probably already exhaustively familiar with the mole, but in the following sections, we provide you with a comprehensive review of the mole and its usefulness in case you forgot any of the basics along the way.

Counting your particles: The mole

Chemists routinely deal with hunks of material containing trillions of trillions of atoms, but ridiculously large numbers can induce migraines. For this reason, chemists count particles (like atoms and molecules) in multiples of a quantity called the Mole. Initially, counting particles in moles can be counterintuitive, similar to the weirdness of Dustin Hoffman’s character in the movie Rainman, When he informs Tom Cruise’s character that he has spilled 0.41 fraction of a box of 200 toothpicks onto the floor. Instead of referring to 82 individual toothpick particles, he refers to a fraction of a larger unit, the 200-toothpick box. A mole is a very big box of toothpicks — 6.022 x 1023 toothpicks, to be precise.

If 6.022 x 1023 strikes you as an unfathomably large number, then you’re thinking about it correctly. It’s larger, in fact, than the number of stars in the sky or the number of fish in the sea, and is many, many times more than the number of people who have ever been born throughout all of human history. When you think about the number of particles in something as simple as, say, a cup of water, all your previous conceptions of "big numbers" are blown out of the water, as it were.

The number 6.022 x 1023, known as Avogadro’s number, Is named after the 19th century Italian scientist Amedeo Avogadro. Posthumously, Avogadro really pulled one off in giving his name to this number, because he never actually thought of it. The real brain behind Avogadro’s number was that of a French scientist named Jean Baptiste Perrin. Nearly 100 years after Avogadro had his final pasta, Perrin named the number after him as an homage. Ironically, this humble act of tribute has misdirected the resentment of countless hordes of high school chemistry students to Avogadro instead of Perrin.

Avogadro’s number is the conversion factor used to move between particle counts and numbers of moles:

1 mole / 6.022 x 1023 particles

Like all conversion factors, you can invert it to move in the other direction, from moles to particles.

Assigning mass and volume to your particles

Chemists always begin a discussion about moles with Avogadro’s number. They do this for two reasons. First, it makes sense to start the discussion with the way the mole was originally defined. Second, it’s a sufficiently large number to intimidate the unworthy.

Still, for all its primacy and intimidating size, Avogadro’s number quickly grows tedious in everyday use. More interesting is the fact that one mole of a pure Monatomic Substance (in other words, a substance that always appears as a single atom) turns out to possess exactly its atomic mass’s worth of grams. In other words, one mole of monatomic hydrogen weighs about 1 gram. One mole of monatomic helium weighs about 4 grams. The same is true no matter where you wander through the corridors of the periodic table. If we define Molar mass As the mass of one mole of any substance, then the number listed as the atomic mass of an element equals that element’s molar mass If the element is monatomic. Guess what — chemists actually set up the value of the mole to make that happen, though it took years of arguments and an international commission to get the final details down to the last decimal place just right.

Of course, chemistry involves the making and breaking of bonds, so talk of pure monatomic substances gets you only so far. How lucky, then, that calculating the mass per mole of a complex molecule is essentially no different than finding the mass per mole of a monatomic element. For example, one molecule of glucose (C6H12O6) is assembled from 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. To calculate the number of grams per mole of a complex molecule (such as glucose), simply do the following:

1. Multiply the number of atoms per mole of the first element by its atomic mass.

In this case, the first element is carbon, and you’d multiply its atomic mass, 12, by the number of atoms, 6.

2. Multiply the number of atoms per mole of the second element by its atomic mass.

Here, you multiply hydrogen’s atomic mass of 1 by the number of hydrogen atoms, 12.

3. Multiply the number of atoms per mole of the third element by its atomic mass. Keep going until you’ve covered all the elements in the molecule.

The third element in glucose is oxygen, so you multiply 16, the atomic mass by 6, the number of oxygen atoms.

4. Finally, add the masses together.

In this example, (12g mol-1 x 6) + (1g mol-1 x 12) + (16g mol-1 x 6) = 180g mol-1.

This kind of quantity, called the Gram molecular mass, Is exceptionally convenient for chemists, who are much more inclined to measure the mass of a substance than to count all of millions or billions of individual particles that make it up.

If chemists don’t try to intimidate you with large numbers, they may attempt to do so by throwing around big words. For example, older chemists may distinguish between the molar masses of pure elements, molecular compounds, and ionic compounds by referring to them as the Gram atomic mass, gram molecular mass, And Gram formula mass, Respectively. Don’t be fooled! The basic concept behind each is the same: molar mass. Fortunately these other terms are never used by those modern folks who write AP chemistry exams.

It’s all very good to find the mass of a solid or liquid and then go about calculating the number of moles in that sample. But what about gases? Let’s not engage in phase discrimination; gases are made of matter, too, and their moles have the right to stand and be counted. Fortunately, there’s a convenient way to convert between the moles of gaseous particles and their Volume. Unlike gram atomic/molecular/formula masses, this conversion factor is constant No matter what kinds of molecules make up the gas. Every gas assumed to behave ideally has a volume of 22.4 liters per mole, regardless of the size of the gaseous molecules.

Before you start your hooray-chemistry-is-finally-getting-simple dance, however, understand that certain conditions apply to this conversion factor. For example, it’s only true at Standard temperature and pressure (STP), Or 0° Celsius and 1atm. Also, the figure of 22.4L mol-1 applies only to the extent that a gas resembles an ideal gas, one whose particles have zero volume and neither attract nor repel one another. Ultimately, no gas is truly ideal, but many are so close to being so that the 22.4L mol-1 conversion is very useful.

Giving credit where it’s due: Percent composition

Chemists are often concerned with precisely what percentage of a compound’s mass consists of one particular element. Lying awake at night, uttering prayers to Avogadro, they fret over this quantity, called Percent composition. Percent composition was mentioned in Chapter 5, but now that you have reviewed the basics of stoichiometry, we will explain in detail how to calculate it.

Follow these three simple steps:

1. Calculate the molar mass of the compound, as we explain in the previous section.

Percent compositions are completely irrelevant to molar masses for pure monatomic substances, which by definition have 100 percent composition of a given element.

2. Multiply the atomic mass of each element present in the compound by the number of atoms of that element present in one molecule.

3. Divide each of the masses calculated in Step 2 by the total mass calculated in Step 1. Multiply each fractional quotient by 100%. Voila! You have the Percent composition By mass of each element in the compound.

Checking out empirical formulas

What if you don’t know the formula of a compound? Chemists sometimes find themselves in this disconcerting scenario. Rather than curse Avogadro (or perhaps After Doing so), they analyze samples of the frustrating unknown to identify the percent composition. From there, they calculate the ratios of different types of atoms in the compound. They express these ratios as an Empirical formula, The lowest whole-number ratio of elements in a compound.

To find an empirical formula given percent composition, use the following procedure:

1. Assume that you have 100g of the unknown compound.

The beauty of this little trick is that you conveniently gift yourself with the same number of grams of each elemental component as its contribution to the percent composition. For example, if you assume that you have 100g of a compound composed of 60.3 percent magnesium and 39.7 percent oxygen, you know that you have 60.3g of magnesium and 39.7g of oxygen.

2. Convert the assumed masses from Step 1 into moles by using gram atomic masses.

3. Divide each of the element-by-element mole quantities from Step 2 by the lowest among them.

This division yields the mole ratios of the elements of the compound.

4. If any of your mole ratios aren’t whole numbers, multiply all numbers by the smallest possible factor that produces whole-number mole ratios for all of the elements.

For example, if there is one nitrogen atom for every 0.5 oxygen atoms in a compound, the empirical formula is not N1O0.5. Such a formula casually suggests that an oxygen atom has been split, something that would create a small-scale nuclear explosion. Though impressive-sounding, this scenario is almost certainly false. Far more likely is that the atoms of nitrogen to oxygen are combining in a 1:0.5 Ratio, But do so in groups of 2 x (1:0.5) = 2:1. The empirical formula is thus N2O.

Because the original percent composition data is typically experimental, expect to see a bit of error in the numbers. For example, 2.03 is probably within experimental error of 2.

5. Write the empirical formula by attaching these whole-number mole ratios as subscripts to the chemical symbol of each element. Order the elements according to the general rules for naming ionic and molecular compounds (described in Chapter 5).

Differentiating between empirical formulas and molecular formulas

Many compounds in nature, particularly compounds made of carbon, hydrogen, and oxygen, are composed of atoms that occur in numbers that are multiples of their empirical formula. In other words, their empirical formulas don’t reflect the actual numbers of atoms within them, but only the ratios of those atoms. What a nuisance! Fortunately, this is an old nuisance, so chemists have devised a means to deal with it. To account for these annoying types of compounds, chemists are careful to differentiate between an empirical formula and a Molecular formula. A molecular formula uses subscripts that report the actual number of atoms of each type in a molecule of the compound (a Formula unit Accomplishes the same thing for ionic compounds).

Molecular formulas are associated with molar masses that are simple whole-number multiples of the corresponding Empirical formula mass. In other words, a molecule with the empirical formula CH2O has an empirical formula mass of 30.0g mol-1 (12 for the carbon + 2 for the two hydrogens + 16 for the oxygen). The molecule may have a molecular formula of CH2O, C2H4O2, C3H6O3, and so on. As a result, the compound may have a molar mass of 30.0g mol-1, 60.0g mol-1, 90.0g mol-1, and so on.

You can’t calculate a molecular formula based on percent composition alone. If you attempt to do so, Avogadro and Perrin will rise from their graves, find you, and slap you 6.022 x 1023 times per cheek (Ooh, that smarts!). The folly of such an approach is made clear by comparing formaldehyde with glucose. The two compounds have the same empirical formula, CH2O, but different molecular formulas, CH2O and C6H12O6, respectively. Glucose is a simple sugar, the one made by photosynthesis and the one broken down during cellular respiration. You can dissolve it into your coffee with pleasant results. Formaldehyde is a carcinogenic component of smog. Solutions of formaldehyde have historically been used to embalm dead bodies. You are not advised to dissolve formaldehyde into your coffee. In other words, molecular formulas differ from empirical formulas, and the difference is important.

To determine a molecular formula, you must know the molar mass of the compound as well as the empirical formula (or enough information to calculate it yourself from the percent composition; see the previous section for details). With these tools in hand, calculating the molecular formula involves a three-step process:

1. Calculate the empirical formula mass.

2. Divide the molar mass by the empirical formula mass.

3. Multiply each of the subscripts within the empirical formula by the number calculated in Step 2.

Keeping the See-Saw Straight: Balancing Reaction Equations

Equations that simply show reactants on one side of the reaction arrow and products on the other are Skeleton equations, And are perfectly adequate for a qualitative description of the reaction: who are the reactants and who are the products. But if you look closely, you’ll see that those equations just don’t add up quantitatively. As written, the mass of one mole of each of the reactants doesn’t equal the mass of one mole of each of the products. The skeleton equations break the Law of Conservation of Mass, Which states that all the mass present at the beginning of a reaction must be present at the end. To be quantitatively accurate, these equations must be Balanced So the masses of reactants and products are equal.

In chemistry this is easy — the requirement is met as mentioned earlier by assuring that the numbers of the same types of atoms are equal on each side. All the equations we have so far written have met this balance requirement. (The crime of losing or splitting atoms is punishable by an Avogadro number of lashes!).

How to use coefficients

To balance an equation, you use Coefficients To alter the number of moles of reactants and/or products so the mass on one side of the equation equals the mass on the other side. A Coefficient Is simply a number that precedes the symbol of an element or compound, multiplying the number of moles of that Entire Compound within the equation. Coefficients are different from Subscripts, Which multiply the number of atoms or groups within a compound. Consider the following:

4Cu(NO3)2

The number 4 that precedes the compound is a coefficient, indicating that there are four moles of copper (II) nitrate. The subscripted 3 and 2 within the compound indicate that each nitrate contains three oxygen atoms, and that there are two nitrate groups per atom of copper. Coefficients and subscripts multiply to yield the total number of moles of each atom:

4mol Cu(NO3)2 x (1mol Cu/mol Cu(NO3)2) = 4mol Cu

4mol Cu(NO3)2 x (2mol NO3/mol Cu(NO3)2) x 1mol N/mol NO3 = 8mol N

4mol Cu(NO3)2 x (2mol NO3/mol Cu(NO3)2) x 3mol O/mol NO3 = 24mol O

When you balance an equation, You change only the coefficients. Changing subscripts alters the chemical compounds themselves. If your pencil were equipped with an electrical shocking device, that device would activate the moment you attempted to change a subscript while balancing an equation.

Here’s a simple recipe for balancing equations:

1. Given a skeleton equation (one that includes formulas for reactants and products), count up the number of each kind of atom on each side of the equation.

If you recognize any polyatomic ions, you can count these as one whole group (as if they were their own form of element). See Chapter 5 for information on recognizing polyatomic ions.

2. Use coefficients to balance the elements or polyatomic ions, one at a time.

For simplicity, start with those elements or ions that appear only once on each side.

3. Check the equation to ensure that each element or ion is balanced.

Checking is important because you may have "ping-ponged" several times from reactants to products and back — there’s plenty of opportunity for error.

4. When you’re sure the reaction is balanced, check to make sure it’s in lowest terms.

For example,

4H2(g) + 2O2(g) —4H2O(l) should be reduced to

2H2(g) + O2(g) — 2H2O(l)

You can’t begin to wrap your brain around the unimaginably large number of possible chemical reactions. It’s good that so many reactions can occur, because they make things like life and the universe possible. From the perspective of a mere human brain trying to grok all these reactions, we have another bit of good news: A few categories of reactions pop up over and over again. After you see the very basic patterns in these categories, you’ll be able to make sense of the majority of reactions out there.

Conversions

Mass and energy are conserved. It’s the law. Unfortunately, this means that there’s no such thing as a free lunch, or any other type of free meal. Ever. On the other hand, the conservation of mass makes it possible to predict how chemical reactions will turn out.

Balancing equations can seem like a chore, like taking out the trash. But a balanced equation is far better than any collection of coffee grounds and orange peels because such an equation is a useful tool. After you’ve got a balanced equation, you can use the coefficients to build Mole-mole conversion factors. These kinds of conversion factors tell you how much of any given product you get by reacting any given amount of reactant. This is one of those calculations that makes chemists particularly useful, so they needn’t get by on looks and charm alone.

Consider the following balanced equation for generating ammonia from nitrogen and hydrogen gases:

N2(g) + 3H2(g) — 2NH3(g)

For every mole of nitrogen molecules reactanting, a chemist expects two moles of ammonia product. Similarly, for every three moles of hydrogen molecules reacting, the chemist expects two moles of ammonia product. These expectations are based on the coefficients of the balanced equation and are expressed as mole-mole conversion factors as shown in Figure 13-1.

2mol NH,

Figure 13-1:

Building mole-mole conversion factors from a balanced equation.

N2 + 3 H2 —*- 2 NH,

2 2 3

3mol H

The mole is the beating heart of stoichiometry, the central unit through which other quantities flow. Real-life chemists don’t have magic mole vision, however. A chemist can’t look at a pile of potassium chloride crystals, squint her eyes, and proclaim: "That’s 0.539 moles of salt." Well, she could proclaim such a thing, but she wouldn’t bet her pocket protector on it. Real reagents (reactants) tend to be measured in units of mass or volume, and occasionally even in actual numbers of particles. Real products are measured in the same way. So, you need to be able to use Mole-mass, mole-volume, And Mole-particle conversion factors To translate between these different dialects of counting. Figure 13-2 summarizes the interrelationship between all these things and serves as a flowchart for problem solving. All roads lead to and from the mole.

Figure 13-2:

A problem-solving flowchart showing the use of mole-mole, mole-mass, mole-volume, and mole-particle conversion factors.

Particles

Mole Particle

Mole

Mass

Volume

Reactants -

Mole Mole

Mole

Volume

Particle Mole

P roducts

Mass Mole

Volume Mole

Particles

Mass

Volume

Mass

*JJIBE*

Getting rid of mere spectators: Net ionic equations

Chemistry is often conducted in aqueous solutions. Soluble ionic compounds dissolve into their component ions, and these ions can react to form new products. In these kinds of reactions, sometimes only the cation or anion of a dissolved compound reacts. The other ion merely watches the whole affair, twiddling its charged thumbs in electrostatic boredom. These uninvolved ions are called Spectator ions.

Because spectator ions don’t actually participate in the chemistry of a reaction, you don’t need to include them in a chemical equation. Doing so leads to a needlessly complicated reaction equation. So, chemists prefer to write Net ionic equations, Which omit the spectator ions. A net ionic equation doesn’t include every component that might be present in a given beaker. Rather, it includes only those components that actually react.

Here is a simple recipe for making net ionic equations of your own:

1. Examine the starting equation to determine which ionic compounds are dissolved, as indicated by the (aq) symbol following the compound name.

Zn(s) + HCl(aq) — ZnCl2(aq) + H2(g)

2. Rewrite the equation, explicitly separating dissolved ionic compounds into their component ions.

This step requires you to recognize common polyatomic ions, so be sure to familiarize yourself with them.

Zn(s) + H+(aq) + GT(aq) — Zn2+(aq) + 3GT(aq) + H2(g)

3. Compare the reactant and product sides of the rewritten reaction. Any dissolved ions that appear in the same form on both sides are spectator ions. Cross out the spectator ions to produce a net reaction.

Zn(s) + H+(aq) + Cl – (aq) — Zn2+(aq) + 2Cl – (aq) + H2(g)

Net reaction:

Zn(s) + H+(aq) — Zn2+(aq) + H2(g)

As written, the preceding reaction is imbalanced with respect to the number of hydrogen atoms and the amount of positive charge.

4. Balance the net reaction for mass and charge.

Zn(s) + 2H+(aq) — Zn2+(aq) + H2(g)

If you wish, you can balance the equation for atoms and charge first (at Step 1). This way, when you cross out spectator ions at Step 3, you’ll be crossing out equivalent numbers of ions. Either method produces the same net ionic equation in the end. Some people prefer to balance the starting reaction equation, but others prefer to balance the net reaction because it is a simpler equation.

Running Out Early: Limiting Reagents

In real-life chemistry, not all of the reactants present convert into product. That would be perfect and convenient. Does that sound like real life to you? More typically, one reagent is completely used up, and others are left behind, perhaps to react another day.

The situation resembles that of a horde of Hollywood hopefuls lined up for a limited number of slots as extras in a film. Only so many eager faces react with an available slot to produce a happily (albeit pitifully) employed actor. The remaining actors are in excess, muttering quietly all the way back to their jobs as waiters. In this scenario, the slots are the limiting reagent.

Those standing in line demand to know, how many slots are there? With this key piece of data, they can deduce how many of their huddled mass will end up with a gig. Or, they can figure out how many will continue to waste their film school degrees serving penne with basil and goat cheese to chemists on vacation.

Chemists demand to know, which reactant will run out first? In other words, which reactant is the Limiting reagent? Knowing that information allows them to calculate how much product they can expect, based on how much of the limiting reagent they’ve put into the reaction. Also, identifying the limiting reagent allows them to calculate how much of the excess reagent they’ll have left over when all the smoke clears. Either way, the first step is to figure out which is the limiting reagent.

In any chemical reaction, you can simply pick one reagent as a candidate for the limiting reagent, calculate how many moles of that reagent you have, and then calculate how many grams of the other reagent you’d need to react both to completion. You’ll discover one of two things. Either you have an excess of the first reagent or you have an excess of the second reagent. The one you have in excess the the excess reagent. The one that is not in excess is the limiting reagent.

Having Something to Show for Yourself: Percent Yield

In a way, reactants have it easy. Maybe they’ll make something of themselves and actually react. Or maybe they’ll just lean against the inside of the beaker, flip through a back issue of People Magazine, and sip a caramel macchiato.

Chemists don’t have it so easy. Someone is paying them to do reactions. That someone doesn’t have time or money for excuses about loitering reactants. So you, as a fresh-faced chemist, have to be concerned with just how completely your reactants react to form products. To compare the amount of product obtained from a reaction with the amount that should have been obtained, chemists use Percent yield. You determine percent yield with the following formula:

Percent yield = 100% x (actual yield) / (theoretical yield)

Lovely, but what is an actual yield and what is a theoretical yield? An Actual yield Is. . . well. . . the amount of product actually produced by the reaction. A Theoretical yield Is the amount of product that could have been produced had everything gone perfectly, as described by theory — in other words, as predicted by your painstaking calculations.

Things never go perfectly. Reagents stick to the sides of flasks. Impurities sabotage reactions. Chemists attempt to dance. None of these ghastly things are Supposed To occur, but they do. So, actual yields fall short of theoretical yields.

In This Chapter

^ Going over important points to remember

^ Putting in some practice with example questions

^ Seeing how your answers stack up

/f we had to name one single thing that is key to acing the AP chemistry exam, a solid understanding of stoichiometry would be it. Stoichiometry is chemistry’s bread and butter. Make sure that you understand all of the concepts outlined in Chapter 13 in detail and you will score some major points on exam day. To ensure you nail the concepts of stoichiometry, we not only include the major points that are essential to your success, but we also include practice questions near the end of the chapter so you can sharpen your test-taking skills on stoichiometry.

Making Sure You React to Questions With the Right Answers

Stoichiometry and the skills that surround it might seem mundane, but they are central skills that let you grapple with the very heart of chemistry: reactions. It’s well worth your time to make sure you know these basics back and forth.

Knowing reactions when you see them

These tips and reminders are all about recognition — knowing how to decode chemical formulas and reaction equations to understand what reaction is taking place, and even how to predict the products of a reaction when all you’re given is the reactants.

Reaction equations use symbols to indicate details of the reaction. Review the symbols in Chapter 13 in Table 13-1 and make sure that you’re familiar with all of them.

Make sure that you’re familiar with the seven common reaction types and are able to identify them and predict products if given only reactants. It is not so important to memeorize the names of the types as it is to be able to predict products and balance

Equations. Many reactions on tests expect "familiar" compounds as products, not truly exotic ones, so don’t get hung up on memorizing lots of exotic compounds!

• Synthesis/Combination reactions: A + B — C. Two or more reactants combine to form a single product.

• Decomposition reactions: A — B + C. Four types of decomposition reaction are seen quite commonly on the AP chemistry exam. These are

Metal carbonate — metal oxide + CO2(g)

Metal chlorate — metal chloride + O2(g)

Metal hydroxide — metal oxide + H2O

Acid — nonmetal oxide + H2O

• Single replacement reactions: A + BC — B + AC. Single replacement reactions are all redox reactions! The likelihood of single replacement reactions occurring relies on the reactivity series of metals outlined in Table 13-2. A will only displace B if it is a more-reactive species, which means it appears higher in the reactivity series.

• Double replacement reactions: AB + CD — AD + CB. A and D will not necessarily combine in the same ratios as A and B did. Take careful account of the charges of each species and make sure to balance the equation in the end.

• Combustion reactions: These contain oxygen among the reactants. The most common types of combustion reactions are those where hydrocarbons are burned to form water and carbon dioxide as products, but a familiar trap is to forget that substances that already contain oxygen, such as ethanol (C2H6O), also "burn" to form carbon dioxide and water. All combustion reactions are also redox reactions.

• Neutralization, or acid/base reactions: Acid + Base — Water + Salt. Chapter 17 explains these in greater detail.

• There are also a number of acid-base reaction types in addition to neutralization reactions. These are

Nonmetal oxide + H2O — acid

Metal oxide + H2O — base

Metal + H2O(l) — base + H2(g)

Active metal + acid — salt + H2(g)

Carbonate + acid — CO2(g) + H2O(l) + salt

Base + salt — metal hydroxide precipitate + salt

Strong base + salt containing NH4 — NH3 + H2O + salt

• Redox reactions are outlined in detail in Chapter 19. They typically involve the transfer of electrons between one reactant and the other. The species that loses the electrons is oxidized and that which gains the electrons is reduced. The mnemonics "LEO and GER" or "OIL RIG" will help you remember this information.

Dealing with the numbers that flow from stoichiometry

Coming up with accurate formulas and reaction equation isn’t the end of the story — it’s usually just the beginning. Once you’ve got an equation, you’ve got to make sure that it’s balanced, and then use the correct stoichiometry to solve problems. These tips deal with that process.

*e Avogadro’s number, 6.022 x 1023, is equivalent to the number of particles in one mole of a substance. By "particles" we mean atoms, ions, or molecules — whichever is relevant for the circumstance.

*e The gram atomic, molecular, or formula mass of a substance is calculated by adding up the atomic masses of all of its components (multiplying by the number of atoms of each substance present where applicable). It’s equivalent to the mass in grams of one mole of particles of that substance.

*e At standard temperature and pressure, one mole of an ideal gas would occupy 22.4 liters volume. No gases are truly ideal (see Chapter 9 on gas laws for further details). However, this approximation is a very good one in many cases and is very useful in converting volume to moles of a gas and vice versa.

E The percent composition of a substance takes into account the mass ratios of the atoms that make it up as well as the ratios in which they combine. To calculate a percent composition of a molecular or ionic compound, calculate the gram molecular or gram formula mass of the compound. Then, calculate the mass of each type of atom in the substance (by multiplying the mass by the number of atoms present in each molecule) and divide by the total mass of the compound.

E You can also use percent compositions to find the formula of a compound, though it is a somewhat complicated process. If you follow the five following steps, you’ll get it every time.

1. Assume that you have 100g of the unknown compound.

2. Convert the assumed masses from Step 1 into moles by using gram atomic masses.

3. Divide each of the element-by-element mole quantities from Step 2 by the lowest among them.

4. If any of your mole ratios aren’t whole numbers, multiply all numbers by the smallest possible factor that produces whole-number mole ratios for all of the elements.

5. Write the empirical formula by attaching these whole-number mole ratios as subscripts to the chemical symbol of each element. Order the elements according to the general rules for naming ionic and molecular compounds (described in Chapter 5).

E Given an empirical formula and either a formula or molar mass for a compound, you can also determine its molecular formula by dividing the gram formula mass by the empirical formula mass and then by taking this whole number value and multiplying the subscripts for each atom in the empirical formula by that number.

E Before you take the AP exam, make sure you’ve mastered the method for balancing equations. Unless the AP exam tells you that an equation is balanced, always check. To check that an equation is properly balanced, simply count the total number of each type of atom on each side of the equation and make sure that they match. Also be sure to check that any total charges on each side are balanced.

E Always consider the possibility of limiting reagents in reactions. Limiting reagents determine theoretical yields. Percent yield is 100 percent times the actual yield divided by the theoretical yield.

Testing Your Knowledge

Now that you’ve grounded yourself in a solid knowledge of reactions and stoichiometry, see if you can keep your balance through these practice questions.

Multiple choice

1. Which of the following represents a process in which a species is reduced?

(A) Mg — Mg2+

(B) Cl2 — Cl-

(C) Ni3+— Ni4+

(D) CO — CO2

(E) NO2- — NO3-5%

2. A sample of Li2SO4 (molar mass 110g) is reported to be 10.2% lithium. Assuming that none of the impurities contain any lithium, what percentage of the sample is pure lithium sulfate?

3. When the following equation for the acid base reaction above is balanced and all of the coefficients are reduced to lowest whole-number terms, the coefficient on the H2O is

_H2SO4 +_Ca(OH)2 —_CaSO4 +_H2O

4. What is the simplest formula for a compound containing only carbon and hydrogen and containing 16.67% H?

5. A substance has an empirical formula of CH2O and a molar mass of 180.0. What is its molec -

6. In the reaction

2Fe2O3 + 3C — 4Fe + 3CO2 if 500.g of Fe2O3 reacts with 75.0g of C, how many grams of Fe will be produced?

(A) 56.4g

(B) 75.0g

(C) 316.g

(D) 349.g

(E) 465.g

7. The mass of 5 atoms of lead is

(A) 3.44 x 10-20g.

(B) 1.72 x 10-21g.

(C) 8.60 x 10-21g.

(D) 1.20 x 10-23g.

(E) 6.02 x 10-23g.

8. AgNO3(aq) + KCl(aq) —_+_

What are the missing products?

(A) AgNO3(aq) + KCl(aq) (no reaction)

(B) AgCl2(s) + K2NO3(aq)

(C) AgCl(s) + KNO3(aq)

(D) Ag2+(aq) + Cl-(aq) +KNO3 (aq)

(E) AgCl2(s) + K+(aq) + Cl – (aq)

9. HCl + NaOH — NaCl + H2O

If 30.0g of HCl is mixed with excess NaOH according to the reaction above and 41.0g of NaCl is produced, what is the difference between the actual and the theoretical yields for this reaction?

(A) 7.1g

(B) 8.3g

(C) 15.5g

(D) 41.0g

(E) 48.1g

Free response

10. Do each of the following conversions based on the reaction below. SiO2 (s) + 6HF(g) — H2SiF6(aq) + 2H2O(l)

(a) 5.25g HF to moles

(b) 6.00 moles SiO2 to grams of H2SiF6

(c) 10.0 moles HF to liters HF (at STP)

(d) 3.20 mol SiO2 to molecules H2SiF6

11. Write the formulas to show the reactants and the products for the reactions described below. Assume that the reaction occurs and that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit spectator ions or molecules. You do not need to balance the equations.

(a) Iron (II) sulfide is added to a solution of hydrochloric acid.

(b) The combustion of methane in large excess of air.

(c) Calcium carbonate is strongly heated (hint: carbon dioxide is one of the products).

(d) A solution of potassium hydroxide is added to solid ammonium chloride.

(e) A strip of iron is added to a solution of copper (II) sulfate.

(f) Chlorine gas is bubbled into a solution of sodium fluoride.

(g) Dilute sulfuric acid is added to a solution of strontium nitrate.

(h) Copper ribbon is burned in oxygen.

12. For each of the following three reactions, in part one for each reaction write a balanced equation for the reaction and in part two for each reaction answer the question about the reaction. In the balanced equation, coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction.

(a) Excess nitric acid is added to solid sodium bicarbonate.

(i) Balanced Equation:

(ii) What is the minimum mass of sodium bicarbonate that must be added in order to react completely with 25.0g of nitric acid?

(b) A solution containing the magnesium ion (an oxidizing agent) is mixed with a solution containing lead(II) (a reducing agent).

(i) Balanced Equation:

(ii) If the contents of the reaction mixture described above are filtered, what sub-stance(s), if any, would remain on the filter paper?

(c) A solution of sodium hydroxide is added to a solution of copper (II) sulfate.

(i) Balanced Equation:

(ii) What is the percent composition of oxygen in sodium hydroxide?

13. For each of the following, use appropriate chemical principles to explain the observation.

(a) Lithium metal will react with water, but copper will not.

(b) The human body is made of 60% to 70% water, but hydrogen is only the third most abundant element in the body by mass.

(c) Water can act as both an acid and a base.

14. A compound is 36.5% sodium, 25.4% sulfur, and 38.1% oxygen.

(a) What is its empirical formula?

(b) If the compound in (a) has a molar mass of 366.3g, what is its molecular formula?

15. Calculate the percent composition of each element in caffeine C8H10N4.

Checking Your Work

Still on your feet? Good. Now check with the judges and receive your score. If you faltered, find out why. If you triumphed, enjoy it.

1. (B). This reaction is the only one where electrons are gained by the reactant to form the product, a sure sign of a reduction reaction. The negative sign on the chlorine is an indicator that extra electrons are present, and the fact that the reactant is of neutral charge cements the reaction as reduction.

2. (D). Begin by finding the percent composition of lithium in pure lithium sulfate, 14.0g * 110g = 12.7%. Impurities, however, have driven the lithium concentration down to 10.2% lithium. Dividing this percentage by the first will give you the percent of the pure compound in the sample. 10.2 * 12.7 = 80.3%, which is closest to answer choice D.

3. (B). There are two tricks to balancing an equation like this. The first is to begin by balancing the polyatomic ions, in this case SO4 and OH-. But where is the OH- among the products? This point is the second important one. In balancing equations, particularly neutralization reactions like this one, it is often extremely useful to write H2O as HOH. The fully balanced equation here is H2SO4 + Ca(OH)2 — CaSO4 + 2H2O. That’s right. Water is the only component that takes a subscript greater than 1, and that subscript is 2.

4. (C). Here you’re given a percent composition and are asked to derive an empirical formula. Follow the steps outlined in Chapter 13 to accomplish this task. Or, since the answers are right in front of you and percent composition is really a fairly easy calculation to make, you can simply calculate the percent composition of hydrogen in each of the compounds among the answer choices and see which one matches the question. You will find that this answer is choice C, pentane.

5. (E). Calculate the empirical formula mass of CH2O (30g per mole) and divide the molar mass by it, giving you 180 * 30 = 6. Multiply each of the subscripts in the empirical formula to give you the molecular formula C6H12O6, answer choice E.

6. (D). First, identify the limiting reagent in the reaction. Converting grams of iron (III) oxide to grams of carbon reveals that 56.4g of carbon are needed to react completely with 500.g of iron (III) oxide.

500g Fe2O3 1mol Fe2O3

3molC x 12.0gC

1

159.7gFe2O3 2mol Fe2O3 1mol C

56.4gC

This discovery means that there is excess carbon among the reactants and iron oxide should be used as the limiting reagent. With this established, all that remains is to convert grams of iron (III) oxide to grams of the iron product according to the calculation

500gFe2O3 x 1moFe2O3 x 4molFe x 55.8gFe = 3494gFe 1 159.7gFe2O3 2mol Fe2O3 1mol Fe

This matches answer choice D.

7. (B). Remember that you must convert to moles first and cannot convert directly from particles to grams. Set up your three conversion factors as follows and you get an answer matching answer choice B.

5atoms Pb x_1mol Pb_x 207.2g Pb = 1 72 x 10-21 „

1 6.022 x 1023 atoms Pb 1mol Pb

8. (C). This choice represents a double replacement reaction. The first thing that you will need to do is deduce the charge on the silver. As a diligent AP chemistry student, you should already have memorized the charges on the common polyatomic ions and know that NO3 carries a charge of -1. Your excellent powers of deduction should then lead you to the conclusion that you are dealing with Ag+, whose charge balances that of NO3- in a 1:1 ratio. When silver and potassium switch places, the silver and chlorine and potassium and nitrate will all combine in 1:1 ratios, giving the products AgCl and KNO3, which is consistent with answer choice C.

9. (A). By telling you that NaOH is present in excess, the problem saves you from having to do a limiting reagent calculation. All you need to do now is to convert grams of HCl, your limiting reagent, to grams of the NaCl product as in the calculation

30:0gHCl x x 1m±NaCi x 58.5gNaCl — 48.1gNaCl

1 36.5gHCl 1mol HCl 1mol NaCl

Be careful, though, that you’ve read the problem carefully. It does not ask for the theoretical yield or the percent yield, but the Difference Between the theoretical and actual yields. Subtract the two to get 48.1g – 41.0g = 7.1g, answer choice A.

10. For all of the conversions in this problem, be careful to set up your conversion factors so that the units cancel correctly.

Here is the answer for (a):

5.25gHF x j^HF = 0.263molHF 1 20.0gHF

Here is the answer for (b):

6.00mol SiO2 x 1mol H2SiF6 x 144.1gH2SiF6 = h SiF 1 1molSiO2 1molH2SiF6 g 2 6

Here is the answer for (c):

10.0mol HF x 22.4LHF — 224i HF 1 1mol HF

Here is the answer for (d):

3.20molSiO2 1molH2SiF6 6.022 x 1023molecules H2SiF6 = 3 1024 l l H

-1-x —"- — x-"- — -— 1.93 x 10 molecules H2SiF6

1 1mol SiO 2 1mol H 2SiF6

11. (a) The complete, balanced reaction here is FeS(s) + 2HCl(aq) — FeCl2(aq) + H2S(g). However, the problem has asked you to eliminate spectators and also allows you to ignore balancing. Chlorine is the only spectator here, so eliminating it from both sides leaves you with FeS(s) + H+(aq) — Fe2+(aq) + H2S(g).

(b) This reaction does not take place in solution, so there will not be any spectators. Recall that all combustion reactions require the reactant O2 and that the products are always water and carbon dioxide. With this knowledge, you are left with CH4 + O2 — CO2 + H2O, which you do not need to balance. In minimal air, some CO will form.

(c) Since this substance is "strongly heated" you should be able to identify it as a decomposition reaction immediately. If one of the products is CO2, the remaining product must be CaO, giving you the reaction CaCO3 — CaO + CO2.

(d) This is a double replacement reaction of the pattern. KOH (aq) + NH4Cl(s) — NH4OH(aq) + KCl(aq). You know that both of the products are soluble because one contains ammonium and the other contains an alkali metal, salts of both of which are always soluble. Eliminating spectators, this leaves you with the equation NH4Cl(s) — NH4+ (aq) + Cl- (aq).

(e) This is both a single replacement reaction and a redox reaction. Iron is more reactive than copper and so displaces it in solution. (Recall that you are to assume that a reaction always occurs in this type of problem, so even if you didn’t remember that iron was higher than copper on the reactivity series, you could still treat it as such.) Eliminating spectators, your net equation is Fe(s) + Cu2+(aq) — Cu(s) + Fe2+(aq).

(f) Nothing happens here — fluorine is a stronger oxidizing agent than chlorine.

(g) This double replacement reaction follows the equation H2SO4(aq) + Sr(NO3)2(s) — HNO3(aq ) + SrSO4(s), where you know that strontium sulfate is a precipitate because it is insoluble. Eliminating spectators, you are left with the equation SO42-(aq) + Sr(NO3)2(s) — NO3-(aq) + SrSO4(s).

(h) This is a combustion/combination reaction in which a metal is burned to create a metal oxide according to the equation. Cu(s) + O2(g) — CuO(s).

12. (a) (i) This is one of those seven acid/base reactions outlined earlier in the chapter in which an acid and carbonate mix to form three products: water, carbon dioxide, and a salt. In this case, the reaction is therefore NaHCO3(s) + HNO3(aq) — CO2(g) + H2O(l) + NaNO3(aq). Both the nitric acid and the sodium nitrate product will be significantly dissociated in water, leaving you with the equation NaHCO3(s) + H+(aq) + NO3- (aq) — CO2(g) + H2O(l) + Na+(aq) + NO3-(aq). NO3- is the only spectator ion here, so the final equation is NaHCO3 + H+ — CO2 + H2O + Na+. Note that you do not need to include phases in your final answer.

(ii) Convert mass of nitric acid to mass of sodium bicarbonate as follows. 25.0gHNO3 1molHNO3 1molNaHCO3 84.0gNaHCO3 „„ „ … ,,„„

_2_Iv_3 V_ly_2_L— XX Xrf NaT-IPO

(b) (i) Magnesium is the oxidizing agent, so it must be reduced (gain electrons) in the reaction. Lead (II) is the reducing agent, so it must be oxidized (lose electrons) in the reaction. This leaves you with the net reaction. Mg2+ + Pb2+ — Mg + Pb4+.

(ii) As a solid, magnesium will be on the paper. If the lead salt formed is insoluble then it may also be on the paper. Either way, Mg(s) will precipitate.

1

63.0g HNO

1mol HNO

1mol NaHCO

(c) (i) This reaction also follows one of the seven acid/base reaction types listed earlier, this time the base + salt — metal hydroxide precipitate + salt. In other words, this is a relatively simple double replacement reaction. The complete, balanced reaction is 2NaOH(aq) + CuSO4(aq) — Cu(OH)2(s) + Na2SO4(aq). Sodium and sulfate both appear in aqueous solution on both sides of the equation, so eliminating them as spectators, you are left with 2OH – + Cu2+ — Cu(OH)2.

(ii) 40%. The percent composition of oxygen can be calculated by dividing the total number of grams per mole of oxygen by the molecular mass of sodium hydroxide. This results in 16g440gx100 = 40%.

13. (a) Lithium is a very reactive metal and is significantly higher than hydrogen on the activity series and is therefore bound to displace it in a reaction. Copper, however, is lower than hydrogen on the activity series and therefore will not replace it.

(b) Both carbon and oxygen are more abundant in the human body than hydrogen by mass because even though there are many more hydrogen atoms than either carbon or oxygen atoms, hydrogen is the lightest of the elements. Carbon is 12 times more massive and oxygen is 16 times more massive than hydrogen.

(c) Simply speaking, water can act as both an acid and a base because it contains both an acid’s characteristic hydrogen and the hydroxide ion common to most bases. Another way to look at it is that water can accept or donate a proton.

14. (a) Na2SO3.To find the empirical formula from the percent composition, follow the steps outlined in Chapter 13. Begin by assuming that you have a total of 100g of the substance, which means that you have 36.5g sodium, 25.4g sulfur and 38.1g oxygen. Divide each of these by their gram atomic masses as follows:

36.5gNa x 1mol Na = 1 59mol Na 1 23.0gNa

25.4gS 1mol S —7s— X "7 „ = 0.79mol S 1 32.1gS

38lgOximoLO = 2.38molO 1 16.0gO

Next, divide each of these by the lowest among them (1.12), giving 1.59mol Na 4 0.79 = 2.0mol Na, 0.79mol S 4 0.79 = 1.0mol S, and 2.38mol O 4 0.79 = 3.0mol O. Attach each of these as subscripts on your empirical formula Na2SO3. This is your empirical formula.

(b) Na6S3O9. You are given the molar mass and will need to divide it by the empirical formula mass in order to find your conversion factor from the empirical to molecular formulae. The empirical formula mass of Na2SO3 is 122.1g per mole. 366.3 divided by 122.1 is 3. Multiply each of the subscripts in the empirical formula by this number to arrive at the molecular formula of Na6S3O9.

15. 59.3% C, 6.2%H and 34.6%N. Begin all percent composition calculations by finding the mass per mole of each type of atom in the substance by multiplying the number of each atom present by its gram atomic mass. In this case, you get 8 12.0g = 96g C, 10 1.0g = 10.0g H, and 4 14.0g = 56.0g N. Next calculate the molar mass of the compound by adding these three values together (96.0 + 10.0 + 56.0 = 162.0g). Finally, divide each of the individual masses by this molar mass and multiply by 100 to give you the percent composition. You should get 59.3% C, 6.2% H, and 34.6% N.

Favorable AH cannot overcome unfavorable TAS

Favorable TAS cannot overcome unfavorable AH

AH

TAS

But wait! The second law of thermodynamics states that all spontaneous processes occur with a positive change in entropy — how can a chemical reaction be driven forward by enthalpy alone? Good question. Answer: The second law refers to the entropy of the Universe. The Gibbs equation refers to the enthalpy and entropy of the System. A decrease in the enthalpy of a system means an increase in the entropy of the surroundings, which means an increase in the entropy of the universe. The second law holds true. What a relief.

Under standard conditions (typically, 1atm, pure substances at 1M concentration), the state functions are said to be in standard state and the state functions of the Gibbs equation are annotated with the ° symbol:

AG°= AH° – TAS°

However, while most free energy changes are quoted at 298K, technically this is not defined by the "standard conditions." Texts will provide tables where the temperature is clearly indicated, as in AGo298. Some texts may omit this, in which case 298K temperature must be assumed.

Changes in standard free energy, standard enthalpy, and standard entropy for a reaction can be calculated directly from the stoichiometry of the reaction along with tabulated values. The tabulated values are those for the Standard free energies, enthalpies, or entropies of formation Of the reactants and products, AG/1, AHf°and AS/1, respectively:

AG°= X aAGf°(products) – X bAGf°(reactants) AH° = X aAHf°(products) – X bAHf°(reactants) AS° = X aAS/°(products) – X bASf°(reactants)

Where X refers to the sum of products or reactants and where A And B Refer to the stoichio-metric coffiecients of each product or reactant. Standard enthalpies and entropies of formation are the enthalpies and entropies associated with forming a compound under standard conditions (1atm, 298K, 1mol L-1) from its component elements, in their elemental forms under standard conditions.

The relationship between the standard free energy and the free energy under nonstandard conditions is:

AG = AG°+ RT Ln Q

Where Q Is the reaction quotient (see Chapter 15). Compare this equation to Keq = E ~hGIBJ And be sure you understand the differences.

Measuring Heat: Thermochemistry and Calorimetry

Energy shifts between many forms. It may be tricky to detect, but energy is always conserved. Sometimes energy reveals itself as heat. Thermodynamics explores how energy moves from one form to another. Thermochemistry Investigates changes in thermal energy that accompany chemical reactions. To understand how thermochemistry is done, you need to first understand how the particular form of energy called thermal energy and the elusive term "heat" fit into the overall dance of energy and matter. In this section, we discuss how thermal energy fits into the overall energy picture, and we describe how you can measure heat transfer during chemical reactions by using a concept called heat capacity and a technique called calorimetry.

Seeing heat within the bigger energy picture

You’re probably familiar with the idea that objects can have greater or lesser amounts of energy associated with them. One of the kinds of energy an object can possess is thermal energy. In this section, we show how thermal energy relates to the overall energy of a system (like a chemical system), and how thermal energy relates to heat.

Energy itself can be divided into

Potential energy (PE) Is energy due to position. Chemical energy Is a kind of potential energy, arising from the positions of particles within systems.

Kinetic energy (KE) Is the energy of motion. Thermal energy Is a kind of kinetic energy, arising from the movement of particles within systems.

The total Internal energy Of a system (E) Is the sum of its potential and kinetic energies. When a system moves between two states (as it does in a chemical reaction), the internal energy may change as the system exchanges energy with the surroundings. The difference in energy (AE) Between the initial and final states derives from heat (q) Added to or lost from the system, and from work (w) Done by the system or on the system. "Heat" in physics and chemistry is strictly defined as a quantity of thermal energy being transferred between one system and another. As a result nothing can "contain" heat!

We can summarize these energy explanations with the help of a couple of handy formulas:

Etotal = KE + PE

AE = Efinal – Einitial = Q + W

What kind of "work" can atoms and molecules do in a chemical reaction? Remember (maybe) from your physics class that work is defined in terms of motion against a force. No motion, no work! One kind of work that is easy to understand is Pressure-volume work. Consider the following reaction:

CaCO3(s) — CaO(s) + CO2(g)

Solid calcium carbonate decomposes into solid calcium oxide and carbon dioxide gas. At constant pressure (P), This reaction proceeds with a change in volume (V). The added volume comes from the production of carbon dioxide gas. As gas is made, it expands, pushing against the surroundings. The carbon dioxide gas molecules do work as they push into a greater volume:

W = -PAV

The negative sign in this equation means that the system loses internal energy due to the work it does on the surroundings. If the surroundings did work on the system, thereby decreasing the system’s volume, then the system would gain internal energy. Note that if there were no pressure (resistive force), then P = 0, so W = 0 . . . so there is no work! So, expanding into a vacuum a gas does no work!

So, pressure-volume work can partly account for changes in internal energy during a reaction. When pressure-volume work is the only kind of work involved, any remaining changes come from heat. As mentioned in the previous section, Enthalpy (H) Corresponds to the thermal energy content of a system at constant pressure. An Enthalpy change (AH) In such a system corresponds to heat. The enthalpy change equals the change in internal energy minus the energy used to perform pressure-volume work:

AH = AE – (-PA V) = AE + PAV

Although E, P, V, And H Are state functions, heat (q) Is Not A state function, but is simply a quantity of thermal energy that transfers from a warmer object to a cooler object.

Now breathe. The practical consequences of all this theory are the following: W Chemical reactions involve an amount of heat, Q.

W Chemists monitor changes in thermal energy by measuring changes in temperature.

W At constant pressure, the change in thermal energy content equals the change in enthalpy, AH.

W Knowing AH values helps to explain and predict chemical behavior.

Using heat capacity in calorimetry

Heat is a quantity of thermal energy that transfers from warmer objects to cooler objects. But how much thermal energy can an object hold? If objects have the same thermal energy content, does that mean they are the same temperature? You can measure temperature changes, but how do these temperatures relate to heat? These kinds of questions revolve around the concept of Heat capacity. Heat capacity is the amount of heat required to raise the temperature of a system by 1K (or 1°C). In this section, we describe how the concept of heat capacity is used in an experimental technique called calorimetry.

It takes longer to boil a large pot of water than a small pot of water. With the burner set on high, the same amount of heat transfers into each pot, but the larger pot of water has a higher heat capacity. So, it takes more heat transfer to increase the temperature of the larger pot.

You’ll encounter heat capacity in different forms, each of which is useful in different scenarios. Any system has a heat capacity. To best compare heat capacities between chemical systems you use one of the following:

W Molar heat capacity, Which is the heat capacity of 1 mole of a substance

W Specific heat capacity, Or just Specific heat, Which is the heat capacity of 1 gram of a substance

How do you know whether you’re dealing with heat capacity, molar heat capacity, or specific heat capacity? Look at the units:

W Heat capacity: Energy / K

W Molar heat capacity: Energy / (molK)

W Specific heat capacity: Energy / (g°K)

Fine, but what are the units of energy? Well, that depends. The SI unit of energy is the Joule (J) (see Chapter 3 for more about the International System of units), but the units of Calorie (cal) And Liter-atmosphere (Latm) May also be used. Here’s how the joule, the calorie, and the liter-atmosphere are related (AP will stick to joules!):

1 J = 0.2390 cal 101.3 J = 1 Latm

Note that in everyday language about nutrition, a "calorie" actually refers to a "Calorie" or a kilocalorie — a calorie of cheesecake is 1,000 times larger than you think it is. Why do you think the American food industry wants to keep it that way? In Europe and Canada, candy bars are rated in kilojoules, so a 100 calorie bar is 418 kJ of energy.

Calorimetry Is a family of techniques that puts all this thermochemical theory to use. When chemists do calorimetry, they initiate a reaction within a defined system, and then measure any temperature change that occurs as the reaction progresses. There are a few variations on the theme of calorimetry, each measuring heat transfer under different conditions:

W Constant-pressure calorimetry Directly measures an enthalpy change (AH) For a reaction because it monitors heat transferred at constant pressure: AH= qP.

Typically, heat is observed through changes in the temperature of a reaction solution. If a reaction warms a solution, then that reaction must have released thermal energy into the solution. In other words, the change in thermal energy content of the reaction (qReaction) has the same magnitude as the change in thermal energy for the solution (qSolution) so the heat into one is the same as the heat out of the other, but has opposite

Sign: qsolution = – qreaction.

So, measuring QSolution allows you to calculate QReaction, but how can you measure QSolution? You do so by measuring the difference in temperature (AT) Before and after the reaction:

Qsolution = (mass of solution) x (specific heat of solution) xAT

In other words,

Q = MCPAT

Here, "m" is the mass of the solution and CP is the specific heat capacity of the solution at constant pressure. AT Is equal to TFinal – TInitial.

When you use this equation, be sure that all your units match. For example, if your CP has units of J g-1 K-1, don’t expect to calculate heat flow in kilocalories.

W Constant-volume calorimetry Directly measures a change in internal energy, AE (not AH) For a reaction because it monitors heat flow at constant volume. Often, AE and AH are very similar values, especially if no gases are involved to do pressure-volume work.

A common variety of constant-volume calorimetry is Bomb calorimetry, A technique in which a reaction (often, a combustion reaction) is triggered within a sealed vessel called a bomb. The vessel is immersed in a water bath of known volume. The vessel and water together are considered the "calorimeter" and are enclosed by an insulated container. The temperature of the water is measured before and after the reaction. Because the heat capacity of the calorimeter (CCal) is known, you can calculate heat from the change in temperature:

Q = -CCal xAT

Dealing with Heat by Using Hess’s Law

So, as you find out in the previous section, you can monitor heat by measuring changes in temperature. But what does any of this have to do with chemistry? Chemical reactions transform both matter and energy. Though reaction equations usually list only the matter components of a reaction, you can consider thermal energy as a reactant or product as well. When chemists are interested in heat appearing during a reaction (and when the reaction is run at constant pressure), they may list an enthalpy change (AH) To the right of the reaction equation. As we explain in the previous section, at constant pressure, heat equals AH:

QP = AH= Hfinal – Hinitial

If the AHListed for a reaction is negative, then that reaction releases heat as it proceeds — the reaction is Exothermic. If the AHListed for the reaction is positive, then that reaction absorbs heat as it proceeds — the reaction is Endothermic. In other words, exothermic reactions release heat as a product, and endothermic reactions consume heat as a reactant.

The sign of the AH tells us the direction of heat, but what about the magnitude? The coefficients of a chemical reaction represent molar equivalents. So, the value listed for the AH refers to the enthalpy change for one molar equivalent of the reaction. Here’s an example:

CH4(g) + 2O2(g) — CO2(g) + 2H2O(g) AH = -802 kJ

The reaction equation shown describes the combustion of methane, a reaction you might expect to release heat. The enthalpy change listed for the reaction confirms this expectation: For each mole of methane that combusts, 802 kJ of heat are released. The reaction is highly exothermic. Based on the stoichiometry of the equation, you can also say that 802 kJ of heat are released for every 2 moles of water produced.

So, reaction enthalpy changes (or reaction "heats") are a useful way to measure or predict chemical change. But they’re just as useful in dealing with physical changes, like freezing and melting, evaporating and condensing, and others. For example, water (like most substances) absorbs heat as it melts (or Fuses) And as it evaporates:

Molar enthalpy of fusion: AHfus = 6.01 kJ

Molar enthalpy of vaporization: AHvap = 40.68 kJ

The same sorts of rules apply to enthalpy changes listed for chemical changes and physical changes. Here’s a summary of the rules that apply to both:

W The heat absorbed or released by a process is proportional to the moles of substance that undergo that process. Two moles of combusting methane release twice as much heat as does one mole of combusting methane.

W Running a process in reverse produces heat flow of the same magnitude but of opposite sign as running the forward process. Freezing one mole of water releases the same amount of heat that is absorbed when one mole of water melts.

Yikes. Earlier in the chapter we described that chemical reactions can be complicated, multi-step sorts of things. Now we’re telling you that both chemical and physical processes can be associated with heat. How can you possibly keep track of all these heats? The answer is simple: Hess’s Law.

Imagine that the product of one reaction serves as the reactant for another reaction. Now imagine that the product of the second reaction serves as the reactant for a third reaction. What you have is a set of coupled reactions, connected in series like the cars of a train:

A — B and B — C and C — D

So,

A —B —C —D

You can think of these three reactions adding up to one big reaction, A — D. What is the overall enthalpy change associated with this reaction, AHAD? Here’s the good news:

AHAD = AHAB + AHBC + AHCD

Enthalpy changes are additive. Hess’s Law declares that the total enthalpy change for the overall reaction is the sum of the enthalpy changes for each of the steps, as summarized in Figure 21-8.

But the good news gets even better. Imagine that you’re trying to figure out the total enthalpy change for the following multistep reaction:

Multi-step reaction: A — B — C — D

Equivalent to Figure 21-8:

Schematic Single-step reaction: illustration a — D of Hess’s Law.

AH.,

A

AH„,

AH„,

Here’s a wrinkle: For technical reasons, you can’t measure this enthalpy change, AHXZ, directly, but must calculate it from tabulated values for AHXY and AHYZ. No problem, right? You simply look up the tabulated values and add them. But here’s another wrinkle: When you look up the tabulated values, you find the following:

AHXY = -37.5 kJmol"1 AHZY = -10.2 kJmol-1

Gasp! You need AHYZ, but you’re provided only AHZY! Relax: Reactions are reversible. The enthalpy change for a reaction has the same magnitude and opposite sign as the enthalpy change for the reverse reaction. So, if AHZY = -10.2 kJmol-1, then AHYZ = 10.2 kJmol-1. It really is that simple. So,

AHXZ = AHXY + (-AHZY) = -37.5 kJmol-1 + 10.2 kJmol-1 = -27.3 kJmol-1 Thanks be to Hess.

In This Chapter

^ Pointing out what you need to know

^ Getting in some practice

^ Going through the answers and explanations

I\ Cids and bases are extremely common participants of chemical reactions, and even have their own special class of reaction (neutralization). Because acidity/basicity is such an important property of aqueous solutions, this topic travels with a whole entourage of concepts and equations. There are different ways to define acids and bases. There are different ways to express acid and base concentrations. There are special quantities for expressing just how acidic or basic a particular compound is. This chapter summarizes key points and skills from Chapter 17, and provides practice questions that test your ability to put acid-base concepts to use.

Soaking Up the Main Points

The acid-base nitty-gritty comes down to two major ideas. First, what are acids and bases, and how can you recognize them? Second, how do acids and bases react with each other? This section organizes highlights of Chapter 17 around these two categories.

Knowing and measuring acids and bases

There are several common ways to define acids and bases, all overlapping to some extent. Having defined them, you can measure their concentrations and express those measurements in different ways, as summarized in the following points.

Three important methods for defining acids and bases are as follows:

• Arrhenius acids are those that dissociate to form hydrogen ions, while bases dissociate to form hydroxide ions in solution.

• Bronsted-Lowry acids donate protons (hydrogen ions), which are accepted by the Bronsted-Lowry base or proton acceptor.

• Lewis bases donate a pair of electrons to form a covalent bond with a Lewis acid, which accepts the pair of electrons to form a coordinate covalent bond.

The acid dissociation constant Ka gives an indication of the acidity of a substance through the expression

K a

[ H3O +]x[ A -]

I HA I

Similarly, the base dissociation constant Kb indicates the basicity of a substance through the expression

[H + ][CH3COO ] [CH 3COOH ]

The pH scale ranges from 0 to 14 and measures acidity, with low numbers indicating acidic solutions, high numbers indicating basic solutions, and 7 indicating a perfectly neutral solution. It is calculated using the equation pH = – log[H+].

The pOH scale also ranges from 0 to 14, but measures basicity instead of acidity. Low numbers mean basic solutions and high numbers mean acidic solutions. It is calculated using the equation pOH = – log[OH-].

I PH and pOH are related through the simple expression pH + pOH = 14. In a related way, [H+] x [OH-] = 10-14.

_

Reacting acids and bases

Split H2O and you can get H+ and OH-, acid and base respectively. Add acid to base, and you regain water, neutralizing the acid and base reactants. The idea is simple, but neutralization reactions can be tricky. The following points lay out the core principles.

Neutralization reactions occur when acids and bases are mixed together. The amount of acid needed to neutralize a base or vice versa depends not only on the molarity of the acidic solution, but on the number of hydrogen atoms that the acid can donate to neutralize the hydroxide ions of the base.

Equivalents of acid and base are considered rather than molarities, because different compounds can contribute different numbers of H+ or OH-. Equivalents are calculated by multiplying the number of moles of compound by the number of hydrogen or hydroxide ions being contributed by each molecule of compound.

I Titrations utilize neutralization reactions in order to determine the concentration of a mystery acid or base. They’re accomplished through the following six steps:

1. Measure out a small volume of the mystery acid or base.

2. Add a pH indicator such as phenolphthalein (see Chapter 28 for details about how to choose the best pH indicator for a reaction). Phenolphthalein, for example, is colorless in acidic solutions but pink or purple in basic solutions.

3. Neutralize by dropping the acid or base of known concentration into the solution until the indicator shows that it’s neutral (by changing color), keeping careful track of the volume added.

4. Calculate the number of moles added by multiplying the number of liters of acid or base added by the molarity of that acid or base to get the number of moles added.

5. Account for equivalents.

6. Solve for molarity. Divide the number of moles of the mystery acid or base by the number of liters measured out in Step 1, giving you the molarity.

IU Buffered solutions are made of a weak acid and its conjugate base, or a weak base and its conjugate acid, and resist changes in pH as long as only a small amount of acid or base is added. The best buffers are those that have a pKa value equal or nearly equal to the desired pH of the solution. Note that pKa = – logKa and pKb = – logKb.

PH and pKa are related through the Henderson-Hasselbach equation as follows:

Nnom 0.315mol CH3COOH 1molNaOH 1L N mi M R>u

0.080Lx-;—3-x–, _„ x -,,, „„ = 0.10LNaOH

L 1mol CH3COOH 0.250mol NaOH

3

Practice Questions

With so many concepts and equations swirling about, its clear that acids and bases just aren’t the kind of chemistry you simply read about and leave be. You are guaranteed to encounter them on the AP exam, and you are equally guaranteed to be confused by them — unless you practice, that is. Here are some targeted questions to help give you that practice.

Multiple choice

1. A solution prepared by mixing 15mL of 1.0M H2SO4 and 15mL of 2.0M KOH has a pH of

(A) 6.

(B) 7.

(C) 8.

(D) 9.

(E) 10.

2. H2O < NH3 < Ca(OH)2 < LiOH < KOH < OH-

Six bases are listed above in order of increasing base strength. Which of the following reactions must have an equilibrium constant with a value less than 1?

(A) Ca(OH)2 + K2CO3 — CaCO3 + 2KOH

(B) NH3 + H2O — NH4 + OH-

(C) 2LiOH + CO2 — Li2CO3 + 3H2O

(D) LiOH + H2O — LiH + KOH

(E) KOH + H2O — KH + OH-

3. Which of the following equations represents the reaction between solid magnesium hydroxide and aqueous hydrochloric acid?

(A) Mg(OH)2(s) + 2HCl(/) — MgCl2(a<7) + H2O(I)

(B) MgOH(s) + HCl(a<7) — MgCl(ag) + 2H2O(I)

(C) Mg(OH)2(s) + 2HCl(a<7) — MgCl2(ag) + 2H2O(I)

(D) Mg(OH)2(s) + HCl(a<7) — MgCl2(ag) + H2O(I)

(E) MgOH(s) + 2HCl(ag) — MgCl(ag) + 2H2O (I)

4. What is the ideal pKa for an indicator in a titration when the pOH at the equivalence point is 9.8?

(A) 2.1

(B) 4.2

(C) 4.9

(D) 9.8

(E) 10

Free response

5. CH3COOH(a< ) — H+(a< ) + CH3COO-(a< )

Acetic acid, CH3COOH, is a weak acid that dissociates in water according to the equation above.

(a) Calculate the pH of the solution if the H+ concentration is 3.98×10-4M.

(b) A solution of NaOH is titrated into a solution of CH3COOH.

(i) Calculate the volume of 0.250M NaOH needed to reach the equivalence point when titrated into an 80mL sample of 0.315M CH3COOH.

(c) Calculate the number of moles of CH3COO-Na+ that would have to be added to 150mL of 0.025M CH3COOH to produce a buffered solution with [H+] = 4.25 x 10-7M if the Ka Of acetic acid is 1.8×10-5. Assume that the volume change is negligible.

6. C5H5N(a<7) + H2O(I) — C5H5NH+(a<7) + OH"(aq)

Pyridine (C5H5N), a weak base, reacts with water according to the reaction above.

(a) Write the equilibrium constant expression, KB, for the reaction above.

(b) A sample of pyridine is dissolved in water to produce 50mL of a 0.20M solution. The pH of the solution is 8.50. Calculate the equilibrium constant, KB, for this reaction.

Answers with Explanations

Have any doubts about your answers to the practice questions? Neutralize those doubts by inspecting the answers given here. Even if you find that you’ve answered all the questions correctly, it’s a good idea to read the explanations to solidify your acid-base genius and because, well, you might have been simply lucky on one or two.

1. (B). Problems like this are really just limiting reagent problems. The acid and base will react with one another until one or the other is used up and the amount of remaining reac-tant will determine the pH. In this case, however, the 1.0M diprotic sulfuric acid and the 2.0M potassium hydroxide exactly balance one another:

0.015L H2SO4 1.0mo/ H2SO4 2H + nnon, ,,+

-t^2—1 x—, T,, *—1 x, 2 " = 0.030mo/ H+

1 1L H2SO4 1H2SO4

0.015L KOH x 2.0mo/ K°H x |_OH_ = 0.030mo/ OH-1 1L KOH 1 KOH

The acid has two protons to donate per molecule and the base is twice as concentrated. They’re mixed in equal amounts, so there is no leftover acid or base. The resulting pH is perfectly neutral.

2. (C). If the equilibrium constant is less than 1, then at equilibrium the concentration of reactants must be greater than the concentration of products. Because the information given is regarding base strength, the problem must be dealing with the tendency of dissociation (and therefore concentration) to increase with the strength of the base. You’re looking for a reaction where the stronger base appears on the lefthand side of the reaction arrow. The only reaction where this is the case is C.

3. (C). You’re looking for a balanced reaction, which shows the proper phases for all of the products and reactants. C is the only reaction that fulfills both of those criteria and contains molecules with atoms combining in the proper ratios for charge balance.

4. (B). The ideal PKA Is equal to the pH at the equivalence point. Use the given pOH to determine this value using the equation pH + pOH = 14. This equation gives you 14 – 9.8 = 4.2.

5. (a) Plug the given H+ concentration into the equation pH = – log[H+], which gives you pH = 3.40.

^uaaaoai 0.315molCH3COOH 1molNaOH 1L Aiaim^h

(b) 0.080Lx-R—3-x, x „ »,.,,.,,,,„„ = 0.10LNaOH

L 1mol CH3COOH 0.250mol NaOH

3

(c) 159mL. You’re asked to solve for the number of moles of CH3COO-Na+ to be added. The easiest way to do this is to go through the CH3COO- concentration, which is equivalent to the CH3COO-Na+ concentration because it dissociates in water. Begin by writing out the expression for the acid dissociation constant.

Then, solve for the concentration of CH3COO-:

0080, x 0.315 MolCH3COOH x 1 mol NaOH x 1L = 0i0iNaOH

0.080Lx-l-x 1 mol CH3COOH x 0.250 molNaOH = 0.10LNaOH

Finally, multiply this molarity by the volume to get the number of moles added.

6. (a) The equilibrium constant is the concentration of the products over the concentration of the reactants, excluding water. This results in the expression

[C5H5NH+]x[0H]

KB =-CH N-

(b) 2.3 x 10-9. Use the given pH to find the pOH and from there the OH- concentration. Because pH = 9.33, the pOH = 14 – 9.33 = 4.67. Use this to solve for [OH-].

[OH-]= 10 467 = 2.14x 10 5

This concentration should be equivalent to the [C5H5NH+] concentration. Plug these and the given C5H5N concentration into the equation derived in part (a) to get your KB, giving you

(2.14 x 10-5 )2 Kb =±-0~20— = x 10 9

In This Chapter

Going through what you can’t forget ^ Trying out some sample questions ^ Checking out your answers

Chapter 21 gave a broad overview of the relationships between differences in energy and the rates and equilibria observed for chemical reactions. These kinds of relationships are fundamental to chemistry and are ripe pickings for the AP chemistry exam. Do yourself a favor: Be sure you know the principles at work behind kinetics and thermodynamics and can apply those principles to problems. The best way to ensure that you know what you need to know about kinetics and thermodynamics is to review the tips and suggestions in this chapter and then practice the questions included.

Picking through the Important Points

For many students, the most difficult thing about kinetics and thermodynamics is telling the two apart. Both subjects describe things about chemical reactions. Both use a wealth of mathematical equations. Both have something to do with energy. Before launching into this summary of the highlights of Chapter 21, make sure you understand that kinetics and thermodynamics, though interrelated, deal with entirely separate issues:

Kinetics Deals with the rate at which reactants convert to products and how that rate depends on concentrations during a reaction.

Thermodynamics Deals with differences In different types of energy (enthalpy and entropy) Between reactants and products and how those differences relate to concentrations at equilibrium.

Important points about rates

The rules for measuring and describing the speed of a reaction are expressed in equations. Luckily, the equations are pretty simple and easy to interpret. Seeing all of them together helps make clear how all the equations relate to each other. Instead of spending time trying to memorize all these equations, spend time trying to understand how they all relate to one another.

In summary, here are the important points about rate made in Chapter 21:

The rate of a reaction does Not Determine the final extent of a reaction, Nor Does the final extent of a reaction determine the rate.

For the reaction: A + B — C

• Average rate = – A[A]/At = – A[B]/At = A[C]/At.

• Instantaneous rate = -d[A]/dt = -d[B]/dt = D[C]/dt. I For the reaction: A + 2B — C

• Instantaneous rate = – d[A]/dt = -1/2d[B]/dt = d[C]/dt.

You can extract rates from the shapes of reaction progress curves, which plot the concentration of a reactant or product versus time.

I Concentrations of reactants or products can be measured spectrophotometrically, using Beer’s law: A = Abc

I For a reaction with two reactants in its rate-determining step: Rate = K [reactant A]m[reactant B]n.

• K Is the Rate constant And depends on reaction conditions.

• Exponents M And N Are Reaction orders And depend on reaction conditions.

• The sum, M + N, Is the Overall reaction order. Zero-order reactions: Rates don’t depend on any concentration.

• Rate = K.

First-order reactions: Rates depend on the concentration of a single species.

• Rate = K[A].

Second-order reactions: Rates may depend on the concentration of one or two species:

• Rate = K[A]2 or Rate = K[A][B].

Integrated rate equations Describe the rate of a reaction with respect to the initial concentration of a reactant.

IU Half-life Describes the time it takes for the concentration of a reactant to drop to one half of its initital value.

• For a first-order reaction: t1/2 = (ln 2) / k.

I Reaction rates tend to increase with temperature.

I Increasing concentration tends to increase reaction rate.

I Catalysts increase reaction rates, but do not alter concentrations or final amounts produced at equilibrium.

• Catalyts decrease Activation energy, The energetic hill reactants must climb to reach a Transition state.

The Arrhenius equation relates the rate constant, k, to activation energy, Ea, And temperature, T, and a constant, A (the "frequency factor"), related to reactant orientation and frequency of collisions:

For a first-order reaction: [A] = [A]0-e"kt. or equivalently ln[A] = – kt + lr

Ln[A]0.

K = Ae

Ea /RT

I The slowest elementary step in a reaction mechanism is the Rate-determining step.

Important points about thermodynamics

As described in Chapter 21, thermodynamics is about big ideas and (usually) small equations. Thermodynamic concepts can seem a bit abstract, so as you peruse the highlights in this summary, anchor your brain to the idea that thermodynamics has nothing whatever to do with how fast a reaction proceeds — only how far it goes. Thermodynamics cares only about beginning and end states.

Re-energize your grasp of thermodynamics by perusing these points:

Laws of Thermodynamics:

• First Law: In any process, the total energy of the universe remains constant. Universe = system + surroundings.

• Second Law: In any spontaneous process, the overall entropy of the universe increases.

• Third Law: As temperature approaches absolute zero, the entropy of a system approaches a constant minimum.

The equilibrium constant, Keq, is related to the difference in free energy, AG, between products and reactants:

AG = – RT ln Keq

This equation can be rearranged to the following form:

K = e -AG /RT

Favorable (spontaneous) reactions possess negative values for AG and unfavorable (nonspontaneous) reactions possess positive values for AG.

The Gibbs equation Relates the difference in free energy (AG) between products and reactants to the differences in enthalpy (AH) And entropy (AS) And the temperature of the system in Kelvin:

• AG = AH – TAS

Under standard conditions (usually latm, pure substances at 1Mconcentration), the Gibbs equation reflects "standard state" quantities:

• AG°= AH° – TAS°

IU Nonstandard AG relates to standard AG° via corrections concentration (as reflected in Q, the reaction quotient):

• AG = AG°+ RT Ln Q

Heat capacity Is the amount of heat required to raise the temperature of a system by 1K.

• Molar heat capacity Is the heat capacity of 1 mole of a substance.

• Specific heat capacity Is the heat capacity of 1 gram of a substance.

Constant-pressure calorimetry Directly measures an enthalpy change (AH) for a reaction because it monitors heat flow at constant pressure: AH = qP.

• q = mCpAT

Constant-volume calorimetry Directly measures a change in internal energy, AE (not AH) For a reaction because it monitors heat flow at constant volume.

Q = – Ccal xAT

Exothermic reactions Release heat as a product, AH<0. Endothermic reactions Consume heat as a reactant, AH>0.

I Hess’s Law (enthalpies are additive):

• For the set of coupled reactions,

A — B — C — D the following is true:

AHAD = AHAB + AHBC + AHCD

• Before adding reaction equations, you may need to manipulate those equations, and doing so has consequences on the associated AH values:

Reaction: A — B Enthalpy change: AHAB compare that to:

Reaction: 2A — 2B Enthalpy change: 2AHAB compare that to:

Reaction: B — A Enthalpy change: – AHAB

Testing Your Knowledge

Concentration, rate, and equilibrium. Free energy, enthalpy, and entropy. Heat, mass, and temperature. Time spent on these kinds of problems is never wasted.

Multiple choice

Questions 1 through 4 refer to the following reactions: Reaction 1: A + B — C Rate = K[A]2

Reaction 2: D + E — F + 2G Rate = K[D][E]

1. What are the overall reaction orders for Reaction 1 and Reaction 2, respectively?

(A) First-order, Second-order

(B) Second-order, First-order

(C) First-order, First-order

(D) Second-order, Second-order

(E) Not enough information

2. For Reaction 1, how will the initial rate change if the concentration of A is doubled and the concentration of B is halved?

(A) Twofold increase

(B) Threefold increase

(C) Fourfold increase

(D) Twofold decrease

(E) Depends on the concentration of C

3. For Reaction 2, how will the initial rate change if the concentration of D is doubled and the concentration of E is tripled?

(A) Twofold increase

(B) Threefold increase

(C) Fivefold increase

(D) Sixfold increase

(E) Depends on the concentrations of F and G

4. For Reaction 2, what is the relationship between the rates of change in [D], [E], [F], and [G]?

(A) Rate = d[D]/dt = d[E]/dt = – d[F]/dt = -2d[G]/dt

(B) Rate = d[D]/dt = d[E]/dt = – d[F]/dt = 2d[G]/dt

(C) Rate = d[D]/dt = d[E]/dt = – d[F]/dt = -0.5d[G]/dt

(D) Rate = d[D]/dt = d[E]/dt = – d[F]/dt = 0.5d[G]/dt

(E) Rate = 2d[D]/dt = 2d[E]/dt = – d[F]/dt = -0.5d[G]/dt

5. Methane combusts with oxygen to yield carbon dioxide and water vapor: CH4 + 2O2 — CO2 + 2H2O

If methane is consumed at 2.79mol s-1, what is the rate of change in the concentrations of carbon dioxide and oxygen?

(A) +2.79mol s-1 CO2 and +5.58mol s-1 O2

(B) -2.79mol s-1 CO2 and +5.58mol s-1 O2

(C) +5.58mol s-1 CO2 and -5.58mol s-1 O2

(D) +2.79mol s-1 CO2 and -2.79mol s-1 O2

(E) +2.79mol s-1 CO2 and -5.58mol s-1 O2

6. You study the following reaction: D + E — F + 2G

You vary the concentration of reactants D and E, and observe the resulting rates:

[D], M [E], M Rate, M s-1 Trial 1: 2.7 x 10-2 2.7 x 10-2 4.8 x 106 Trial 2: 2.7 x 10-2 5.4 x 10-2 9.6 x 106 Trial 3: 5.4 x 10-2 2.7 x 10-2 9.6 x 106

At what rate will the reaction occur in the presence of 1.3 x 10-2 M Reactant D and 9.2 x 10-3 M Reactant E?

(A) 7.9 x 105 MS-1

(B) 1.2 x 10-4 MS-1

(C) 6.6 x 109 MS-1

(D) 8.6 x 107 MS-1

(E) 6.1 x 107 MS-1

7. For the reaction: A + 2B 2C

If the KEq = 1.37 x 103, what is the free energy change for the reaction at 25 °C? (Note: R = 8.314 J K-1mol-1)

(A) -1.50 x 103 Jmol-1

(B) +1.50 x 103 Jmol-1

(C) -1.79 x 104 Jmol-1

(D) +1.79 x 104 Jmol-1

(E) Not enough information

8. Which of the following statements is correct?

(I) The entropy of the system can decrease during a spontaneous reaction.

(II) The free energy of the system can increase during a spontaneous reaction.

(III) The enthalpy of the system can increase during a spontaneous reaction.

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I and III only

9. Which of the following statements is incorrect?

(I) Rates typically vary with time during a reaction.

(II) Rate constants typically vary with time during a reaction.

(III) Activation energies typically vary with time during a reaction.

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) II and III only

Questions 10 through 14 refer to the following choices:

10. Chemical bonds store this due to the relative positions of electrons and nuclei

11. Sum of work and heat flow

12. Equal and opposite to pressure multiplied by a change in volume

13. Translation, vibration, and rotation give evidence of this

14. Remains constant in the universe

15. A 375g plug of lead is heated and placed into an insulated container filled with 0.500L water. Prior to the immersion of the lead, the water is at 293K. After a time, the lead and the water reach the same maximum temperature, 297K. The specific heat capacity of lead is 0.127 J g-1 K-1 and the specific heat capacity of water is 4.18 J g-1 K-1. How hot was the lead before it entered the water? (Note: Density of water = 1.00 kg / L)

16. What is the molar reaction enthalpy for the following reaction? C(s) + H2O(g) — CO(g) + H2(g) Use the following data:

Reaction 1: C(s) + O2(g) — CO2(g); AH = -605 kJ Reaction 2: 2CO(g) + O2(g) — 2CO2(g); AH = -966 kJ Reaction 3: 2H2(g) + O2(g) — 2H2O(g); AH = -638 kJ

(A) 999 kJ

(B) 197 kJ

(C) -1407 kJ

(D) 394 kJ

(E) 500 kJ

Answers with Explanations

Having worked through all those problems on rates, energy, work, and heat, you may be feeling a bit depleted of energy yourself. Hang in there long enough to check your answers.

1. (D). Rate laws include a concentration factor for each reactant whose concentration affects the rate. Each concentration factor has an exponent, even if that exponent is simply 1. The exponent on each factor is the individual reaction order for that component. The overall reaction order is the sum of the individual reaction orders.

2. (C). The rate observed for Reaction 1 depends on the rate law for the reaction: Rate = K[A]2. Component B doesn’t even show up in the rate law, so cutting in half the concentration of B has no effect on the rate. However, doubling the concentration of A has a major effect on the rate because the reaction is second order in A (meaning that the concentration factor for A has an exponent of two). You can calculate the rate change associated with doubling A by substituting in simple numbers for A within the rate law. If the original concentration of A is 1 and the doubled concentration is 2, then the original and final rates are K And 4K, Respectively. So, doubling the concentration of A increases the rate fourfold.

3. (D). As described in the answer to question 2, the rate observed for Reaction 2 depends on the rate law for the reaction: Rate = K[D][E]. The reaction is first order in D and first order in E (and second order overall). You can calculate the effect of changing the concentrations of D and E by substituting simple numbers. If the original concentrations of D and E are 1 and 1, respectively, then the new concentrations are 2 and 3, respectively. So, the original rate is K(1)(1) = K, And the new rate is K(2)(3) = 6K. The changes in concentration result in a sixfold increase in rate.

4. (C). You can figure out the relationship between the rates of change of all the reactants and products by inspecting the balanced reaction equation. D + E — F + 2G. First, because D and E are reactants and F and G are products, it’s clear that the rates of change in D and E have opposite sign to those of F and G — as D and E decrease in concentration, F and G increase in concentration. Second, you must account for the stoichiometric coefficients in the reaction equation. Two moles of product G are made for every one mole made of product F and for every one mole consumed of reactants D and E. So, the rate of change in D, E, and F is one half the rate of change in G, as indicated by the coefficient of 0.5. Note that in answer C, whether each term is positive or negative has nothing whatever to do with whether a reac-tant or product is gained or lost during the reaction; positive versus negative values simply indicate that the rates occur in opposite directions.

5. (E). The reaction equation makes clear that for each mole of methane consumed, 2 moles of oxygen gas are consumed and 1 mole of carbon dioxide is produced. So

D[CH4]/dt = 0.5d[O2]/dt = – d[CO2]/dt

This means that the disappearance of 2.79mol s-1 of methane corresponds to the appearance (positive change) of 2.79mol s-1 of CO2 and the disappearance (negative change) of 2 -2.79 = 5.58mol s-1 of O2. The concentration of methane changes at half the rate of oxygen, so the concentration oxygen changes at twice the rate of methane.

6. (A). Doubling the concentration of either reactant (D or E) doubles the rate. So, the data are consistent with the rate law, Rate = K[D][E]. Solve for K By substituting known values of rate [D] and [E] from any of the trial reactions:

K = Rate / ([D][E]) = 4.8 x 106 M S-1 / (2.7 x 10-2 M X 2.7 x 10-2 M) = 6.6 x 109 M -1 s-1.

Use this calculated value of K To determine the rate in the presence of 1.3 x 10-2 MReactant D and 9.2 x 10-3 M Reactant E:

Rate = (6.6 x 109 M -1 s-1)(1.3 x 10-2 M)(9.2 x 10-3 M) = 7.9 x 105 M S-1.

7. (C). You can calculate the free energy change directly from the equilibrium constant and the temperature by using the equation: AG = -RTLnKEq. But to use the equation properly, you must make sure that all your units match. Specifically, notice that the temperature given to you in the problem has units of Celsius degrees, while the gas constant given has temperature units of Kelvin. Convert the temperature to Kelvin first: 25 °C = (25 + 273)K = 298K. Then, substitute into AG = -RTLnKEq. You are told that KEq = 1.37 x 103, so:

AG = (-1)(8.314 J K-1mol-1)(298K)ln(1.37 x 103) = -1.79 x 104 J mol1

8. (E). Although the free energy of a system must decrease during a spontaneous process (otherwise the process would not be spontaneous), both etropy and enthalpy can either decrease or increase during that process. The reason for this flexibility in entropy and enthalpy is made clear by the Gibbs equation:

AG = AH – TAS

Decreases in enthalpy are favorable and increases in entropy are favorable, it is true. However, an unfavorable change in enthalpy can be compensated by a sufficiently favorable change in entropy and vice versa, such that the overall free energy change for the process is favorable, and the process is spontaneous. Note that, although entropy may decrease within a system, entropy always increases for the universe as a whole (system + surroundings = universe).

9. (E). Rates can certainly change during the course of a reaction because rates are frequently dependent on concentrations, as expressed by rate laws. Because concentrations change during the reaction, so do the rates. Rate constants and activation energies are a different matter, however. For a given reaction at a given temperature, the rate constant and activation energy are related by the Arrhenius equation:

K = Ae – Ea /RT

The Arrhenius equation contains no terms for concentration of reactants or products, so the rate constant and activation energy typically remain constant for a given reaction.

10. (C). Potential energy is the portion of total energy due to the position of particles. Chemical energy is essentially potential energy because energy is stored within chemical bonds as a result of the favorable position of electrons between adjacent nuclei.

11. (B). The change in total energy of a system is the sum of heat flow and work:

AE = Efinal – Einitial = Q + W

Positive heat flow (+q) is heat flow into the system, which increases the total energy of the system. Positive work (+w) is work done on the system (not By The system), and also increases the total energy of the system.

12. (E). Pressure-volume work is one kind of work that can be done on a system or by a system, especially by expanding or contracting gases: W = -PAV. If gas within a system expands against the surroundings (increasing the system’s volume), then the system has done work on the surroundings and the sign of work is negative. If the surroundings expand against the system (decreasing the system’s volume), then the surroundings have done work on the system and the sign of work is positive.

13. (D). Kinetic energy is the portion of total energy due to the motion of particles. In an ideal gas, kinetic energy consists entirely of translating particles. In real substances, other forms of motion contribute to the kinetic energy, like vibration and rotation. The average kinetic energy of the particles of a system is proportional to the temperature of that system.

14. (A). Transform as it may between different forms and slip as it may between system and surroundings, the total energy of the universe remains constant.

15. (A). The key to setting up this problem is to realize that whatever heat flows out of the lead flows into the water, so: QLead = -qWater. Calculate each quantity of heat by using Q = mCPAT. Recall that AT = Tfinal – Tinitial. The unknown in the problem is the initial temperature of lead.

Qlead = (375g)(0.127J g-1 K-1)(297K – Tinitial)

To calculate QWater, you must first calculate the mass of 0.500L water by using the density of water: 0.500L x (1.00kg L-1) = 0.500kg = 500g:

QWater = (500g)(4.18J g-1 K-1)(297K – 293K) = 8.36 x 103J

Setting QLead equal to -qWater and solving for TInitial yields 473K: (375g)(0.127J g-1 K-1)(297K – TInitial) = -8.36 x 103J 297K – TInitial = -176K

Tinitial = 473K

16. (B). Reverse Reaction 2 and divide it by 2 to get Reaction 2′ (AH = +483kJ). Reverse Reaction 3 and divide it by 2 to get Reaction 3′ (AH = +319kJ). Add Reactions 1, 2′, and 3′, yielding AH = + 197kJ. Reactions 2 and 3 need reversing to put the right compounds on the reactant and product sides of the equation. Reaction 1 already has the right orientation. Reactions 2 and 3 must be divided by 2 so that the stoichiometry of the final equation matches the stoi-chiometry of the target equation (namely, one mole each of H2O, CO and H2). Even though this division temporarily creates noninteger coefficients for O2 in these equations, the non-integer coefficients cancel out when you add reactions 1, 2′, and 3′.

AG

T

TAS

In This Chapter

^ Getting comfy with the difference between kinetics and thermodynamics ^ Using rate laws to describe reaction kinetics

^ Using state functions and heat to describe reaction thermodynamics Keeping tabs on heat flow by using calorimetry and Hess’s Law

Ost people don’t like waiting. And nobody likes waiting for nothing. Research has tentatively concluded that chemists are people, too. It follows that chemists don’t like to wait, and if they must wait, they’d prefer to get something for their trouble.

To address these concerns, chemists study things such as

Kinetics Is concerned with the rates of things, such as chemical reactions. Kinetics answers the question, "How much product will I get in the next minute?"

Thermodynamics Is concerned with the relationship between energy and chemical systems. Thermodynamics answers questions like, "How far will my reaction go?" and "How hot will my beaker get?"

Kinetics and thermodynamics are separate. There’s no simple connection between how long it takes for a reaction to proceed and how productive a reaction can be. In other words, chemists have good days and bad days, like everyone else. At least they have a little bit of science to help them make sense of these things. In this chapter, you get an overview of this science. Don’t wait. Read on.

Getting There Rapidly or Slowly: Kinetics and Reaction Rates

So, you’ve got this beaker, and a reaction is going on inside of it. Is the reaction a fast one or a slow one? How fast or how slow? How can you tell? These are questions about rates. You can measure a reaction rate by measuring how fast a reactant disappears or by measuring how fast a product appears. If the reaction occurs in solution, the molar concentration of reactant or product changes over time, so rates are often expressed in units of molarity per second (M s"1).

For the following reaction, A + B — C

You can measure the reaction rate by measuring the decrease in the concentration of either reactant A (or B) or the increase in the concentration of product C over time. So, by measuring changes in concentration over a defined change in time, you can measure Average rate, The rate of the reation averaged over a specific time period:

Average rate = – A[A]/At = A[C]/At

Note that A means "average change in." Here, the expression for average rate is given in terms of reactant A or in terms of product C. The expression in terms of A includes a minus sign because that reactant disappears in the course of the reaction. No sign in the C expression means "+" as [C] increases with time. Note that if the reaction were reversed so that [C] decreased and [A] increased over time, the signs of [A] and [C] would still be opposite, but reversed.

As the period of time gets smaller (that is, as At approaches zero), the average rate approaches the Instantaneous rate, The rate of the reaction at a given instant:

Instantaneous rate = -d[A]/dt = D[C]/dt

In these kinds of equations, if you have done calculus you’ll smile down upon those less-fortunate mortals, but "d" Is only math-speak for a change in the amount of something at any given moment. If you plot the concentration of product against the reaction time, for example, you might get a curve like the one shown in Figure 21-1. Each tiny point on the graph corresponds to a slope that is an instantaneous value of D[product]/dt.

Reactions usually occur most quickly at the beginning of a reaction, when the concentration of products is the lowest and the concentration of reactants is highest. Chemists usually measure "initial rates" under these conditions and use these rates to characterize the reactions.

Time

The stoichiometry of a reaction helps determine the relative rates at which reactant and product concentrations change. For example, in the reaction

A + 2B — C

The following is true:

Instantaneous rate = – d[A]/dt = -1/2d[B]/dt = d[C]/dt

Because two moles of reactant B are consumed for every mole of reactant A, moles of B disappear at twice the rate of moles of A.

Measuring and Calculating with Rates

How can chemists follow the concentration of reactants and products as a reaction proceeds, anyway? One common method is by using Absorption spectroscopy, Which measures the concentration of substances in a sample by their ability to absorb electromagnetic radiation. If the substance you want to measure absorbs green light, you shine green light through the sample and measure how much green light passes through to the other side — less transmitted light means more absorbed light, which means a greater concentration of sample. Beer’s Law Is a useful equation that expresses this idea:

A = Abc

Where A Is the light absorbed by the sample, A Is the molar absorptivity of the substance (a property that varies from one substance to another), and B Is the length of the path traveled by the radiation through the sample. To make this easy in practice, solutions are often put into flat sided cells, so the length of the path is easily measured. The variable C Is the concentration of the substance.

Equations that relate the rate of a reaction to the concentration of some species (reactant or product) in solution are called Rate laws. The exact form a rate law assumes depends on the reaction involved. Countless research studies have described the intricacies of rate laws in chemical reactions. Here, we focus on rate laws for only the simplest reactions.

In general, rate laws take the form

Reaction rate = K [reactant A]"[reactant B]n

The rate law shown describes a reaction whose rate depends on the concentration of two reactants, A and B. Other rate laws for other reactions may include factors for greater or fewer reactants. In this equation, K Is the Rate constant, A number that must be experimentally measured for different reactions and reaction conditions (that is, different temperatures, solvents, etc.). The exponents " And N Are called Reaction orders, And must also be measured for different reactions. A reaction order reflects the impact of a change in concentration in overall rate. If M > N, Then a change in the concentration of A affects the rate more than does ^jjjABEfl Changing the concentration of B. The sum, M + N, Is the overall reaction order.

J Some simple kinds of reaction rate laws crop up frequently, so they’re worth your notice:

Zero-order reactions: Rates for these reactions don’t depend on the concentration of any species, but simply proceed at a characteristic rate:

Rate = K

First-order reactions: Rates for these reactions depend linearly on the concentration of a single species:

Rate = K[A]

Second-order reactions: Rates for these reactions may depend on the concentration of one or two species (or some intermediate combination):

Rate = k[A]2

Or

Rate = k[A][B]

Figure 21-2 illustrates how changing the reaction order changes the progress of a reaction; the more the rate of a reaction depends on the concentration of a reactant, the more severely the reaction slows down as that reactant is used up.

Figure 21-2:

Disappearance of reactant in zero-order, first-order, and second-order reactions.

Zero Order First Order Second Order

Time

Because the initial concentration of reactants may vary from one reaction to another, chemists often find it useful to employ Integrated rate equations, Which describe the reactant concentration at time T ([A]t) with respect to the initial concentration of a reactant, usually designated [A]0.

Another useful mathematical tool is Half-life, The time it takes for the concentration of a reac-tant to drop to one half of its initital value. This value is usually given the symbol T1/2.

Figure 21-3 summarizes the rate laws, integrated rate laws, rate constant units, and half-life equations for the three most common reaction types. Other types are not expected in AP.

Figure 21-3:

Summary of expressions for the kinetics of zero-, first-, and second-order reactions.

Changing Rates: Factors That After Kinetics

Despite the impression given by their choice in clothing, chemists are finicky, tinkering types. They often want to change reaction rates to suit their own needs. What can affect rates, and why? Temperature, concentration, and catalysts influence rate. In this section, we describe how these factors can change reaction rates, and we show how those changes are expressed mathematically in the Arrhenius equation, an important equation that relates reaction rate to the activation energy of a chemical reaction. Finally, we give a brief summary of how reaction rates are used to help figure out chemical reaction mechanisms, the nitty-gritty details about how exactly chemical reactions occur.

Seeing rate changes in terms of collisions

The effects of these changes can be understood in light of the Collision model, In which chemical reactions occur when reactatnts slam into one another with sufficient energy and in just the right way, As shown in Figure 21-4, if the collision isn’t "just right" the reactants will bounce away and no new products are formed.

A)

Figure 21-4:

The collision model of nonproductive and productive chemical reactions.

B)

9 «9

Here is a summary of the major factors that affect reaction rates:

Reaction rates tend to increase with temperature. This trend results from the fact that reactants must collide with one another to have the chance to react. If reactants collide with the right orientation and with enough energy, the reaction can occur. So, the greater the number of collisions, and the greater the energy of those collisions, the more actual reacting takes place. An increase in temperature corresponds to an increase in the average kinetic energy of the particles in a reacting mixture — the particles move faster, colliding more frequently and with greater energy. So an increase in temperature provides a "double whammy" effect — increased number of colissions (per second) and increased average energy per collision. Make sure you include both of these in your AP exam answer.

Ii Increasing concentration or surface area tends to increase reaction rate. The reason for this trend also has to do with collisions. Higher concentrations mean there are more reactant particles to undergo more collisions per second and have a greater chance of reacting. Increasing the concentration of reactants may mean dissolving more of those reactants in solution. Some reactants aren’t completely dissolved, but come in larger, undissolved particles. In these cases, smaller particles lead to faster reaction. Smaller particles have a greater fraction of the reactant molecules (like Compound A) on the surface where they can more easily react with molecules of Compound B, making a greater portion of the particles available for reaction.

Catalysts increase reaction rates. Catalysts don’t themselves undergo overall chemical change, and they don’t alter the amount of product a reaction can eventually produce (the Yield). However, they do enhance the probablility of a reaction occurring. Most catalysts form temporary complexes with one or both reactant molecules, holding them in more appropriate orientation for reaction. Then the product is released, leaving the catalyst to "fight another day" or engage more reactants over and over. Thus catalysts are often only needed in very small amounts to increase the rate of a reaction. Chemists (who like to define things, if you didn’t realize that by now), separate catalysts into two types. Homogenous catalysts Are of the same phase (solid, liquid, gas) as the reacting molecules, and Heterogeneous catalysts Are in a different phase. Catalysts operate mostly in the way described, but all of those ways have to do with decreasing Activation energy, The hill in chemical potential energy that reactants must climb to reach a Transition state, The highest-energy state along a reaction pathway. Lower activation energies mean faster reactions. Figure 21-5 shows a Reaction progress diagram, The energetic pathway reactants must traverse to become products. The figure also shows how a catalyst affects reactions, by lowering the activation energy without altering the energies of the reactants or products. If you think this diagram looks much like a roller coaster, you are right! In that case, the energy is gravitational potential energy, rather than chemical potential energy — but the effect is the same. Too little kinetic energy or too big a hill — no ride or no reaction.

Transition state

Figure 21-5:

A reaction progress diagram highlighting the effect of a c atalyst on activation energy.

Reaction Progress

Relating rate to activation energy

Concentration, temperature, and the the activation energy, Ea, Help determine the rate of a reaction. The influence of concentration is clearly evident in the rate laws shown in Figure 21-3. But where is the influence of temperature and activation energy? These two factors help determine the rate constant, K, As shown in the Arrhenius equation:

In the Arrhenius (yes the same Arrhenius who worked out what acids are doing) equation, K Is calculated by raising the natural logarithm (base number e, approximately 2.718) to a rational exponent. The exponent, – Ea / RT, Is the negative of the activation energy divided by the product of the gas constant, R, And the temperature, T. This whole power is multiplied by a prefactor (sometimes called the frequency factor), A. This factor has the same units as K And depends on the reaction order. It is hard to calculate theoretically, though it relates to the effectiveness of the collisions in "going over the hill," and is often an experimental measurement.

The Arrhenius equation can be rearranged into different forms that are more or less convenient for different purposes. If you know a variety of values for K And T, And would like to determine EA, then you may want to use this form:

E, , Ln K = —+ ln A

The previous equation corresponds to a linear plot of ln K Versus 1/T. The slope of the line is equal to -EA / RT And the y-intercept equals ln A.

If you know the rate constant at one temperature and would like to predict it at another temperature, then you may want to use this form:

Ln, =—fT K2 R

T, T

Using rates to determine mechanisms

Measuring reaction rates can help you figure out Mechanism, The molecular details by which a reaction takes place. Chemical reactions can be broken down into a series of Elementary steps, Simple events involving only one or two molecules that occur with characteristic rates. Each elementary step has a certain Molecularity, The number of molecular bits that must collide within that step. Unimolecular steps require one molecule, bimolecular steps require two molecules, and termolecular steps require three molecules. Termolecular steps are rare, as are collisions at one intersection of three cars (except in the movies).

Most reaction mechanisms consist of several elementary steps. Only a few of these steps consume or produce chemical players that you see in the formal reaction equation. The other players, the ones on the path between the formal reactants and products, are called Intermediates. Much chemical research on reaction mechanisms is concerned with establishing the important elementary steps and detecting and defining intermediates.

Each elementary step in a multireaction has its own rate law. The slowest step acts as a bottleneck and determines the overall rate of the reaction. For this reason, the slow step is called the Rate determining step. If an intermediate step is the slow step, chemists often assume that the previous, faster steps come to equilibrium during all the time afforded to them by the bottleneck.

Whatever the details, two things are always true:

I For a mechanism to be true, the rate law proposed by that mechanism must agree with the observed rate law.

I Mechanisms can never be proved correct directly by kinetics, but incorrect mechanisms can be disproved.

Getting There at All: Thermodynamics and State Functions

Figure 21-6 shows a reaction progress diagram like the one shown in Figure 21-5, but highlights the difference in energy between reactants and products. This difference is completely independent of activation energy, which is discussed in the previous section. Although activation energy controls the rate of a reaction, the difference in energy between reactants and products determines the extent of a reaction — how much reactant will have converted to product when the reaction is complete.

Transition state

Reaction Progress

A reaction that has produced as much product as it ever will is said to be at Equilibrium. We discuss equilibrium at length in Chapter 15. You may want to refer to that chapter for a quick refresher on that important concept, particularly on how equilibrium is quantified by the equilibrium constant, KEq. Here, we are primarily concerned with the Thermodynamics Of equilibrium — how the equilibrium position of a reaction relates to energy changes as a reaction proceeds.

^jjjMj|E# The difference in energy between reactants and products is built into the rules that determine the Keq for a reaction. The particular form of energy important in this relationship is called Free energy, G. You may sometimes see this called the Gibbs free energy (hence the G), in honor of one of the earliest American chemists, Josiah Willard Gibbs. According to Gibbs every chemical possesses a quantity of "free energy" G Depending on what it is and what the conditions (phase, temperature, pressure) are. While it is extremely difficult to find out what this is, changes in this value can easily be tracked by smart chemists. The relationship between free energy and equilibrium is

AG = – RT ln Keq

Or

K = e _AG /RT

In these equations, R Is the gas constant and T Is the temperature in Kelvin degrees. Favorable (spontaneous) reactions possess negative values for AG And unfavorable (nonspontaneous) reactions possess positive values for AG. Energy must be added to drive an unfavorable reaction forward. If the AG for a set of reaction conditions is 0, the reaction is at equilibrium. But be careful! These values only apply at a specific set of conditions. A reaction that is non-spontaneous under one set of conditions may be spontaneous if those conditions change. For

Example, most people think of hydrogen and oxygen reacting pretty spontaneously to form water, but heat water to thousands of degrees and it will split into hydrogen and oxygen!

Free energy is an example of a state function, a property of a system that depends only on its characteristics in the given moment. State functions don’t care about how you got here, only about who you are. But they change as conditions change! Your state function is different in bed at home as compared to sitting at your desk in school. In addition to free energy, other state functions include Enthalpy, H, And Entropy, S. At constant pressure, enthalpy equals the the thermal energy content of a system. Entropy is a measure of the energy of disorder in a system.

Energy is a pretty abstract concept. Before looking further into how energy governs chemical reactions, it might be useful to remind yourself of three things that are always true — the laws of thermodynamics:

First Law: In any process, the total energy of the universe remains constant. Remember, chemists divide the universe into two parts, the system and the surroundings. Universe = system + surroundings. The system and the surroundings can exchange energy; the universe cannot.

Second Law: In any spontaneous process, the overall entropy of the universe increases. Things that happen on their own do so with an increase in disorder. If you focus only on the system, the increased disorder may not be apparent, as it may occur in the surroundings. In fact, complex chemical reactions (such as the biochemical ones in our bodies) often increase order at the expense of the disorder in the surroundings.

Third Law: As temperature approaches absolute zero, the entropy of a system approaches a constant minimum. If the system is a pure, perfectly ordered crystalline solid, then that minimum of entropy is also zero.

Chemical reactions are processes that occur in the universe. So, chemical reactions obey the laws of thermodynamics. If you encounter a chemical reaction that fails to obey these laws, patent it. You’ll probably find some investors willing to lose money on it.

Two other concepts follow closely on the heels of the thermodynamic laws: Reversibility And Irreversibility. In a reversible process, a system can go back to its original state by reversing its steps Along the same path It originally took. When a reaction is at equilibrium, mass (in the form of atoms) moves between reactant and product states reversibly.

In an irreversible process, the system may be able to return to its original state, but must do so along a different path. Spontaneous processes are irreversible, as they irrevocably increase the entropy of the universe. Never fear, irreversible processes and their properties do not show up on AP chemistry exams.

Adding Up the Energies: Free Energy, Enthalpy, and Entropy

Three different kinds of energy are important for determining whether or not a particular change occurs, such as whether or not a chemical reaction goes forward. These three kinds of energy are

Gibbs free energy, G, The portion of a system’s energy that is available to do work

Enthalpy, H, The thermal energy content of a system at constant pressure

Entropy, S, A measure of the portion of a system’s energy that is unavailable to do work; entropy is related to disorder

By tracking changes in these three energies between different states (like the reactant and product states of a chemical reaction), you can determine whether a change will occur on its own, or whether that change needs to be driven forward by the addition of energy. The change in Gibbs free energy, AG, Is negative for spontaneous processes, ones that occur on their own. The change in Gibbs free energy can be seen in terms of changes in enthalpy and entropy in the Gibbs equation:

AG = AH - TAS (where T Is temperature in Kelvin)

The Gibbs equation reveals that spontaneity arises from the interplay of enthalpy and entropy, two state functions that underlie free energy. A lot of AP questions revolve around which is the winning term on the righthand side, so get all your ducks (sorry, arguments) in a row here. Negative changes in enthalpy are spontaneous, as are positive changes in entropy. Note that the effect of entropy changes on the free energy (which are generally 1,000 times smaller than enthalpy changes) is magnified by the temperature. The upshot of this equation is that chemical reactions can be driven forward by spontaneous changes in enthalpy, entropy, or both. If the changes in neither enthalpy nor entropy are spontaneous, then the reaction will not occur spontaneously. These patterns are summarized in Figure 21-7.

+

Spontaneous Reactions:

TAS

Favorable AH overcomes unfavorable TAS