Chapter 6–1

Chapter 6 Energy Changes, Reaction Rates, and Equilibrium Solutions to In-Chapter Problems 6.1 Use conversion factors to solve each problem.

a. (42 J) × (1 cal/4.184 J) = 10. cal c. (326 kcal) × (4.184 kJ/1 kcal) = 1,360 kJ b. (55.6 kcal) × (1000 cal/1 kcal) = 55,600 cal d. (25.6 kcal) × (4.184 kJ/1 kcal) × (1000 J/1 kJ) =

107,000 J 6.2 Use conversion factors to convert kcal to kJ and J.

(11.5 kcal) × (4.184 kJ/1 kcal) = 48.1 kJ (48.1 kJ) × (1000 J/1 kJ) = 48,100 J

6.3 Use conversion factors to determine the number of Calories in 14 g of olive oil.

(14 g fat) × (9 Cal/1 g fat) = 126 Cal, or 100 Cal when rounded to one significant figure 6.4 Calculate the number of Calories as in Example 6.2.

Total Calories = (26 g carb) × (4 Cal/1 g carb) + (6 g protein) × (4 Cal/1 g protein)

Total Calories = 104 Cal + 24 Cal

Total Calories = 128 Cal = rounded to 100 Cal 6.5 Use Table 6.2 to determine the bond dissociation energy for each reaction. Forming a bond is an

exothermic reaction and ∆H is a negative number. Breaking a bond is an endothermic reaction and ∆H is a positive number.

H Br+H Bra. b. H F+ H F H OH H OH+c.

Bond is broken.+119 kcal/molendothermic

Bond is broken.+88 kcal/molendothermic

Bond is formed.–136 kcal/molexothermic

6.6 The higher the bond dissociation energy, the stronger the bond. In comparing bonds to atoms in

the same group of the periodic table, bond dissociation energies and bond strength decrease down a column.

H CH

HI H C

H

HBrora. b. H OH H SHor

higher bond dissociation energy

stronger bond

higher bond dissociation energy

stronger bond

Energy Changes, Reaction Rates, and Equilibrium 6–2

6.7 Use Table 6.3 to answer the questions for a reaction with ∆H = +22.0 kcal/mol.

a. Heat is absorbed. b. The bonds in the reactants are stronger. c. The reactants are lower in energy. d. The reaction is endothermic.

6.8 Use a conversion factor to solve the problem.

C3H8(g) 5 O2(g) 3 CO2(g) 4 H2O(l)+ + ΔH = –531 kcal/molpropane

531 kcal

5 mol O21.00 mol O2

kcal–molconversion factor

x = 106 kcal of energy released

Moles cancel.Answer

6.9 Use conversion factors to solve the problems.

16 kcal1 mol C6H12O6

6.0 mol C6H12O6 x = 96 kcal of energy releasedAnswer

2 C2H6O(l) 2 CO2(g)C6H12O6(s) + ΔH = –16 kcal/molglucose ethanol

16 kcal

2 mol C2H6O1.00 mol C2H6O x = 8.0 kcal of energy released

Answer

x = 1.8 kcal of energy releasedAnswer

a.

b.

c. 20.0 g C6H12O61 mol C6H12O6

180.2 g C6H12O6x

16 kcal1 mol C6H12O6

6.10 A high energy of activation means a high energy barrier (a large hill) that separates reactants and

products. When ∆H is positive, the products are higher in energy than the reactants.

ΔH = +20 kcal/molEa

Reaction coordinate

transition state

reactants

products

Ener

gy

Chapter 6–3

6.11 Increasing the concentration of the reactants increases the number of collisions, so the reaction rate increases. Increasing the temperature increases the reaction rate. a. Increasing the concentration of O3 increases the rate of the reaction. b. Decreasing the concentration of NO decreases the rate of the reaction. c. Increasing the temperature increases the rate of the reaction. d. Decreasing the temperature decreases the rate of the reaction.

6.12

Ea uncatalyzed

Reaction coordinate

reactants

products

Ener

gy

Ea catalyzed

uncatalyzed reactioncatalyzed reaction

6.13 The forward reaction proceeds from left to right as drawn.

The reverse reaction proceeds from right to left as drawn.

2 SO2(g) O2(g)+ 2 SO3(g)a.

b. N2(g) 2 NO(g)+ O2(g)

c. C2H4O2 CH4O C3H6O2 H2O+ +

forward reaction

2 SO3(g) 2 SO2(g) O2(g)+ reverse reaction

forward reaction

2 NO(g) N2(g) + O2(g) reverse reaction

forward reaction

C3H6O2 H2O+ C2H4O2 CH4O+ reverse reaction 6.14 To write an expression for the equilibrium constant multiply the concentration of the products

together and divide this number by the product of the concentrations of the reactants. Each concentration term is raised to a power equal to its coefficient in the balanced chemical equation.

Energy Changes, Reaction Rates, and Equilibrium 6–4

6.15 Since K = 1, there are equal amounts of A and B at equilibrium. A is represented with grey spheres, whereas B is the black spheres.

6.16 When K > 1, the equilibrium favors the products.

When K < 1, the equilibrium favors the reactants. When K ≈ 1, both the reactants and products are present at equilibrium. a. 5.0 × 10–4, K < 1, reactants favored b. 4.4 × 105, K > 1, products favored c. 350, K > 1, products favored d. 0.35, K ≈ 1, reactants and products present

6.17 Write the expression for the equilibrium constant as in Answer 6.14.

K =a.[H2S]

[H2]2[S2]

2

b. K > 1, products favored c. ΔH would be negative because K > 1. d. The products are lower in energy because ∆H is negative. e. One cannot predict the rate of reaction without knowing the energy of activation.

6.18 First write an expression for K using the balanced equation. Then substitute the equilibrium

concentrations of all substances in the expression to calculate K.

K = = (0.0164)(0.0164)(0.0236)(0.00240)

4.75=Answer

a. [CO2][H2][CO][H2O]

K = = (1.26)(0.0760)2

218=Answer

b. [N2O4][NO2]2

= 1.260.0760 x 0.0760

6.19 Substitute the number of molecules of reactants and products in the expression for the equilibrium

constant to see if the system is at equilibrium. If the value is smaller than K, the reaction proceeds to the right. If the value is larger than K, the reaction proceeds to the left.

K = = 10=a. [C][D]

[A][B]=(4C)(5D)

(1A)(2B)202

The system is not at equilibrium. b. Since the value is greater than K, there is more product than would be present at equilibrium,

so the reaction proceeds to the left to form more reactants.

Chapter 6–5

6.20 Use Le Châtelier’s principle to predict the effect of a change in concentration on equilibrium. Adding more reactant or removing product drives the equilibrium to the right. Adding more product or removing reactant drives the equilibrium to the left.

a. increase [H2] (reactant) = right c. decrease [Cl2] (reactant) = left b. increase [HCl] (product) = left d. decrease [HCl] (product) = right

6.21 When temperature is increased, the reaction that removes heat is favored. When temperature is decreased, the reaction that adds heat is favored. a. The equilibrium shifts to the right when the temperature is increased. b. The equilibrium shifts to the left when the temperature is decreased.

6.22 When temperature is increased, the reaction that removes heat is favored. When temperature is decreased, the reaction that adds heat is favored.

a. The equilibrium shifts to the left when the temperature is increased. b. The equilibrium shifts to the right when the temperature is decreased.

6.23 When pressure is increased, the equilibrium shifts in the direction that decreases the number of

moles. When pressure is decreased, the equilibrium shifts in the direction that increases the number of moles. a. The equilibrium shifts to the right when the pressure is increased. b. The equilibrium shifts to the left when the pressure is decreased.

6.24 a. False. The reaction is exothermic, so increasing the temperature shifts the equilibrium to the

left. b. True. c. False. Increasing the pressure drives the equilibrium to the left, the side with fewer moles. d. True.

Solutions to End-of-Chapter Problems 6.25 Potential energy is stored energy, whereas kinetic energy is the energy of motion. A stationary

object on a hill has potential energy, but as it moves down the hill this potential energy is converted to kinetic energy.

6.26 A calorie (cal) is the amount of energy needed to raise the temperature of 1 g water 1 °C. A

Calorie (Cal) is also a unit of energy. One Calorie is equal to 1,000 calories. 6.27 Use conversion factors to solve the problems.

a. (563 Cal/h) × (1000 cal/1 Cal) = 563,000 cal/h b. 563 Cal/h = 563 kcal/h c. (563 Cal/h) × (1000 cal/1 Cal) × (4.184 J/1 cal) = 2.36 × 106 J/h d. (563 Cal/h) × (1000 cal/1 Cal) × (4.184 J/1 cal) × (1 kJ/1000 J) = 2,360 kJ/h

Energy Changes, Reaction Rates, and Equilibrium 6–6

6.28 Use conversion factors to solve the problems.

a. (704 Cal/h) × (1000 cal/1 Cal) = 704,000 cal/h b. 704 Cal/h = 704 kcal/h c. (704 Cal/h) × (1000 cal/1 Cal) × (4.184 J/1 cal) = 2.95 × 106 J/h d. (704 Cal/h) × (1000 cal/1 Cal) × (4.184 J/1 cal) × (1 kJ/1000 J) = 2,950 kJ/h

6.29 Use conversion factors to solve the problems.

a. (50 cal) × (1 kcal/1000 cal) = 0.05 kcal b. (56 cal) × (4.184 J/1 cal) × (1 kJ/1000 J) = 0.23 kJ c. (0.96 kJ) × (1 cal/4.184 J) × (1000 J/1 kJ) = 230 cal d. (4,230 kJ) × (1 cal/4.184 J) × (1000 J/1 kJ) = 1.01 × 106 cal

6.30 Use conversion factors to solve the problems.

a. (5 kcal) × (1000 cal/1 kcal) = 5,000 cal b. (2,560 cal) × (4.184 J/1 cal) × (1 kJ/1000 J) = 10.7 kJ c. (1.22 kJ) × (1 cal/4.184 J) × (1000 J/1 kJ) = 292 cal d. (4,230 J) × (1 cal/4.184 J) × (1 kcal/1000 cal) = 1.01 kcal

6.31 Calculate the number of Calories as in Example 6.2. Total Calories = 1(6 g fat) × (9 Cal/1 g fat) + (7 g carb) × (4 Cal/1 g carb) + (16 g protein) × (4 Cal/1 g

protein)

Total Calories = 144 Cal + 28 Cal + 64 Cal

Total Calories = 236 Cal, rounded to 200 Cal 6.32 Calculate the number of Calories as in Example 6.2. Total Calories = (2 g fat) × (9 Cal/1 g fat) + (19 g carb) × (4 Cal/1 g

carb) + (4 g protein) × (4 Cal/1 g

protein)

Total Calories = 18 Cal + 76 Cal + 16 Cal

Total Calories = 110 Cal, rounded to 100 Cal 6.33 Use a conversion factor to solve the problem.

(120 Cal) × (1 g carbohydrate/4 Cal) = 30 g carbohydrates 6.34 Calculate the number of Calories as in Example 6.2 and then compare.

Total Calories salmon =

(5 g fat) × (9 Cal/1 g

fat)

+ (17 g protein) × (4 Cal/1 g protein)

Total Calories = 45 Cal + 68 Cal

Total Calories 3 oz salmon =

113 Calories

Chapter 6–7

Total Calories chicken =

(3 g fat) × (9 Cal/1 g

fat)

+ (20 g protein) × (4 Cal/1 g protein)

Total Calories = 27 Cal + 80 Cal

Total Calories 3 oz chicken =

107 Calories

If we take significant figures into account, then both answers round to 100 Calories.

6.35 Calculate the number of Calories as in Example 6.2.

Total Calories = (8 g fat) × (9 Cal/1 g fat)

+ (11 g carb) × (4 Cal/1 g carb)

+ (8 g protein) × (4 Cal/1 g protein)

Total Calories = 72 Cal + 44 Cal + 32 Cal Total Calories = 148 Cal, rounded to 100 Cal

6.36 Calculate the number of Calories as in Example 6.2.

Total Calories = (14 g fat) × (9 Cal/1 g fat)

+ (40 g carb) × (4 Cal/1 g carb)

+ (16 g protein) × (4 Cal/1 g protein)

Total Calories = 126 Cal + 160 Cal + 64 Cal Total Calories = 350 Cal, rounded to 400 Cal

6.37 Use the common element colors on the inside back cover to identify the atoms. In comparing

bonds to atoms in the same group of the periodic table, bond dissociation energies and bond strength decrease down a column.

a. Cl2 has a stronger bond than Br2, because Br is below Cl in the periodic table. b. Cl2 has a stronger bond than I2, because I is below Cl in the periodic table. c. HF has a stronger bond than HBr, because Br is below F in the periodic table.

6.38

H CH

HHH C

H

HC

highest bond dissociation energy

strongest bond

H H C C H

H

H

lowest bond dissociation energy

weakest bond

98 kcal/mol 104 kcal/mol 125 kcal/mol

6.39 Use Table 6.3 to answer the questions. a. ∆H is a negative value = exothermic. b. The energy of the reactants is lower than the energy of the products = endothermic. c. Energy is absorbed in the reaction = endothermic. d. The bonds in the products are stronger than the bonds in the reactants = exothermic.

Energy Changes, Reaction Rates, and Equilibrium 6–8

6.40 Use Table 6.3 to answer the questions.

a. ∆H is a positive value = endothermic. b. The energy of the products is lower than the energy of the reactants = exothermic. c. Energy is released in the reaction = exothermic. d. The bonds in the reactants are stronger than the bonds in the products = endothermic.

6.41 Use conversion factors to solve the problems.

C(s) O2(g) CO2(g)+ ΔH = –94 kcal/mol

94 kcal1 mol C

2.5 mol C x = 240 kcal of energy releasedAnswer

a.

94 kcal1 mol O2

3.0 mol O2 x = 280 kcal of energy releasedAnswer

b.

x = 2.0 x 102 kcal of energy releasedAnswer

c. 25.0 g C 1 mol C12.01 g C

x94 kcal1 mol C

6.42 Use conversion factors to solve the problems.

2 NH3(g) N2(g)3 H2(g) + ΔH = +22.0 kcal/mol

22.0 kcal1 mol N2

1 mol N2 x = 22.0 kcal of energy absorbedAnswer

a.

22.0 kcal2 mol NH3

1 mol NH3 x = 11.0 kcal of energy absorbedAnswer

b.

x = 2.26 kcal of energy absorbedAnswer

c. 3.50 g NH31 mol NH3

17.04 g NH3x

22.0 kcal2 mol NH3

6.43 Use conversion factors to solve the problems.

6 CO2(g) 6 H2O(l)C6H12O6(aq) 6 O2(g) ++ ΔH = –678 kcal/molglucose

(molar mass 180.2 g/mol)

a. stronger bonds in products

678 kcal

1 mol C6H12O64.00 mol C6H12O6 x = 2,710 kcal

Answerb.

Chapter 6–9

678 kcal6 mol O2

3.00 mol O2 x = 339 kcalAnswer

c.

x = 37.6 kcalAnswer

10.0 g C6H12O61 mol C6H12O6

180.2 g C6H12O6x 678 kcal

1 mol C6H12O6d.

6.44 Use conversion factors to solve the problems.

H2O(l)+ ΔH = –9.0 kcal/molC CH H

HHethylene

C2H6O

ethanol

9.0 kcal

1 mol C2H4

3.5 mol C2H4 x = 32 kcalAnswer

a.

9.0 kcal

1 mol H2O0.50 mol H2O x = 4.5 kcal

Answerb.

x = 4.81 kcalAnswer

15.0 g C2H41 mol C2H4

28.06 g C2H4

x9.0 kcal

1 mol C2H4

c.

x = 0.49 kcal

Answer2.5 g C2H6O

1 mol C2H6O

46.08 g C2H6Ox

9.0 kcal

1 mol C2H6Od.

6.45 Label the points on the energy diagram.

Ener

gy

Reaction coordinate

X

Y

Zreactants

transition statehighest energy

productslowest energy

Ea

∆H

a. X b. Z c. Y d. X, Y e. X, Z f. Y g. Z

Energy Changes, Reaction Rates, and Equilibrium 6–10

6.46

Ener

gy

Reaction coordinate

A

D

C

E

B

Ea

Ea

∆H

∆H

a. Reaction A to B is endothermic. Reaction A to C is exothermic. b. Reaction A to C is faster. c. Reaction A to C generates the product lower in energy. d. D and E correspond to the transition states. e. Ea for each reaction is labeled above. f. ΔH for each reaction is labeled above.

6.47 Draw an energy diagram to fit each description.

∆H > 0Ea

Reaction coordinate

transition state

reactants

products

a.

Ener

gy

Ener

gy

Ea

Reaction coordinate

transition state

reactantsproductsb.

∆H < 0

Ener

gy Ea

Reaction coordinate

transition state

reactantsproducts

c. ∆H < 0

Chapter 6–11

6.48 Draw an energy diagram to fit each description.

∆H < 0

Ea

Reaction coordinate

transition state

reactants

productsa.

Ener

gy

Ener

gy

Ea

Reaction coordinate

transition state

reactants

products

b.∆H > 0

Ener

gy

Ea

Reaction coordinate

transition state

reactantsproductsc.

∆H < 0

6.49 Draw an energy diagram that fits the description.

Ener

gy

transition state

ΔH = –12 kcal/mol

Ea = 5 kcal

Reaction coordinate

A2 + B2

2 AB

The reaction is exothermic.

Energy Changes, Reaction Rates, and Equilibrium 6–12

6.50 Draw an energy diagram that fits the description.

Ener

gy

Ea = 21 kcal

Reaction coordinate

transition state

A + B∆H = +13 kcal/mol

C

The products are higher in energy. The reaction is endothermic.

6.51 Collision orientation affects the rate of reaction because reacting molecules must have the proper

orientation for new bonds to form. 6.52 A high energy of activation causes a reaction to be slow because few molecules have enough

energy to cross the energy barrier. 6.53 Increasing temperature increases the number of collisions, and thereby increases the reaction rate.

Since the average kinetic energy of the colliding molecules is larger at higher temperatures, more collisions are effective at causing reaction.

6.54 Decreasing the concentration decreases the number of collisions that occur, which decreases the

rate of the chemical reaction. 6.55 a. The reaction with Ea = 1 kcal will proceed faster because the energy of activation is lower.

b. K doesn’t affect the reaction rate, so we cannot predict which reaction is faster. c. One cannot predict which reaction will proceed faster from the value of ΔH.

6.56 a. The reaction with Ea = 0.10 kcal will proceed faster because the energy of activation is lower.

b. K doesn’t affect the reaction rate, so we cannot predict which reaction is faster. c. One cannot predict which reaction will proceed faster from the value of ΔH.

6.57 Energy of activation (b) and temperature (c) affect the rate of reaction. K (a) doesn’t affect the

reaction rate. 6.58 Concentration (a) and the energy difference between the reactants and transition state (c) affect the

rate of reaction. ΔH (b) doesn’t affect the reaction rate. 6.59 A catalyst increases the reaction rate (a) and lowers the Ea (c). It has no effect on ΔH (b), K (d), or

the relative energies of the reactants and products (e). 6.60 A catalyst increases the reaction rate and can be recovered unchanged in a reaction. Enzymes are

biological catalysts that bind to a reactant that can then undergo a specific reaction at an increased reaction rate.

Chapter 6–13

6.61 Use the procedure in Answer 6.19.

K = = =a. [C][D][A][B]

=(1C)(2D)(5A)(6B)

230

The system is not at equilibrium.

115

b. Since the value is smaller than the equilibrium constant (K = 4), there are more reactants than

would be present at equilibrium, and the reaction proceeds to the right to form more products. 6.62 Use the procedure in Answer 6.19.

[AB]2

[A2] [B2]K = = (6)2

(2) (2)= 9a.

The system is not at equilibrium. b. Since the value is greater than the equilibrium constant (K = 6, there are more products than

would be present at equilibrium, and the reaction proceeds to the left to form more reactants.

6.63 When K > 1, the equilibrium favors the products. When K < 1, the equilibrium favors the reactants. When K ≈ 1, both the reactants and products are present at equilibrium. When ∆H is positive, the reactants are favored. When ∆H is negative, the products are favored. a. K = 5.2 × 103, so K > 1 and the equilibrium favors the products. b. ∆H = –27 kcal/mol, so the products are favored. c. K = 0.002, so K < 1 and the equilibrium favors the reactants. d. ∆H = +2 kcal/mol, so the reactants are favored.

6.64 When K > 1, the equilibrium favors the products.

When K < 1, the equilibrium favors the reactants. When K ≈ 1, both the reactants and products are present at equilibrium. When ∆H is positive, the reactants are favored. When ∆H is negative, the products are favored. a. K = 5.2 × 10–6, so K< 1 and the equilibrium favors the reactants. b. ∆H = +16 kcal/mol, so the reactants are favored. c. K = 10,000, so K > 1 and the equilibrium favors the products. d. ∆H = –21 kcal/mol, so the products are favored.

6.65 K > 1 is associated with a negative value of ΔH. A K < 1 is associated with a positive value of ΔH. 6.66 The sign and magnitude of ∆H are unaffected by the presence of a catalyst.

Energy Changes, Reaction Rates, and Equilibrium 6–14

6.67 A is represented with black spheres, and B is represented with grey spheres. a. b. c.

[2] K = 1[1] K = 5 [3] K = 0.5 6.68 A is represented with black spheres and B is represented with grey spheres.

K = [B]/[A] = 0.1. The concentration of the product is in the numerator, so there is 10 times as much reactant (A) as product (B) at equilibrium.

A B

6.69 Use the number of molecules of reactants and products to determine K.

[AX][A][X]

K =

4(2)(2)

= = 1

[AY][A][Y]

K =

4(1)(1)

= = 4

[AZ][A][Z]

K =

1(3)(3)

= = 19

largest equilibrium constant

smallest equilibrium constant

6.70 Diagram [3] represents a reaction that has an equilibrium constant of 16.

[AB]2

[A2] [B2]K = = (0)2

(4) (4)= 0[1] [AB]2

[A2] [B2]K = = (4)2

(2) (2)= 4[2]

[AB]2

[A2] [B2]K = = (4)2

(1) (1)= 16[3]

6.71 Write the expression for the equilibrium constant as in Answer 6.14.

b. [HBr]2[CH2Br2][CH4][Br2]2

a. [NO2]2

[NO]2[O2]K = K =

6.72 Write the expression for the equilibrium constant as in Answer 6.14.

b.[H2][Cl2][HCl]2

a.[H3O+][Br–]

K = K =[HBr]

Chapter 6–15

6.73 Work backwards from the expression for the equilibrium constant to write the chemical equation.

A2 A2 + 3 B2 2 AB3b.a.[A2][A]2

K = K = [AB3]2

[A2][B2]32 A

6.74 Work backwards from the expression for the equilibrium constant to write the chemical equation.

2 A + 3 B A2B3b.a.[AB2]2

[A2][B2]2K = K =[A2B3]

[A]2[B]3A2 + 2 B2 2 AB2

6.75

K =a.[HBr]2[Br2][H2]

b. The reactants are favored at equilibrium because K < 1. c. ΔH is predicted to be positive because K < 1. d. The reactants are lower in energy because the reactants are favored at equilibrium. e. You can’t predict the reaction rate from the value of K.

6.76

K =a.[CO][Cl2][COCl2]

b. The products are favored at equilibrium because K > 1. c. ΔH is predicted to be negative because K > 1. d. The products are lower in energy because the products are favored at equilibrium. e. You can’t predict the reaction rate from the value of K.

6.77

K =[CO][H2O][CO2][H2] = K = 4.2a. b. (0.15)(0.30)

(0.090)(0.12) 6.78

K =[H2][I2]

[HI]2 = K = 0.098a. b. (0.27)2

(0.95)(0.78) 6.79 The concentration of reactant A has increased, so the equilibrium is driven to the right to form

more product. 6.80 The value of K for [4] is larger than the value for [3]. The concentration of product AB has

increased, so the reaction will be driven to the left to achieve equilibrium again.

[AB]2

[A]2[B2]K = = (2)2

(2)2 (2)= 1/2[3] [AB]2

[A]2 [B2]K = = (4)2

(2)2 (2)= 2[4]

6.81 Use Le Châtelier’s principle to predict the effect of a change in concentration on equilibrium.

Adding more reactant or removing product drives the equilibrium to the right. Adding more product or removing reactant drives the equilibrium to the left.

a. When O2 is increased, it drives the equilibrium to the right: increases NO and decreases N2. b. When NO is increased, it drives the equilibrium to the left: increases N2 and O2.

Energy Changes, Reaction Rates, and Equilibrium 6–16

6.82 Use Le Châtelier’s principle to predict the effect of a change in concentration on equilibrium. Adding more reactant or removing product drives the equilibrium to the right. Adding more product or removing reactant drives the equilibrium to the left.

a. When H2 is decreased, it drives the equilibrium to the left: increases F2 and decreases HF. b. When HF is increased, it drives the equilibrium to the left: increases H2 and F2.

6.83 In an endothermic reaction, increasing the temperature drives the reaction to the right to form more

product. Diagram [1] has more product, so it corresponds to the higher temperature, 200 °C. 6.84 The forward reaction is exothermic. Increasing the temperature of an exothermic reaction drives

the reaction to the left to form more reactant.

K[3] = (3)(3)/(3) = 3 K[4] = (5)(5)/(1) = 25 6.85 Use Le Châtelier’s principle to predict the effect of each change.

a. decrease [O3], shift to right d. decrease temperature, shift to left b. decrease [O2], shift to left e. add a catalyst, no change c. increase [O3], shift to left f. increase pressure, shift to right

6.86 Use Le Châtelier’s principle to predict the effect of each change.

a. decrease [HI], shift to right d. increase temperature, shift to left b. increase [H2], shift to right e. decrease temperature, shift to right c. decrease [I2], shift to left f. increase pressure, no change (same number of moles

on each side) 6.87 Use Le Châtelier’s principle to predict the effect of each change.

a. increase [C2H4], shift to right d. decrease pressure, shift to left b. decrease [Cl2], shift to left e. increase temperature, shift to left c. decrease [C2H4Cl2], shift to right f. decrease temperature, shift to right

6.88 Use Le Châtelier’s principle to predict the effect of each change.

a. increase [NH3], shift to right d. increase temperature, shift to right b. decrease [N2], shift to right e. decrease temperature, shift to left c. increase [H2], shift to left f. increase pressure, shift to left

6.89

K =a.[C2H6]

[H2][C2H4]K =a.

[C2H6][H2][C2H4]

b. The reactants are higher in energy when ∆H is negative. c. K > 1 because ∆H is negative. d. (20.0 g ethylene) × (1 mol ethylene/28.05 g) × (28 kcal/1

mol ethylene) = 20. kcal e. If ethylene (reactant) concentration is increased, the

reaction rate will increase. f. 1,2,4: favor shift to right; 3: favors shift to left; 5: no

change, but reduces the rate of reaction.

Chapter 6–17

6.90

K =b.[CO2]2[H2O]4

[CH4O]2[O2]3

2 CO2(g) 4 H2O(g)2 CH4O(g) 3 O2(g) ++ ΔH = –174 kcal/mola.

c. The reactants are higher in energy. d. (10.0 g CH4O) × (1 mol CH4O/32.05 g CH4O) × (174 kcal/2 mol CH4O) = 27.1. kcal released e. Although this reaction is exothermic, the reaction is very slow unless a spark or flame initiates the reaction because the reaction has a high energy of activation.

6.91 Lactase is an enzyme that converts lactose, a naturally occurring sugar in dairy products, into the

two simple sugars, glucose and galactose. 6.92 A catalytic converter uses a metal as a surface to catalyze the three reactions shown in Figure 6.4.

The unreacted gasoline molecules and carbon monoxide are oxidized to carbon dioxide and water. Nitrogen monoxide is converted to oxygen and nitrogen. Therefore, three molecules that contribute to air pollution—gasoline, carbon monoxide, and nitrogen monoxide—are removed from the engine exhaust.

6.93 Use conversion factors to solve the problem.

2000 mL100 mL

5 g glucose1 g glucose4 Calories

x x = 400 Calories

6.94 Increasing the concentration of salicylic acid, increasing the concentration of acetic acid,

decreasing the temperature, and removing water (a product) are four ways to drive the reaction to the right to favor products.

6.95 Total Calories = (29 g fat) × (9 Cal/1 g

fat) + (34 g carb) × (4 Cal/1 g

carb) + (32 g protein) × (4 Cal/1 g protein)

Total Calories = 261 Cal + 136 Cal + 128 Cal

Total Calories = 525 Calories

(525 Cal) × (1 h/280 Cal) = 1.9 h rounded to 2 h 6.96 Total Calories = (11 g fat) × (9 Cal/1 g

fat) + (30 g carb) × (4 Cal/1 g

carb) + (12 g protein) × (4 Cal/1 g protein)

Total Calories = 99 Cal + 120 Cal + 48 Cal

Total Calories = 267 Calories

(4.5 h) × (710 Cal/h) = 3,200 Cal or 3,200 kcal because 1 Cal = 1 kcal (3,200 Cal) × (1 piece of pizza)/(267 Cal) = 12 pieces of pizza

Energy Changes, Reaction Rates, and Equilibrium 6–18

6.97

6.98 Use conversion factors to solve the problem.

68.4 kcal x = 33.9 kcal/g H21 mol H2

1 mol H2

2.02 g H2

327 kcal x = 7.09 kcal/g C2H6O1 mol C2H6O1 mol C2H6O46.1 g C2H6O

On a per gram basis, hydrogen is a better source of energy than ethanol.

6.99 Use conversion factors to solve the problem.

4 qt x1 gal

1303 kcal1 mol = 30,200 kcal1 gal gas x 946 mL x

1 qt0.700 g x

1 mL1 mol C8H18 x

114.3 g C8H18 6.100 Total Calories = (4 g fat) × (9 Cal/1 g fat) + (24 g carb) × (4 Cal/1 g

carb) + (12 g protein) × (4 Cal/1 g protein)

Total Calories = 36 Cal + 96 Cal + 48 Cal

Total Calories = 180 Calories

(180 Cal/1 bar) × (2 bars/1 day) × (30 days/1 month) × (1000 cal/1 Cal) × (4.184 J/1 cal) × (1 kJ/1000 J) = 4.5 × 104 kJ