Chapter 1 Matter and Measurement - Riverside City College



Chapter 6 Energy Changes, Reaction Rates, and Equilibrium

Solutions to In-Chapter Problems

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 |

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

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

4. Calculate the number of Calories as in Example 6.2.

|Total Calories = |6 g fat ( (9 Cal/1 g fat) |+ 20 g carb ( (4 Cal/1 g carb) |+ 2 g protein ( (4 Cal/1 g protein) |

|Total Calories = |54 Cal + |80 Cal + |8 Cal |

|Total Calories = |142 Cal = rounded to 100 Cal |

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.

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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.

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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.

8. Use a conversion factor to solve the problem.

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9. Use conversion factors to solve the problems.

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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.

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11. A low energy of activation means a low energy barrier (a small hill) that separates reactants and products. The ∆H is negative, so the products are lower in energy than the reactants.

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12. 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.

13. Yes, H2SO4 is a catalyst because it speeds up the rate of the reaction. It does not affect the relative energies of reactants and products.

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

The reverse reaction proceeds from right to left as drawn.

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15. 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.

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

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17. 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

18. When K < 1, the reaction is endothermic. When K > 1, the reaction is exothermic.

a. K < 1, reactants favored, endothermic

b. K > 1, products favored, exothermic

c. K > 1, exothermic

d. K < 1, (slightly) endothermic

19. Write the expression for the equilibrium constant as in Answer 6.15.

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b. K > 1, products favored

c. (H would be negative since K > 1.

d. The products are lower in energy since ∆H is negative.

e. One cannot predict the rate of reaction without knowing the energy of activation.

20. First write an expression for K using the balanced equation. Then substitute the equilibrium concentrations of all substances in the expression to calculate K.

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21. 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

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 right when the temperature is increased.

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

23. 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.

24. 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.

Solutions to End-of-Chapter Problems

6.25 Potential energy is stored energy while 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

6.28 Use conversion factors to solve the problems.

a. 704 Cal/h ( 1,000 cal/1 Cal = 704,000 cal/h

b. 704 Cal/h = 704 kcal/h

c. 704 Cal/h ( 1,000 cal/1 Cal ( 4.184 J/1 cal = 2.95 ( 106 J/h

d. 704 Cal/h ( 1,000 cal/1 Cal ( 4.184 J/1 cal ( 1 kJ/1,000 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 ( 1,000 cal/1 kcal = 5,000 cal

b. 2,560 cal ( 4.184 J/1 cal ( 1 kJ/1,000 J = 10.7 kJ

c. 1.22 kJ ( 1 cal/4.184 J ( 1,000 J/1 kJ = 292 cal

d. 4,230 J ( 1 cal/4.184 J ( 1 kcal/1,000 cal = 1.01 kcal

6.31 Calculate the number of Calories as in Example 6.2.

|Total Calories = |16 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 Use a conversion factor to solve the problem.

29.7 kJ/g ( 1,000 J/kJ ( 1 cal/4.184 J ( 1 Cal/1,000 cal = 7.10 Cal/g

6.35 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 |

|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, both answers round to the same value, 100 Calories.

6.36 Calculate the number of Calories as in Example 6.2 and then compare.

|Total Calories | | | |

|egg = |6 g fat ( (9 Cal/1 g fat) |+ 6 g protein ( (4 Cal/1 g protein) | |

|Total Calories = | 54 Cal + 24 Cal | |

|Total Calories | |

|one egg = |78 Calories |

|Total Calories | | | |

|nonfat milk = |12 g carb ( (4 Cal/1 g carb) |+ 9 g protein x (4 Cal/1 g protein) | |

|Total Calories = | 48 Cal + 36 Cal | |

|Total Calories | |

|1 c. nonfat milk = |84 Calories |

If we take significant figures into account, both answers round to the same value, 80 Calories.

6.37 When energy is absorbed, the reaction is endothermic and (H is positive (+).

When energy is released, the reaction is exothermic and (H is negative (–).

6.38 (H is the heat of reaction or enthalpy change and describes the amount of energy that is absorbed or evolved during a chemical reaction. The bond dissociation energy describes the amount of energy required to separate the atoms in a bond by breaking the covalent bond(s) between them.

6.39 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.

6.40

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6.41 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.

6.42 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.

6.43 Use conversion factors to solve the problems.

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6.44 Use conversion factors to solve the problems.

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6.45 Use conversion factors to solve the problems.

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6.46 Use conversion factors to solve the problems.

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6.47 The transition state is the unstable intermediate located at the top of the energy hill that separates reactants from products. The difference in energy between the reactants and the transition state is the energy of activation, symbolized by Ea.

6.48 Ea is the difference in energy between the reactants and the transition state, whereas ∆H is the amount of energy that is absorbed or evolved during a chemical reaction. That is, ΔH is the difference in energy between the reactants and the products of the reaction.

6.49 Label the points on the energy diagram.

|[pic] |a. X |

| |b. Z |

| |c. Y |

| |d. X, Y |

| |e. X, Z |

| |f. Y |

| |g. Z |

6.50

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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.

6.51 Draw an energy diagram to fit each description.

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6.52 Draw an energy diagram to fit each description.

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6.53 Draw an energy diagram that fits the description.

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6.54 Draw an energy diagram that fits the description.

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6.55 Collision orientation affects the rate of reaction because reacting molecules must have the proper orientation for new bonds to form.

6.56 A high energy of activation causes a reaction to be slow because few molecules have enough energy to cross the energy barrier.

6.57 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.

6.58 Decreasing the concentration decreases the number of collisions that occur and therefore decreases the rate of the chemical reaction.

6.59 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. We cannot predict which reaction will proceed faster from the value of (H.

6.60 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. We cannot predict which reaction will proceed faster from the value of (H.

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

6.62 Concentration (a) and the energy difference between the reactants and transition state (c) affect the rate of reaction. (H (b) doesn’t affect the reaction rate.

6.63 A catalyst increases the reaction rate (a) and lowers the Ea (c). It has no effect on (H (b), K (d), or the relative energies of the reactants and products (e).

6.64 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.

6.65 The forward reaction proceeds from left to right as drawn and the reverse reaction proceeds from right to left as drawn.

6.66 A reversible reaction can occur in either direction (from reactants to products or products to reactants). The reverse reaction proceeds from right to left as written (from products to reactants).

6.67 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, K > 1, the equilibrium favors the products.

b. ∆H = –27 kcal/mol, the products are favored.

c. K = 0.002, K < 1, the equilibrium favors the reactants.

d. ∆H = +2 kcal/mol, the reactants are favored.

6.68 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, K< 1, the equilibrium favors the reactants.

b. ∆H = +16 kcal/mol, the reactants are favored.

c. K = 10,000, K > 1, the equilibrium favors the products.

d. ∆H = –21 kcal/mol, the products are favored.

6.69 K > 1 is associated with a negative value of (H. A K < 1 means (H has a positive value.

6.70 The sign and magnitude of ∆H are unaffected by the presence of a catalyst.

6.71 A is represented with black spheres, and B is represented with grey spheres.

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6.72 A is represented with black spheres, and B is represented with grey spheres.

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6.73 Write the expression for the equilibrium constant as in Answer 6.15.

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6.74 Write the expression for the equilibrium constant as in Answer 6.15.

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

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

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6.77

|[pic] |b. The reactants are favored at equilibrium since K < 1. |

| |c. (H is predicted to be positive since K < 1. |

| |d. The reactants are lower in energy since the reactants are favored at |

| |equilibrium. |

| |e. You can’t predict the reaction rate from the value of K. |

6.78

|[pic] |b. The products are favored at equilibrium since K > 1. |

| |c. (H is predicted to be negative since K > 1. |

| |d. The products are lower in energy since the products are favored at |

| |equilibrium. |

| |e. You can’t predict the reaction rate from the value of K. |

6.79

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6.80

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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.

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 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

6.84 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

6.85 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

6.86 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

6.87

|[pic] |b. The reactants are higher in energy when ∆H is negative. |

| |c. K > 1 since ∆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. |

6.88

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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.

6.89 Lactase is an enzyme that converts lactose, a naturally occurring sugar in dairy products, into the two simple sugars, glucose and galactose.

6.90 A catalytic converter uses a metal as a surface to catalyze 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.

6.91 Use conversion factors to solve the problem.

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6.92 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.

6.93

|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

6.94

|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 = 3200 Cal or 3200 kcal since 1 Cal = 1 kcal

3200 Cal ( 1 piece of pizza/267 Cal = 12 pieces of pizza

6.95 Use conversion factors to solve the problem.

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6.96 Use conversion factors to solve the problem.

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

6.97 Use conversion factors to solve the problem.

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6.98

|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 |

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