Chem 121 Fritsch

Ch 14 and 15 Practice Problems - KEY The following problems are intended to provide you with additional practice in preparing for the exam. Questions come from the textbook, previous quizzes, previous exams, and other sources. A solutions manual is supplied in a separate document. 1. When the concentrations of reactant molecules are increased, the rate of reaction increases. The best explanation for this phenomenon is that as the reactant concentration increases, A. the average kinetic energy of molecules increases. B. the frequency of molecular collisions increases. C. the rate constant increases. D. the activation energy increases. E. the order of reaction increases. Answer: B 2. With respect to the reaction profile diagram, which choice correctly identifies all the numbered positions?

2 3 4

A. transition state intermediate product B. transition state intermediate product C. transition state activation energy catalyst D. intermediate transition state product E. intermediate intermediate transition state

Answer: B 3. What is the equilibrium constant for the overall reaction? overall reaction SO3(g) + NO2(g) ↔ SO2(g) + NO3(g) Kc = ? elementary reaction 2 SO2(g) + O2(g) ↔ 2 SO3(g) Kc1 = 1.89 x 1013 elementary reaction NO2(g) + ½O2(g) ↔ NO3(g) Kc2 = 7.14 x 102

Chem 121 Fritsch

4. In the General Chemistry Laboratory you and your lab partners seek to determine overall rate law for the reaction of iodine monochloride with hydrogen which has the following overall reaction:

2ICl + H2 → I2 + 2HCl Determine the rate law and calculate the value of the rate constant, k.

Expt [ICl]0 (M) [H2]0 (M) Initial Rate (M•s-1)

1 0.1004 0.1001 1.48 x 10-3

2 0.2002 0.0998 6.04 x 10-3

3 0.0999 0.0493 1.51 x 10-3

4 0.3009 0.1997 1.37 x 10-2

Chem 121 Fritsch

5. Use the data at the right that were collected in a study of the isomerization of cis-butene to trans-butene. Analysis of the data gave a straight line relationship when the ln [cis-butene] was plotted versus time.

Calculate the concentration at 95 s.

6. For the following reaction

2SO3(g) 2SO2(g) + O2(g), Kc = 1.3 x 10-5 at 500 K. If [SO3] = [SO2] = [O2] = 1.75 × 10−3 M at 500 K, which one of the following statements is true? A. The system is at equilibrium, thus no concentration changes will occur. B. The concentrations of SO3 and O2 will increase as the system approaches equilibrium. C. The concentration of SO3 will increase as the system approaches equilibrium.

time (s) [cis-butene] (M) 0 0.0894 5 0.0765

10 0.0621 15 0.0527 20 0.0446 25 0.0381 30 0.0317 40 0.0231

Chem 121 Fritsch

D. The concentrations of SO2 and SO3 will fall as the system moves toward equilibrium. E. The concentrations of SO2 and O2 will increase as the system approaches equilibrium. Answer:_____C____ 7. The degradation of iodide by hypochlorous acid relies on a two-step mechanism where the first elementary step is an equilibrium followed by a rate determining step as shown below.

1

fast, Kc = 2.98 x 10-4

2

slow

a. Write the rate laws for each of the elementary step.

b. Using a fast, pre-equilibrium analysis, show that the rate law for the production of chloride is overall first order.

Chem 121 Fritsch

8. The reaction below is at equilibrium, how will the equilibrium will shift to the left, the right, or remain unchanged when the system is modified. Place an X in the appropriate box. Use same starting equilibrium conditions for each individual change.

SO2(g) + NO3(g) SO3(g) + NO2(g) ∆Hºrxn = -15 kJ/mol

Equilibrium will shift to the

Change to the system Left Unchanged Right

SO2 is added. X

The volume is increased. X

NO2 is added. X

The temperature is decreased. X

9. For the following chemical equations, balance the equations and then write the equilibrium expression (Kc) and the reaction quotient (Q). a. NO(g) + O2(g) ↔ N2O3(g) b. SF6(g) + SO3(g) ↔ SO2F2(g) c. C2H6(g) + O2(g) ↔ CO2(g) + H2O(g)

Chem 121 Fritsch

10. Beginning with the rate law for first-order kinetics, derive the equation for half-life for first-order kinetics. Be explicit with each step.

Start here End here

Chem 121 Fritsch

11. Answer the questions below regarding reaction profiles A and B.

Which profile has more transition states? A

Which profile will have the slower first reaction step? A

In which profile will the return of products to reactants occur at the fastest rate?

A

Which profile has the slowest step converting an intermediate to final products?

B

Which profile is appropriate for an endothermic reaction? A

12. In a 5.0 L flask, 0.25 atm HBr(g), 0.15 atm H2(g), and 0.025 atm Br2(g) were combined. The temperature was raised to 425 °C and allowed to come to equililbrium. What are the equilibrium partial pressures of HBr, H2, and Br2?

2HBr(g) ↔ H2(g) + Br2(g) Kp = 4.18 x 10-9 at 425 °C

Chem 121 Fritsch

13. The data at the right were collected in a study of the photochemical degradation of ozone, O3, by light. Analysis of the data gave a straight line relationship when the [O3]-1 was plotted versus time.

a. Calculate the concentration at 523 s.

time (s) [O3] (M)

0 2.98 x 10-3 50 2.40 x 10-3

100 1.90 x 10-3 150 1.60 x 10-3

200 1.35 x 10-3 250 1.18 x 10-3 300 1.07 x 10-3 400 8.65 x 10-4

Chem 121 Fritsch

[A]t = 7.101 x 10-4 M

14. The formation of NO(g) from its elements is not favored Kc = 4.10 x 10-3 at 2000 °C. What will be the equilibrium concentrations of N2 and O2 after 0.264 mol of NO are added to a 1.5L flask and the temperature raised to 2000 °C? What are the masses in grams of N2 and NO in the flask at equilibrium?

N2(g) + O2(g) ↔ 2NO(g)

Chem 121 Fritsch

15. The reaction below was studied in aqueous solution and the data in the table were collected. The concentrations were held constant between experiments, [C2H5I]0 = 0.00203 M and [OH-]0 = 1.413M. (The equation below represents the one elementary step in the reaction.)

C3H7I(aq) + OH-(aq) C3H7OH(aq) + I-(aq)

What is the activation energy, Ea, for the reaction in kJ/mol?

Expt # T (°C) k (s-1)

1 12.2 0.0616

2 23.7 0.200

3 42.5 1.20

4 68.8 10.8

Chem 121 Fritsch

16. During the combustion of H2, hydroxyl radical (•OH) is formed by the reaction, H(g) + ½O2(g) ↔ •OH(g) Kc = ?

Use the following reactions to determine Kc for the above reaction. ½H2(g) + ½O2(g) ↔ •OH(g) Kc = 0.58

H2(g) ↔ 2H(g) Kc = 1.6 x 10-3

Chem 121 Fritsch

17. Find the activation energy, Ea, and collision frequency, A, from the data provided below:

Answer: Ea = 69.3 kJ/mol

A = 1.24 x 1013 s-1 18. For the reaction below, at a particular time, −∆[BrO3

−]/∆t = 1.5 × 10−2 M/s. What is −∆[Br−]/∆t at the same instant?

BrO3− + 5Br−+ 6H+ → 3Br2 + 3H2O

Answer: 7.5 × 10−2 M/s 19. For the overall chemical reaction shown below, which one of the following statements can be rightly assumed?

Chem 121 Fritsch

2H2S(g) + O2(g) → 2S(s) + 2H2O(l)

A. The reaction is third-order overall. B. The reaction is second-order overall. C. The rate law is, rate = k[H2S]2 [O2]. D. The rate law is, rate = k[H2S] [O2]. E. The rate law cannot be determined from the information given. Answer: E 20. Dinitrogen tetraoxide (N2O4, 32.56 g, 92.01 g/mol) was added to a 10 L flask and the system was allowed to come to equilibrium according to the reaction below at 430 °C. At equilibrium [NO2]eq = 0.0117 M.

20a. Calculate the equilibrium constant.

20b. The gas was compressed to ½ its original volume with no temperature change. Find the [N2O4]eq after compression.

Chem 121 Fritsch

Chem 121 Fritsch

21. When the concentrations of reactant molecules are increased, the rate of reaction increases. The best explanation for this phenomenon is that as the reactant concentration increases, A. the average kinetic energy of molecules increases. B. the frequency of molecular collisions increases. C. the rate constant increases. D. the activation energy increases. E. the order of reaction increases. Answer: B 22. The thermal decomposition of acetaldehyde, CH3CHO → CH4 + CO, is a second-order reaction. The following data were obtained at 518°C inside a 1.164 L flask. Based on the data given, what is the half-life for the disappearance of acetaldehyde?

time (s) Pressure CH3CHO (mm Hg)

0 364 42 330

105 290

720 132

Chem 121 Fritsch

23. Use the following data to determine the rate law for the reaction

2NO + H2 → N2O + H2O

Expt # [NO]0 (M) [H2]0 (M) Initial rate (M/s) (10-2)

1 0.205 0.0649 1.461 2 0.210 0.2621 1.459 3 0.419 0.0651 5.839

24. A first-order reaction has a rate constant of 7.5 × 10−3 s-1. Find the time required for the reaction to convert 60% of A to B.