201B Work 1 Kinetics

Chem 201B Dr. Lara Baxley

Kinetics Worksheet 1. For the reaction below, if substance A is disappearing at a rate of 1.82 x 10–2 mol/Ls, at what

rate is C appearing?

3 A + 3 B → 5 C + 2 D

2. Ozone (O3) is produced in the stratosphere by the chemical reaction shown below. If at a given instant, molecular oxygen (O2) is reacting at a rate of 2.17 x 10–5 mol/Ls, at what rate is ozone being produced?

3 O2 (g) → 2 O3 (g)

3. For the reaction, H2O2 (aq) + 3 I− (aq) + 2 H+ (aq) → 2 H2O (l) + I3− (aq)

The rate law was experimentally determined to be Rate = k[H2O2][I− ]

a. What is the reaction order in terms of H2O2? b. What is the reaction order in terms of I−? c. What is the reaction order in terms of H+? d. What is the overall reaction order of this reaction?

4. If a reaction is second order in B and the concentration of B increased from 0.0850 M to

0.2975 M, what should happen to the rate?


5. The data below were collected for the reaction: BrO3– + 5 Br– + H3O

+ → 3 Br2 + 9 H2O

Exp # Initial Concentration of Reactants (mol/L) Initial Rate (mol/L•s) BrO3

– Br– H3O+

1 0.10 0.10 0.10 1.2 2 0.20 0.10 0.10 2.4 3 0.10 0.30 0.10 3.6 4 0.20 0.10 0.15 5.4

a. Determine the rate law for this reaction. b. Calculate the value of k for this reaction and express it with the correct units.

6. The data below were collected for the reaction at 327 °C: H2 (g) + I2 (g) → 2 HI (g)

Experiment Initial [H2], M Initial [I2], M Initial Rate, mol/L•s 1 0.113 0.110 3.01 x 10–4 2 0.220 0.330 1.76 x 10–3 3 0.550 0.110 1.47 x 10–3

a. Determine the rate law for this reaction. b. Calculate the value of k for this reaction and express it with the correct units.


7. The data below was collected for the hypothetical reaction 2 A → B + C.

Time (s) [A]t

0 0.1 40 0.0768 80 0.059

120 0.0453 160 0.0348

a. In Excel, create both a first-order and a second-order integrated plots. b. Based on your graphs, is the reaction zero, first or second order? Explain your reasoning.

c. Determine the value for the rate constant with correct units.

8. The rate constant for the reaction below is 6.2 x 10−5 mol L−1s−1. If the initial concentration of A is 0.0500 M, what is its concentration after 115 s?

A → B + C

9. Hydrogen iodide decomposes according to the equation shown below. The second order rate constant for this reaction is 1.6 x 10–3 L mol–1 s–1 at 700 ºC. If the initial concentration of HI in a container is 5.1 x 10–2 M, how many minutes will it take for the concentration to be reduced to 4.9 x 10–3 M at 700 ºC?

2 HI(g) → H2(g) + I2(g)


10. What is the half-life of a reaction where k = 10 s−1? Why is it that this question cannot be answered for a reaction where k = 10 L mol−1s−1?

11. The half-life of a certain first order reaction is 430 seconds. What is the rate constant for this

reaction? 12. The graph below shows rate constant data at different temperatures for the second order

reaction, HI(g) + CH3I(g) → CH4(g) + I2(g).

a. Use the equation for the line (not the Arrhenius equation) to determine the rate constant at 300 ºC.

b. Use the equation for the line (not the Arrhenius equation or estimating from graph) to

determine the activation energy of this reaction.

y = -16653x + 30.88

-4.5

-4

-3.5

-3

-2.5

-2

0.00198 0.002 0.00202 0.00204 0.00206 0.00208 0.0021

ln k

1/T (K-1)

Arrhenius Graph for Reaction of HI and CH3I


13. The reaction 2 NOCl → 2 NO + Cl2 has k = 9.3 x 10–5 L mol–1 s–1 at 100 ºC and an activation energy of 99 kJ/mol. a. What is the rate constant at 130 ºC?

b. At what Celcius temperature will the rate constant be 5.1 x 10–7 L mol–1 s–1? 14. The figure below shows a potential energy diagram for a chemical reaction that can occur by

two different reaction pathways.

a. What is the equation for the overall reaction? b. What is the mechanism for the reaction with the higher pathway? c. What is the mechanism for the reaction with the lower pathway? d. On the figure, identify, 1) The activation energy for the reaction with the higher energy

pathway, 2) the activation energy for the reaction with the lower energy path, and 3) ∆H for this reaction.

e. Identify each of the species A to F as a reactant, intermediate, product, or catalyst.


15. Consider the following mechanism for a reaction

C4H9Br → C4H9+ + Br– slow

C4H9+ + H2O → C4H9OH2

+ fast C4H9OH2

+ + H2O → C4H9OH + H3O+ fast

a. Write the chemical equation for the overall reaction b. What are the intermediates in the mechanism? c. Write the rate law for the overall reaction.

16. For the reaction, NO2(g) + CO(g) → NO(g) + CO2(g), the experimentally determined rate law is, Rate = k[NO2]

2.

A suggested mechanism for this reaction is,

step 1: NO2(g) + NO2(g) → NO3(g) + NO(g)

step 2: NO3(g) + CO(g) → NO2(g) + CO2(g)

a. Write the rate law for each elementary step in the reaction. b. Is this a reasonable mechanism for this reaction? Why or why not? c. Which step is the slow step in this mechanism? Why? d. Identify any intermediates in the reaction.

17. A proposed mechanism for a reaction is:

A + B ⇌ C (fast) C + A → B + E (slow) E + F → G + D (fast) a. Write the overall reaction. b. Which species is likely a catalyst of this reaction? c. Which species is an intermediate? d. Keeping in mind that a rate law may include a concentration term for a catalyst, what is the

rate law for this reaction?


18. For the reaction A + B + C → D + E, the rate law is rate = k[A][B][C]. Three possible mechanisms are shown below. Identify whether each one is a likely mechanism for this reaction. If it is not a possible mechanism, explain why not.

a. Mechanism 1:

A + B + C → X + E (slow) X → D (fast)

b. Mechanism 2: Step 1 A + B ⇌ X (fast)

Step 2 X + C → Y + E (slow)

Step 3 Y → D + E (fast) c. Mechanism 3:

Step 1 A + B ⇌ X (fast)

Step 2 X + C → Y + E (slow)

Step 3 Y → D (fast) 19. Consider the following mechanism for a reaction

BrCl + H2 ⇌ HBr + HCl fast

HCl + BrCl → HBr + Cl2 slow a. Write the chemical equation for the overall reaction b. What is/are the intermediate(s) in the mechanism? c. Write the rate law for the overall reaction. (This is trickier than it looks. Be careful)


First Order Integrated Plot

y = -0.0066x - 2.3026

-3.6

-3.4

-3.2

-3

-2.8

-2.6

-2.4

-2.2

-2

0 20 40 60 80 100 120 140 160 180

Time (s)

ln [

A]

Second Order Integrated Plot

0

5

10

15

20

25

30

35

0 20 40 60 80 100 120 140 160 180

Time (s)

1/[

A]

Answer Key 1. 3.03 x 10–2 mol/Ls 2. 1.45 x 10–5 mol/Ls 3. a. 1st

b. 1st c. zero d. 2nd

4. Rate will increase by a factor of 12.3 5. a. Rate = k[BrO3

–][Br–][H3O+]2

note: you will not receive full credit if you only write k[BrO3–][Br–][H3O

+]2, of if you capitalize K (that’s a different constant!), or if you write in the value of k instead of just k. The great thing about a rate law is that it works at any temperature, but k is different at different temperatures.

b. k = 1.2 x 104 L3/mol3!s 6. a. Rate = k[H2][I2]

b. k = 0.0243 L/mol!s 7. a.

b. The reaction is first order because the first order integrated rate plot is linear, while the

second order rate plot is curved. c. m = −k, so k = −m = −(−0.0066 s−1) = 0.0066 s−1

8. 0.0429 M 9. 1.9 x 103 min 10. 0.069 s

The units of k indicate a second order reaction. The half-life of a second order reaction depends on the initial concentration and the initial concentration is not given.

11. 1.6 x 10–3 s–1 12. a. Use the equation for the line, but substitute ln k for y and 1/T for x.

ln k = –16653 K (1/573 K) + 30.88 k = 6.15 L mol–1 s–1 (units are determined by the fact that the question refers to the reaction as being second order)

b. slope = –Ea/R Ea = 138 kJ


13. a. 1.0 x 10–3 L mol–1 s–1 b. 48 ºC

14. a. A + B " C + D b. One step reaction: A + B " C + D c. step 1: A + E " F + C

step 2: F + B " D + E d.

e. A & B are reactants, C & D are products, F is an intermediate, E is a catalyst

15. a. C4H9Br + 2 H2O " C4H9OH + Br− + H3O+

b. C4H9+ and C4H9OH2

+ c. Rate = k[C4H9Br]

16. a. step 1: Rate = k[NO2]2

step 2: Rate = k[NO3][CO] b. Yes, it is reasonable. The steps add up to the overall reaction, the mechanism is consistent

with the rate law, and there are no collisions that are greater than bimolecular collisions. c. The first step is the slow step. Its rate law is the same as the rate law for the overall

reaction. d. NO3 is an intermediate

17. a. 2A + F " G + D b. B, because it reacts (in a fast step) before it is produced (as opposed to an intermediate,

which is produced before it reacts). c. C and E are both intermediates d. Slow step: Rate = k[C][A], but C is an intermediate, so replace C with terms from step 1

Reaction: Rate = k[B][A]2

18. a. Not likely because step 1 is a trimolecular collision, which is so unlikely that the reaction probably wouldn’t occur.

b. Not possible because the overall reaction for this mechanism is A + B + C " D + 2E, which is not the correct overall reaction given above.

c. This is the most likely mechanism. It’s rate law is Rate = k [A][B][C], which is the same as the experimental rate law, and the individual steps add up to the overall reaction.

19. a. 2 BrCl + H2 → 2 HBr + Cl2 b. HCl

c. Rate = k[BrCl]2[H2]

[HBr]

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201B Work 1 Kinetics