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?
3A+3B→5C+2D
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?
Chem 201B
5.
Dr. Lara Baxley
The data below were collected for the reaction: BrO3– + 5 Br– + H3O+ → 3 Br2 + 9 H2O
Exp #
1
2
3
4
Initial Concentration of Reactants (mol/L)
BrO3–
Br–
H3 O+
0.10
0.10
0.10
0.20
0.10
0.10
0.10
0.30
0.10
0.20
0.10
0.15
Initial Rate (mol/L•s)
1.2
2.4
3.6
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
1
2
3
Initial [H2], M
0.113
0.220
0.550
Initial [I2], M
0.110
0.330
0.110
Initial Rate, mol/L•s
3.01 x 10–4
1.76 x 10–3
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.
Chem 201B
Dr. Lara Baxley
7. The data below was collected for the hypothetical reaction 2 A → B + C.
Time (s)
0
40
80
120
160
[A]t
0.1
0.0768
0.059
0.0453
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)
Chem 201B
Dr. Lara Baxley
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).
Arrhenius Graph for Reaction of HI and CH3I
-2
y = -16653x + 30.88
-2.5
ln k
-3
-3.5
-4
-4.5
0.00198
0.002
0.00202
0.00204
1/T (K-1)
0.00206
0.00208
0.0021
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.
Chem 201B
Dr. Lara Baxley
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.
Chem 201B
Dr. Lara Baxley
15. Consider the following mechanism for a reaction
C4H9Br → C4H9+ + Br–
C4H9+ + H2O → C4H9OH2+
C4H9OH2+ + H2O → C4H9OH + H3O+
slow
fast
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
C+A→B+E
E+F→G+D
(fast)
(slow)
(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?
Chem 201B
Dr. Lara Baxley
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
X→D
(slow)
(fast)
b. Mechanism 2:
Step 1
A +B ⇌ X
Step 2
X +C → Y+E
Step 3
Y → D+E
(fast)
(slow)
(fast)
c. Mechanism 3:
Step 1
A +B ⇌ X
Step 2
X +C → Y+E
Step 3
Y → D
(fast)
(slow)
(fast)
19. Consider the following mechanism for a reaction
BrCl + H2 ⇌ HBr + HCl
HCl + BrCl → HBr + Cl2
fast
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)
Chem 201B
Dr. Lara Baxley
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/mol3s
6. a. Rate = k[H2][I2]
b. k = 0.0243 L/mols
7. a.
Second Order Integrated Plot
35
-2
-2.2
-2.4
-2.6
-2.8
-3
-3.2
-3.4
-3.6
30
25
y = -0.0066x - 2.3026
1/[A]
ln [A]
First Order Integrated Plot
20
15
10
5
0
0
20
40
60
80 100 120 140 160 180
Tim e (s)
0
20
40
60
80 100 120 140 160 180
Tim e (s)
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
Chem 201B
13. a.
b.
14. a.
b.
c.
Dr. Lara Baxley
1.0 x 10–3 L mol–1 s–1
48 ºC
A+BC+D
One step reaction: A + B C + D
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
2
c. Rate = k[BrCl] [H2]
[HBr]