Unit 9 – Kinetics
Section 2 – Mechanisms and Catalysts
1. The first-order decomposition of a colored chemical species, X, into colorless products is monitored with a
spectrophotometer by measuring changes in absorbance over time. Species X has a molar absorptivity constant
of 5.00 x 103 cm-1M-1 and the path length of the cuvette containing the reaction mixture is 1.00 cm. The data
from the experiment are given in the table below.
[X]
(M)
? 1.20 x 10-4
4.00 x 10–5
Absorbance
0.600
0.200
Time
(min)
0.0
35.0
3.00 x 10–5
1.50 x 10–5
0.150
44.2
0.075
?
a. Calculate the initial concentration of the colored species.
A = abc  This is from the formula sheet. So 0.600 = (5000 cm -1M-1)(1.00 cm)(x), then x = 1.20 x 10-4 M
b. Calculate the rate constant for the first-order reaction using the values given for concentration and time.
Include units with your answer. You can use any of first three values on the table for initial and final
concentration. I used the first two.
ln X t  ln X 0  kt  1.099  k 35.0 min  k  3.14  10 2 min -1
c.
Calculate the number of minutes it takes for the absorbance to drop from 0.600 to 0.075.
ln X t  ln X 0  kt  2.079  0.0314x min  x  66.2 min
d. Calculate the half-life of the reaction. Include units with your answer.
t1 / 2 
e.
0.693 0.693

 22.1 min
k
0.0314
Experiments were performed to determine the value of the rate constant for this reaction at various
temperatures. Data from these experiments were used to produce the graph below, where T is
temperature. This graph can be used to determine the activation energy, Ea, of the reaction.
i. Label the vertical axis of the graph. ln k
ii. Explain how to calculate the activation energy from this graph.
The slope of the line is related to the activation energy:
Slope = -Ea/R and R = -8.314 J/mol K
2. An environmental concern is the depletion of O 3 in Earth's upper atmosphere, where O3 is normally in
equilibrium with O2 and O. A proposed mechanism for the depletion of O 3 in the upper atmosphere is shown
below.
Step I
O3 + Cl  O2 + ClO
Step II
ClO + O  Cl + O2
a. Write a balanced equation for the overall reaction represented by Step I and Step II above.
O3 + O  2O2
(The highlighted parts cancel out.)
b. Clearly identify the catalyst in the mechanism above. Justify your answer.
Cl is the catalyst because it is a reactant in Step I and a product in the final step. (It is required but does
not get used up.)
c. Clearly identify the intermediate in the mechanism above. Justify your answer.
ClO is the intermediate in the reaction. It is a product in Step I and appears as a reactant in Step II.
d. If the rate law for the overall reaction is found to be rate = k[O3][Cl], determine the following.
i. The overall order of the reaction
The overall order of the reaction is 2 (Add the powers: 1 + 1 = 2)
ii. Appropriate units for the rate constant, k
rate = k[O3][Cl] so [M/time]= k[M][M] The units for k have to be M-1time-1
iii. The rate-determining step of the reaction, along with justification for your answer
Just by looking at the rate law (it contains chlorine and ozone only) the only step with those two
as the reactants is step I.
3. Methane gas reacts with chlorine gas to form dichloromethane and hydrogen chloride, as represented by the
following equation: CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g)
a. A 25.0 g sample of methane is placed in a reaction vessel containing 2.58 mol of Cl 2(g).
i. Identify the limiting reactant when the methane and chlorine gases are combined. Justify your
answer with a calculation.
2.58 mol Cl 2 1 mol CH 2 Cl 2

 1.29 mol CH 2 Cl 2 The Cl2 is the limiting reactant.
1
2 mol Cl 2
25 g CH 4 1 mol CH 4 1 mol CH 2 Cl 2


 1.56 mol CH 2 Cl 2
1
16 g CH 4
1 mol Cl 2
ii. Calculate the total number of moles of CH2Cl2(g) in the container after the limiting reactant has
been totally consumed.
From the above equation, 1.29 mol of CH2Cl2 will be created.
Initiating most reactions involving chlorine gas involves breaking the Cl-Cl bond, which has a bond energy of
242 kJ mol-1.
b. Calculate the amount of energy, in joules, needed to break a single Cl-Cl bond.
242 kJ 1000 J
1 mol


 4.02  10 19 J
23
mol
1 kJ
6.022 x 10
c.
Calculate the longest wavelength of light, in meters, that can supply the energy per photon necessary to
break the Cl-Cl bond.
E
hc

 4.02  10 19 J 
6.63  10
-34

J s 3.0  10 8 m / s

    4.9  10
7
m
The following mechanism has been proposed for the reaction of methane gas with chlorine gas. All species are in
the gas phase.
Step 1
Cl2  2 Cl
fast equilibrium
Step 2
CH4 + Cl → CH3 + HCl
slow Rate determining step! Use this for determining reaction orders!
Step 3
CH3 + Cl2 → CH3Cl + Cl
fast
Step 4
CH3Cl + Cl → CH2Cl2 + H
fast
Step 5
H + Cl → HCl
fast
d. In the mechanism, is CH3Cl a catalyst, or is it an intermediate? Justify your answer.
It is an intermediate; it is produced in step 3 and consumed in step 4.
e.
Identify the order of the reaction with respect to each of the following according to the mechanism. In
each case, justify your answer.
i. CH4(g)
The order of the reaction with respect to CH4 is 1 because the rate law for the slowest step in the
reaction is rate = k[CH4][Cl]
ii. Cl2(g)
Ok… so your book doesn’t do the BEST job of explaining this one, so I’ll give it a stab:
Because the first step is at equilibrium we can use that Keq we learned about in the last unit, and
because the forward and reverse rates are equal to each other, we can write this: Keq =
[Cl]2/[Cl2]. If we solve for [Cl] we get K1/2[Cl2]1/2 Next we will substitute the [Cl] into our previous
rate law (the one for the slowest step: the rate determining step), and we get:
rate = k[CH4](K1/2[Cl2]1/2) That’s a lot of weirdness… but the reaction order for Cl 2 is ½. Not too
common. This is not a nice problem. 