CHM 3400 – Problem Set 8
Due date: Wednesday, April 4th.
Do all of the following problems. Show your work.
"The worst vice is advice." (traditional proverb)
1) For many gas phase chemical reactions the temperature dependence of the rate constant for the reaction is
modeled using modifications of the Arrhenius equation. In those cases, an activation energy for the reaction can still
be defined, using
Ea = - R [d ln(k)/d(1/T)] = RT 2 [ d ln(k)/dT ]
(1.1)
Using equn 1.1, find a general expression for the activation energy for a reaction whose rate constant is given by the
expression
k = AT2 exp(-B/T)
(1.2)
where A and B are constants. Note that the value for Ea will depend on temperature, unlike the original Arrhenius
equation.
2) Because of their harmful effect on the ozone layer, chlorofluorocarbons (CFCs) have been replaced in recent
years by other compounds that are not expected to affect ozone in the stratosphere. Hydrofluorocarbons (HFCs)
have often been used as CFC substitutes. Because of this, there is interest in the gas phase reactions of HFCs in the
lower atmosphere of the Earth.
In a 2003 research paper, Kozlov and coworkers obtained the following data for the reaction
CH3CH2F + OH  "products"
T (K)
k (cm3/molecule.s)
210.
220.
230.
250.
272.
7.94 x 10-14
9.62 x 10-14
10.58 x 10-14
13.48 x 10-14
17.47 x 10-14
(2.1)
a) Assuming the above data obey the Arrhenius equation, give an appropriate plot of the data, and from that
data plot determine the Arrhenius parameters A (pre-exponential factor) and Ea (activation energy), including correct
units.
b) Based on your answer in a, calculate the value of k for the above reaction at T = 300. K.
3) The rate of a chemical reaction is found to double when the temperature is raised from 25.0 C to 35.0 C.
Determine the activation energy for the reaction.
4) For each of the following mechanisms predict the corresponding rate law
a)
Formation of phosgene (COCl2)
stoic.
CO + Cl2  COCl2
step 1
Cl2  2 Cl
fast, reversible (k1, k-1)
step 2
Cl + CO  COCl
fast, reversible (k2, k-2)
step 3
COCl + Cl  COCl2
slow (k3)
b)
Decomposition of NO2Cl
stoic
2 NO2Cl  2 NO2 + Cl2
step 1
NO2Cl  NO2 + Cl
fast, reversible (k1, k-1)
step 2
NO2Cl + Cl  NO2 + Cl2`
slow (k2)
Also do the following problems from Atkins:
11.5 The reaction mechanism
A2  A + A
(fast, reversible)
A+B  P
(slow)
involves an intermediate A. Deduce the rate law for the formation of P.
Solutions.
1) The expression for the rate constant has the general form
k = A T2 exp(-B/T)
So
ln k = ln(A) + 2 ln(T) - B/T
Ea = RT2 d/dT { ln(A) + 2 ln(T) - B/T }
= RT2 { (2/T) + (B/T2) } = 2RT + RB
2)
a) To fit the data to the Arrhenius equation we need to plot ln(k) vs 1/T. The data are presented below
along with the plot
T (K)
k (cm3/molecule.s)
1/T (K-1)
ln(k)
210.
220.
230.
250.
272.
7.94 x 10-14
9.62 x 10-14
10.58 x 10-14
13.48 x 10-14
17.47 x 10-14
0.004762
0.004545
0.004348
0.004000
0.003676
- 30.164
- 29.972
- 29.877
- 29.635
- 29.376
ln(k)
ln(k) vs 1/T
-29.3
-29.4
0.0036
-29.5
-29.6
-29.7
-29.8
-29.9
-30
-30.1
-30.2
-30.3
0.0041
0.0046
1/T (K-1)
Based on the plot, I get
slope = - 705.7 K
intercept = - 26.794
And so Ea = - mR = - (- 705.7 K) (8.314 J/mol.K) = 5870. J/mol
lnA = - 26.794, and so A = e-26794 = 2.3 x 10-12 cm3/molecule.s
b) At T = 300. K
k = (2.3 x 10-12 cm3/molecule.s) exp[ - (5870. J/mol)/(8.314 J/mol.K)(300. K)]
= 21.8 x 10-14 cm3/molecule.s
3) To do this problem we may use the relationship
ln(k2/k1) = - (Ea/R) { (1/T2) - (1/T1) }
or
Ea = - R ln(k2/k1)
= - (8.314 J/mol.K) ln(2)
= 52960. J/mol = 52.96 kJ/mol
{ (1/T2) - (1/T1) }
{ (1/308.2 K) - (1/298.2 K) }
4)
a) The overall rate is equal to the rate of the slow step, so
d[COCl2]/dt = k3[COCl][Cl]
There are pre-equilibrium steps involving both intermediates, and so
k1[Cl2] = k-1[Cl]2
[Cl] = {(k1/k-1)[Cl2]}1/2
k2[Cl][CO] = k-2[COCl]
[COCl] = (k2/k-2) [Cl][CO] = (k2/k-2)(k1/k-1)1/2 [CO][Cl2]1/2
Substituting, we get
d[COCl2] = (k2/k-2)(k1/k-1) k3 [CO][Cl2]
dt
predicting a reaction that is first order in both CO and Cl2.
b) The overall rate is equal to the rate of the slow step, so
d[Cl2]dt = k2[NO2Cl][Cl]
The intermediate is in pre-equilibrium, so
k1[NO2Cl] = k-1[NO2][Cl]
[Cl] = (k1/k-1)[NO2Cl]/[NO2]
Substituting gives
d[Cl2] = (k1k2/k-1) [NO2Cl]2
dt
[NO 2]
predicting a reaction that is second order in NO2Cl and - 1st order in NO2, and so first order overall.
11.5
The overall rate of reaction is the rate of the slow step, and so
d[P]/dT = k2 [A] [B]
A is a reaction intermediate. Since the first step in the mechanism is fast and reversible, we may say
k1 [A2] = k-1 [A] [B] ;
[A] = (k1/k-1) ([A2]/[B])
If we use the above to substitute for [A], we get
d[P]/dT = k2 [B] (k1/k-1) ([A2]/[B]) = (k1k2/k-1) [A2] = koverall [A2]
a rate law that is first order in A2.