CHM 3410 – Problem Set 9
Due date: Monday, November 15th. (NOTE: The third hour exam is Friday, November 19th. It will cover the following material: Chapter 5, section 13; Chapter 6, sections 5-9; Chapter 20, sections 1-4; Chapter 21, all).
Do the following problems. Show your work.
1) Baasandorj and coworkers (J.Phys.Chem.A 114 (2010) 4619-4633) recently examined the gas phase reaction of hydroxyl radical (OH) with CH2=CHF. A summary of their data is given below
T(K) 220. 249. 296. 321. 373.
k(cm3/molecule.s) 7.37 x 10-12 6.31 x 10-12 5.18 x 10-12 4.92 x 10-12 4.06 x 10-12
a) Fit the above data to the Arrhenius equation. Find the values for A and Ea corresponding to the data.
b) If you have done the fitting correctly you will have obtained a negative value for Ea, which is difficult to reconcile with a direct reaction. Consider the following mechanism for the reaction
step 1 OH + CH2=CHF D (OH-CH2=CHF) k1, k-1 (1.1)
step 2 (OH-CH2=CHF) + M ® “product” k2 (1.2)
Find a general expression for the rate law predicted for the above mechanism.
c) Your general expression in b has the following low pressure ([M] ® 0) limiting behavior
d[product]/dt = (k1k2/k-1) [OH][CH2=CHF] (1.3)
Assume that k1, k-1, and k2 each obey the Arrhenius equation. For what conditions, if any, will the observed rate law for the reaction have a negative activation energy?
2) For each of the following mechanisms predict the corresponding rate law
a) Formation of phosgene (COCl2)
stoic. CO + Cl2 ® COCl2
step 1 Cl2 D 2 Cl fast, reversible (k1, k-1)
step 2 Cl + CO D 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 D NO2 + Cl fast, reversible (k1, k-1)
step 2 NO2Cl + Cl ® NO2 + Cl2` slow (k2)
3) Consider fluorescence (light emission from an electronically excited state of a molecule) in the presence of a quenching molecule M. Quenching refers to the removal of energy from an excited electronic state by collision, and so without light emission. A mechanism for such a system can be given as follows:
A + hn ® A* rate = d[A]/dt = ke [A] [I] (3.1)
A* + M ® A + M rate = kq [A*] [M] (3.2)
A* ® A + hn' rate = kf [A*] (3.3)
where the rates of each individual step have been given. In the above expressions ke is the rate constant for formation of electronically excited molecules, [I] is the intensity of the light source used to excite molecules, kq is the rate constant for quenching of molecules, and kf is the rate constant for fluorescence.
a) Find an expression for the steady state concentration of electronically excited molecules, [A*]ss.
b) We may assume that If, the intensity of fluorescence, is proportional to [A*]ss, the steady state concentration of electronically excited molecules. Call the (unknown) constant of proportionality c (so If = c [A*]). Find an expression for If in terms of c and other constants.
c) Based on your answer in b, suggest an experimental method for finding kf/kq, the ratio of the rate constants for fluorescence and quenching.
Solutions.
1) ` a) The data are presented and plotted below
T (K) k(cm3/molecule.s) 1/T (K-1) ln k
220. 7.37 x 10-12 0.004545 - 25.634
249. 6.31 x 10-12 0.004016 - 25.789
296. 5.18 x 10-12 0.003378 - 25.986
321. 4.92 x 10-12 0.003115 - 26.038
373. 4.06 x 10-12 0.002681 - 26.230
Based on the best fit to the data I get ln k = (310.6 K)/T - 27.04
So Ea = - R (slope) = - (8.3145 J/mol.K) (310.6 K) = - 2.58 kJ/mol
A = exp(-27.04) = 1.81 x 10-12 cm3/molecule.s
b) Basedon the mechanism
d[product]/dt = k2 [OH-CH2=CHF]
If we make the steady state approximation for OH-CH2=CHF, then
d[OH-CH2=CHF]/dt @ 0 = k1[OH][CH2=CHF] – k-1[OH-CH2=CHF] – k2[OH-CH2=CHF][M]
[OH-CH2=CHF] = k1[OH][CH2=CHF]/(k-1 + k2[M])
Substituting into our expression for rate gives
d[product] = k1k2[OH][CH2=CHF]
dt (k-1 + k2[M])
c) In the low pressure limit ([M] ® 0) the second term in the denominator becomes negligible, and the rate law reduces to a low pressure limiting law
d]product]/dt = (k1k2/k-1)[OH][CH2=CHF] = k¢[OH][CH2=CHF]
Assume each individual rate constant fits an Arrhenius equation, so
k1 = A exp(-Ea/RT)
k-1 = B exp(-Eb/RT)
k2 = C exp(-Ec/RT)
Then the apparent rate constant in the low pressure limit is
k¢ = [A exp(-Ea/RT)][B exp(-Eb/RT)] = (AB/C) exp[-(Ea + Eb – Ec)/RT]
[C exp(-Ec/RT]
which has the same form as the Arrhenius equation, but with an apparent activation energy
E¢ = Ea + Eb – Ec
If Ec > (Ea + Eb), then E¢ will e negative, as is observed in these experiments.
2) 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)1/2 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 [NO2]
predicting a reaction that is second order in NO2Cl and - 1st order in NO2.
3) a) The steady state concentration of A* may be found using
d[A*]/dt @ 0 = ke [A] [I] - kq [A*] [M] - kf [A*]
Solving for [A*] gives
[A*] = ke [A] [I]
kq [M] - kf
b) Since If = c [A*], we may use the result from part a to get
If = c ke [A] [I]
kq [M] - kf
c) Consider a series of experiments where [A] and [I] are kept constant, but where [M] is allowed to vary. If we define c' = c ke [A] [I] = a constant for the experimental conditions we have chosen, then
If = c'
kq [M] - kf
If we invert both sides of this relationship, we get
1/If = (kq/c') [M] + (kf/c')
Based on the above relationship, if we plot 1/If vs [M], we expect to get a straight line, with
slope = kq/c' intercept = kf/c'
Based on this we may say
kf/kq = (intercept)/(slope)