CHM 3410 – Problem Set 9

Due date: Wednesday, November 9th. (Note - Friday, November 11th, is a holiday, and so no class.)

Do all of the following problems. Show your work. (NOTE: R = 0.082057 L atm/mol K, and R = 8.3145 J/mol K). Unless otherwise stated, you may assume all data and problems are for T = 25.0 °C.

“Reality is how math feels.” - Brian Greene

1) The following experimental data were obtained for a particular irreversible reaction. It is assumed that the reaction obeys a rate law of the form

rate = - d[A]/dt = k [A]p [B]q (1.1)

Trial [A]initial [B]initial (Rate)initial

(mol/L) (mol/L) (mol/L min)

1 0.0100 0.0100 2.9 x 10-8

2 0.0100 0.0200 1.5 x 10-7

3 0.0100 0.0400 5.7 x 10-7

4 0.0100 0.0800 2.4 x 10-6

5 0.0200 0.0200 1.2 x 10-7

6 0.0400 0.0200 1.6 x 10-7

7 0.0800 0.0200 1.4 x 10-7

Based on the above data find p, q, and k (including correct units).

2) Data for a particular irreversible homogeneous reaction are given below.

t (s) [A] (mol/L) t (s) [A] (mol/L)

0. 47.8 x 10-3 200. 13.4 x 10-3

50. 35.1 x 10-3 300. 7.3 x 10-3

100. 25.3 x 10-3 400. 3.8 x 10-3

150. 18.6 x 10-3 500. 2.2 x 10-3

a) Plot ln[A] vs t for the above reaction.

b) Plot (1/[A]) vs t for the above reaction.

c) Based on your results from parts a and b of this problem, determine whether the reaction is first order homogeneous, second order homogeneous, or neither of these. If the reaction is first order or second order homogeneous find k, the rate constant, for the reaction.

3) The rate constant for a chemical reaction is often fit to the expression

k = A T-n exp(-B/RT) (3.1)

where A, B, and n are constants found by fitting experimental data to the equation.

a) Using the general expression for the activation energy for a reaction

Ea = - R (d ln(k)/d(1/T)) (3.2)

find an expression for Ea for the above reaction. Give your answer in terms of A, B, n, R, and/or other constants.

b) As a check on your answer in a, show that your expression reduces to the result found for a reaction obeying the Arrhenius equation for the appropriate choice of parameters, and find the value for Ea for the case of Arrhenius behavior.

4) In a recent paper (M. Antinolo, E. Jimenez, A. Notario, E. Martinez, J. Albaladejo, Atmos. Chem. Phys. (2010) 10, 1911-1922) the co-authors studied the following gas phase chemical reaction:

Cl + CF3CH2CH2CHO ® “products” (4.1)

The rate constant for the above reaction at several different temperatures is given below.

T(K) k (cm3/molecule s) T(K) k (cm3/molecule s)

268. 2.85 x 10-11 316. 4.05 x 10-11

278. 2.98 x 10-11 331. 4.60 x 10-11

288. 3.20 x 10-11 351. 4.79 x 10-11

298. 3.39 x 10-11 371. 5.13 x 10-11

Based on this information, and assuming the data fit the Arrhenius equation, find A and Ea for the above reaction.

5) The gas phase reaction of hydroxyl radical with fluoroethene in synthetic air (80 % N2, 20 % O2) is

CH2=CHF + OH ® “products” (5.1)

The following two step mechanism has been proposed for the reaction

step 1 OH + CH2=CHF D (OH-CH2=CHF) k1, k-1 (5.2)

step 2 (OH-CH2=CHF) + M ® “product” k2 (5.3)

where M represents a molecule of air.

a) Find the general expression for the rate law corresponding to the above reaction. You may make the steady state approximation for any reaction intermediates, but since you do not know the relative values for k1, k-1, and k2, you cannot assume pre-equilibrium occurs.

b) Find the expression for the rate law for the above reaction in the low pressure limit ([M] ® 0).

c) Find the expression for the rate law for the above reaction in the high pressure limit ([M] ® ¥).

6) Consider the following three mechanisms for the formation of COCl2 from CO and Cl2 in a gas mixture of CO/Cl2/air. Note that M = air, and [M] > [CO], [Cl2]

stoic. CO + Cl2 ® COCl2

mechanism 1 mechanism 2

step 1 Cl2 + M D Cl + Cl + M fast step 1 Cl2 + M ® Cl + Cl + M slow

step 2 CO + Cl + M D COCl + M fast step 2 CO + Cl + M D COCl + M fast

step 3 COCl + Cl2 ® COCl2 + Cl slow step 3 COCl + Cl2 ® COCl2 + Cl fast

mechanism 3

step 1 Cl2 + M D Cl + Cl + M fast

step 2 CO + Cl + M ® COCl + M slow

step 3 COCl + Cl2 ® COCl2 + Cl fast

Find the rate law predicted for each of the above three mechanisms. Show your work.

Solutions.

1) To find the reaction orders we need to find the slope in a plot of ln(rate) vs ln(conc) for experiments where all but one of the concentrations is kept constant, and it is the initial rate and initial concentration that are used.

The value for p can be found using trials 2, 5, 6, and 7, and the value for q can be found using trials 1, 2, 3, and 4. The information needed for the plots is tabulated below.

Trial ln[A]initial ln[B]initial ln(rate)initial

1 - 4.605 - 17.356

2 - 4.605 - 3.912 - 15.713

3 - 3.219 - 14.378

4 - 2.526 - 12.940

5 - 3.912 - 15.936

6 - 3.219 - 15.648

7 - 2.526 - 15.782

The data are plotted below.


Based on linear regression, the best fit lines to the data are as follows

Plot of ln[A] vs ln(rate) slope = - 0.02 intercept = - 15.72

Plot of ln[B] vs ln(rate) slope = 2.10 intercept = - 7.59

Therefore the orders of the reaction are p = 0, q = 2.

We can find the best value for k by finding values for each trial and averaging. Note that k is given by the equation

k = (rate)/[B]2

trial k (L/mol min)

1 2.9 x 10-4

2 3.8 x 10-4

3 3.6 x 10-4

4 3.8 x 10-4

5 3.0 x 10-4

6 4.0 x 10-4

7 3.5 x 10-4

______

kave = 3.51 x 10-4 L/mol min

2) a+b )The data needed for the plots are tabulated below

t (s) ln[A] 1/[A] (L/mol)

0. - 3.041 20.92

50. - 3.350 28.49

100. - 3.677 39.53

150. - 3.985 53.76

200. - 4.313 74.63

300. - 4.920 137.0

400. - 5.573 263.2

500. - 6.119 454.5


The data are plotted below.

c) The data for the plot of ln[A] vs t fits a straight line, but the data for a plot of 1/[A] vs t does not (it shows systematic curvature from a line). The reaction is therefore first order homogeneous.

The best fit to the data for ln[A] vs t gives slope = - 6.2 x 10-3 s-1, intercept = - 3.05 . The value for the rate constant for the reaction is therefore k = 6.2 x 10-3 s-1.

3) a) Since k = A T-n exp(-B/RT)

ln(k) = ln(A) - n ln(T) - (B/RT)

Since we are planning to take the derivative of this with respect to 1/T, we can rewrite this as

ln(k) = ln(A) + n ln(1/T) - (B/R)(1/T)

So Ea = - R (d ln(k)/d(1/T)) = - R [ 0 + nT - B/R ]

And so Ea = B - nRT

b) If n = 0 then eq 3.1 has the same form as the Arrhenius equation. This means in that case tat that B = Ea.

4) The data needed to fit to the Arrhenius equation is given below

T(K) k(cm3/molecule s) 1/T (K-1) ln(k)

268. 2.85 x 10-11 0.003731 - 24.281

278. 2.98 x 10-11 0.003597 - 24.237

288. 3.20 x 10-11 0.003472 - 24.165

298. 3.39 x 10-11 0.003356 - 24.108

316. 4.05 x 10-11 0.003165 - 23.930

331. 4.60 x 10-11 0.003021 - 23.802

351. 4.79 x 10-11 0.002849 - 23.762

371. 5.13 x 10-11 0.002695 - 23.693

The data are plotted below.


Based on the best fit of the data to a straight line, we get slope = - 620. K, intercept = - 21.99

Therefore Ea = - R(slope) = - (8.3145 J/mol K)(- 620. K) = 5.15 kJ/mol

ln(A) = - 21.99 A = e-21.99 = 2.82 x 10-10 cm3/molecule s

5) a) rate = k2 [OH-CH2=CHF][M]

d[OH-CH2=CHF] @ 0 = k1 [OH][CH2=CHF] - k-1 [OH-CH2=CHF] - k2 [OH-CH2=CHF][M]

dt

k1 [OH][CH2=CHF] = k-1 [OH-CH2=CHF] + k2 [OH-CH2=CHF][M]

k1 [OH][CH2=CHF] = (k-1 + k2[M])[OH-CH2=CHF]

So [OH-CH2=CHF] = k1 [OH][CH2=CHF]

(k-1 + k2[M])

Therefore rate = k2 [M] k1 [OH][CH2=CHF] = k1k2[OH][CH2=CHF][M]

(k-1 + k2[M]) (k-1 + k2[M])

b) In the limit [M] ® 0, the term k2 [M] can be dropped from the denominator, giving

rate = k1k2 [OH][CH2=CHF][M] = k [OH][CH2=CHF][M] ; where k = k1k2/k-1

k-1

c) In the limit [M] ® ¥, the term k-1 can be dropped from the denominator, giving

rate = k1 [OH][CH2=CHF]

What this means is that as total pressure of air ( [M] ) increases, the reaction goes from being third order to being second order.

6) a) The overall rate is the rate of the slow step, and so

rate = k3 [COCl][Cl2] This is in terms of a reaction intermediate, which must be eliminated.

The second step is fast and reversible, and so

k2 [CO][Cl][M] = k-2 [COCl] [M] [COCl] = k2 [CO][Cl][M] = k2 [CO][Cl]

k-2 [M] k-2

The first step is also fast and reversible, and so

k1 [Cl2][M] = k-1 [Cl]2[M] [Cl]2 = k1 [Cl2][M] = k1 [Cl2]

k-1 [M] k-1

and so [Cl] = (k1/k-1)1/2 [Cl2]1/2

Substituting for [Cl] in the expression for [COCl] gives

[COCl] = k2 [CO] (k1/k-1)1/2 [Cl2]1/2 = (k1/k-1)1/2(k2/k-2) [CO][Cl2]1/2

k-2

Substituting into our expression for the rate of reaction we get our final result

rate = k3 [Cl2] {(k1/k-1)1/2(k2/k-2) [CO][Cl2]1/2} = (k1/k-1)1/2(k2/k-2)k3 [CO][Cl2]3/2 = k [CO][Cl2]3/2

where k = (k1/k-1)1/2(k2/k-2)k3

b) The overall rate is the rate of the slow step, and so

rate = k1 [Cl2][M]

c) The overall rate is the rate of the slow step, and so

rate = k2 [CO][Cl][M] This is in terms of a reaction intermediate, which must be eliminated.

The first step is fast and reversible, and so

k1 [Cl2][M] = k-1 [Cl]2[M] [Cl]2 = k1 [Cl2][M] = k1 [Cl2]

k-1 [M] k-1

and so [Cl] = (k1/k-1)1/2 [Cl2]1/2

Substituting into the expression for the rate of reaction gives the final result

rate = k2 [CO] {(k1/k-1)1/2 [Cl2]1/2} = (k1/k-1)1/2k2 [CO][Cl2]1/2 = k [CO][Cl2]1/2

where k = (k1/k-1)1/2k2

Notice that each of these rate mechanisms leads to a different rate law, and so experimental data could eliminate at least two of these mechanisms from consideration.