CHEMISTRY 156, SPRING, 2002

SECOND MIDETERM EXAMINATION

SOLUTIONS

Our understanding of complex kinetic systems such as photochemical damage to DNA is based on a thorough analysis of simple systems. To this end, consider the solution-phase photochemical transformation of ethylene in a solvent S to form cyclobutane:

2 C2H4 + n ® C4H8 [n represents a UV photon]

The following mechanism has been proposed for this transformation:

(a) The excitation of a molecule of ethylene by the absorption of a photon, n , to form electronically excited ethylene, C2H4*

.

1

C2H4 + n « C2H4*

-1

[The concentration of photons, n , is measured by the intensity of the UV radiation, Ia. The rate constants k1 and k-1 are determined by quantum mechanics; their values will not be addressed in this exam.]

b) The bimolecular reaction of excited ethylene with a second, unexcited ethylene molecule to form the product, cyclobutane.

2

C2H4* + C2H4 « C4H8

-2

[Assume in your analysis that the reverse rate constant, k-2, is too small to make a measurable contribution to the rate law, i.e. k-2 » 0.]

c) The de-excitation of the ethylene by a collision with a solvent molecule S.

3

C2H4* + S « C2H4 + S

-3

[As in step (b), assume that the reverse rate constant k-3 is very small.]

 

Answer the following four questions relating to this mechanism.

1) Use the steady-state approximation and derive an empirical rate law for the time rate of production of the product cyclobutane, d[C4H8]/dt.

Rate = d[C4H8]/dt = k2[C2H4*][C2H4].

d[C2H4*]/dt = k1[C2H4*]Ia –k-1[C2H4*] –k2[C2H4*][C2H4] –k3[C2H4*][S] » 0 (steady state approximation)

Therefore, [C2H4*] = k1[C2H4]Ia/(k-1 + k2[C2H4] + k3[S]) and

Rate = d[C4H8]/dt = (-1/2)d[C2H4]/dt

= k1k2[C2H4]2Ia/(k-1 + k2[C2H4] + k3[S])

2) Verification of your rate law.

a) Simplify your rate law under the experimental conditions of a very low concentration of ethylene.

rate = d[C4H8]/dt = (-1/2)d[C2H4]/dt

= k1k2[C2H4]2Ia/(k-1 + k3[S])

b) Under these experimental conditions, which function of the concentration of ethylene will yield a straight line when plotted versus time?

Second order with respect to ethylene so plot the reciprocoal of the ethylene concentration versus time.

c) Provide an expression for the slope of this plot.

2k1k2[C2H4]2Ia/(k-1 + k3[S])

3) The de-excitation of an electronically excited molecule via collision with a solvent molecule, e.g. step (b) in the mechanism outlined above, is often diffusion controlled. Calculate the value of the rate constant k3 at 298 K under the following conditions and assumptions:

a) The solvent used in the photolysis is 100% ethanol whose viscosity is 1.096 x 10-3 Kg/m-s at 298 K.

b) The de-excitation is diffusion controlled.

c) Treat the ethylene and ethanol molecules as spherical. Their effective radii are 2.31 Å and 2.55 Å, respectively.

k = (2kBT/3h )[(rA + rB)2/rArB]

= [2(1.381 x 10-23 J/K-molecule)(298 K)/(3)(0.001096 Kg/m-s)][(2.31 + 2.55)2/(2.31)(2.55)]

= 1.00 x 10-7 m3/s-molec (times 1000 liter/m3)

= 1.00 x 10-4 liter/molec-s (times Avogadro's number)

= 6.0 x 109 M-1s-1

4) Consider the second step of the mechanism, which is exothermic.

a) Modeling calculations provide the following information on the forward reaction: D H† » 0 and D S† » -110 J/K-mole. Calculate an estimate of k2 at 298 K.

k = (kBT/h)exp(-D G†/RT) = (kBT/h)exp(D S†/RT) since D H† » 0

k = [(1.38 x 10-23 J/K)(298 K)/(6.63 x 10-34 J-s)]exp[(-110 J/K)/(8.314 J/K)(298 K)] = 1.0 x 107 M-1s-1

The units are a consequence of the standard state for solutes.

b) What can be concluded about the activated complex from the values of D H† and D S†?

The overall enthalpy change is large and negative so the enthalpy of activation for the forward reaction is small. The transition state should be much closer to the reactants than two the products. Two reagents combine to form a structured product so the transition state will reflect a decrease in entropy.

c) Does the rate constant k2 depend on applied pressure? If it does, what does the dependence of k2 on applied pressure reveal about the activated complex?.

Yes. ln(k2) depends linearly on the volume of activation.

d) How is the equilibrium constant for the second elementary step related to the forward and reverse rate constants, k2 and k-2?

K2 = k2/k-2

exam2_2002.htm, 29 April 2002, WES