1. Consider the interconversion of the \'boat\' and \'chair\' conformations of c

Question

1. Consider the interconversion of the 'boat' and 'chair' conformations of cyclohexane BoatChair (a) If the equilibrium constant, K, at 305 K is 1 x 102 and the activation energy, Ea, for the conversion of the chair conformer to the boat conformer (reverse reaction) is 35 kJ mol-1 determine the forward and reverse rate constants at 298 K. Assume that the Arrhenius Equation prefactor, A, is 1 x 1012 s-1. (b) Determine the activation energy of the forward reaction. (c) The rate law expression for the production of the chair conformation is d|Chair] dt = kf [Boat]-k, [Chair] If at t 0 the [Boat-Boat]0 and [Chair] law expression for [Chair] is 0, show that the integrated rate kf + kr (d) Determine the integrated rate law expression for Boat] (e) Determine the equilibrium concentrations of the Boat and Chair conformations by using their integrated rate law expressions as t oo (f) If the temperature is suddenly increased to 1000 K, determine the new equilibrium concentrations and the relaxation constant, T.

Explanation / Answer

(a) The Arrhenius equation can be written as shown below.

k = A.e-Ea/RT

Here, the pre-exponential factor A = 1012 s-1, R = 8.314 J mol-1 K-1, Ea = 35 KJ mol-1 and T = 298 K

i.e. k = 1012 s-1 * e-35*1000 J mol-1 / (8.314 J mol-1 K-1 * 298 K)

i.e. k = 7.3*105 s-1 (reverse rate constant, kr)

Formula: The equilibrium constant (K) = kf/kr

Here, K = 102, i.e. kf/7.3*105= 102

i.e. The forward rate constant, kf = 7.3*107

(b) k = A.e-Ea/RT

i.e. Ea = -RTln(k/A)

i.e. Ea = -8.314 J mol-1 K-1 * 298 K * ln(7.3*107/1012)

i.e. The activation energy of the forward reaction, Ea = 23.59 KJ/mol

(c) d[C]/dt = kf[B] - kr[C], where B = boat and C = chair

i.e. d[C]/dt + kr[C] = kf[B]

i.e. d[C]/dt + kr[C] = kf[B]0 e-kft

Multiply with ekrt on both sides.

i.e. {d[C]/dt + kr[C]} ekrt =  kf[B]0 e(kr-kf)t

i.e. {d[C]/dt + kr[C]} e-krt = kf[B]0 e(kr-kf)t

i.e. {d[C]/dt + kr[C]} c= kf[B]0 e(kr-kf)t

i.e. d([C].ekrt)/dt = kf[B]0 e(kr-kf)t

Integrate on both sides, you will get

i.e. [C].ekrt = kf[B]0 {e(kr-kf)t/(kf + kr) - 1/(kf + kr)}

i.e. [C] = kf[B]0 {e-kft/(kf + kr) - e-krt/(kf + kr)}

i.e. [C] = kf[B]0 {e-kft - e-krt) / (kf + kr)}

(d) [B] = [B]0e-Kft

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