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 Boat Chair (a) If the equilibrium constant, K, at 305 K is 1 102 and the activation energy, Ea, for the conversion of the chair conformer to the boat conformer (reverse reaction) is 35 k.J mol1 determine the forward and reverse rate constants at 298 K. Assume that the Arrhenius Equation prefactor, A, is 1 × 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 Beat]-k, [Chair] If at t 0 the BoatBoat o and [Chair law expression for [Chair] is 0, show that the integrated rate kj + 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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