Question

The second order rate constant of the following gas phase reaction at 338°C was found to be 7.5 x 10-4dm-mol-15-1: H2+C2H4 → CHE Determine k2 at 113°C assuming the activation energy and the collision cross sections are constants (ignore the steric factor).

Answer 1

rate constant, k = A*e-(E / RT) E = activation energy R = universal gas constant = 8.314 T = temerature in k

now, ln k = ln A - E / RT

ln k1 = ln A - E / RT1  and ln k2 = ln A - E / RT2 here, T1 = 113 + 273 = 386 K and T2 = 338 + 273 = 611 K and k2 = 7.5 * 10-4

so, ln (k1 / k2 ) = (E / R) * ( 1/T2 - 1/T1 )

hence, k1 / k2 = e (E / R) * ( 1/T2 - 1/T1 )

so, k1= k2 * e [(E / R) * ( 1/T2 - 1/T1 )] = [ 7.5 * 10-4] * e [(E / 8.314) * ( 1/611 - 1/386 )] = [7.5 * 10-4] * e- [(11.47*10^-5 ) * E]