Question

A certain reaction has an activation energy of 64.0 kJ/mol and a frequency factor of A1 = 5.70×10¹² M−1s−1 . What is the rate constant, k , of this reaction at 20.0 ∘C ?

Express your answer with the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot (asterisk) or a dash.

Part B

An unknown reaction was observed, and the following data were collected:

T (K )	k (M−1⋅s−1 )
352	109
426	185

Determine the activation energy for this reaction.

Answer 1

Given the activation energy, Ea = 64.0 KJ/mol = 64000 J/mol

Frequency factor, A1 = 5.70x10¹² M^-1s^-1

Temperature, T = 20 C = 20+273 = 293 K

Rate constant can be calculated from the following formulae

k = A x (exp)^(-Ea/RT) = (5.70x10¹² M^-1s^-1) x (exp)^{(- 64000 J/mol / 8.314 J.K-1.mol-1 x 293K)} = 22.17 M^-1s^-1 (answer)

Part - B:

T1 = 352 K, k1 = 109 M^-1s^-1

T2 = 426 K, k2 = 185 M^-1s^-1

ln(k1 / k2) = - (Ea / R)x(1/T1 - 1/T2) = - (Ea / R)x(T2 - T1) / T1T2

=> ln(109 M^-1s^-1 / 185 M^-1s^-1 ) = - (Ea / R)x(426K - 352K) / 352Kx426K

=> Ea = 8912 J = 8.91 KJ/mol (answer)