Question

The uncatalyzed reaction has an activation energy of 120 kJ/mol, and the temperature of the reaction is 298 K, leading to the rate constant of 4.9 x 10^{–11} s^{–1}. . At what temperature in Kelvin would the uncatalyzed reaction need to be run at to match the rate of the acid catalyzed reaction (7.3 x 10^{+5} s^{–1}

Answer 1

283.5 K.

Explanation:

k = A * exp * (E/(R * T))

where,

k = rate constant

Ea = activation energy

R = gas constant (8.3145 J/K mol)

T = temperature expressed in Kelvin

A = frequency factor, having units of s-1

k1 = A * exp * (E/(R * T1)

k2 = A * exp * (E/(R * T2)

(R/E) * Ln (k2/k1) = 1/T1 - 1/T2

Inputting values,

T2 = 283.5 K.