The activation energy for a given reaction is 56 kj/mol. at what temperature would the rate constant be quadruple what is was at 273k?

The rate equation is given as:

k = A e^ ( - Ea / RT)

Dividing state 1 and state 2:

k1/k2 = e^ ( - Ea / RT1) / e^ ( - Ea / RT2)

k1/k2 = e^[ - Ea / RT1 - ( - Ea / RT2) ]

k1/k2 = e^[ - Ea / RT1 + Ea / RT2) ]

Taking the ln of both sides:

ln (k1/k2) = - Ea / RT1 + Ea / RT2

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

Since k2 = 4k1, therefore k1/k2 = ¼

ln (1/4) = [ - (56,000 J/mol) / (8.314 J / mol K) ] (1/273 K - 1 / T2)

2.058 x 10^-4 = 1/273 - 1/T2

T2 = 289.25 K