Question

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?

Answer 1

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