Question

In going from room temperature 25° to 35 the rate of a reaction doubles calculate the activation energy for the reaction

Answer 1

Answer:

52.9 KJmol-1

Explanation:

From;

log(k2/k1) = Ea/2.303 * R (1/T1 - 1/T2)

The temperatures must be converted to Kelvin;

T1 = 25° C + 273 = 298 K

T2= 35°C + 273 = 308 K

R= gas constant = 8.314 JK-1mol-1

Substituting values;

log 2 = Ea/2.303 * 8.314 (1/298 - 1/308)

Ea = 52.9 KJmol-1