What is the activation energy (in kJ/mol)

Question

What is the activation energy (in kJ/mol) of a reaction whose rate constant increases by a factor of 87 upon increasing the temperature from 308 K to 347 K? R = 8.314 J/(mol • K). Only enter the numerical value as an integer in the answer box below. Do NOT type in the unit (kJ/mol). 

 

Answer 1

Arrhenius Equation:

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

Let k1 = 1

k2 = 87

T1 = 308K

T2 = 347K

R = 8.314 J/mol-K

ln (87/1) = -E_act/8.314 (1/347 - 1/308)

4.47 = -E_act/8.314 (0.00288 - 0.003247)

4.47 = 0.000367_Eact/8.314

E_act = 101,263 J/mol = 101.3 kJ/mol = 101 (without units and to 3 sig. figs)