The temperature at which a reaction is running is increase from 11.052 celsius to 102.567 celsius. If the rate constant increases by a factor of 5.608, what is the activation energy of the reaction in kJ/mol?

Question

The temperature at which a reaction is running is increase from 11.052 celsius to 102.567 celsius. If the rate constant increases by a factor of 5.608, what is the activation energy of the reaction in kJ/mol?

Use the Arrhenius equation.

I AM STILL LOST

The Arrhenius equation is
ln(k2/k1)=(E_a/R)*[(1/T1) - (1/T2)]

You have T1 and T2. You have k1 and that times 5.608 will give you k2. R is 8.314 J/mol*K. Solve for E_a. Plug and chug.

Is the answer 16.733 kJ/mol.

I obtained 16.726 kJ/mol. I used 273.15 to convert to K. Using 8.314 (4 significant figures) would round to 16.73 kJ/mol for both answers.