Question

reaction is found to have a rate constant of 12.5 s-1 at 25.0 Celsius. When you heat the reaction up by ten degrees Celsius, the rate of the reaction exactly doubles. What is the activation energy for this reaction in kJ/mol?

Answer 1

Answer : The activation energy for the reaction is, 52.9 kJ/mol

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$T_1$ = initial temperature = $25.0^oC=273+25.0=298.0K$

$T_2$ = final temperature = $25.0^oC+10=35.0^oC=273+35.0=308.0K$

$K_1$ = rate constant at $25.0^oC$ = $12.5s^{-1}$

$K_2$ = rate constant at $35.0^oC$ = $2\times K_1=2\times 12.5s^{-1}=25.0s^{-1}$

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

Now put all the given values in this formula, we get:

$\log (\frac{25.0s^{-1}}{12.5s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{298.0K}-\frac{1}{308.0K}]$

$Ea=52903.05J/mole=52.9kJ/mol$

Therefore, the activation energy for the reaction is, 52.9 kJ/mol