Question

If a temperature increase from 18.0 ∘C to 37.0 ∘C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

Answer 1

Answer : The activation energy for the reaction is, 43.4 KJ

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,

$K_1$ = rate constant at $18.0^oC$

$K_2$ = rate constant at $37.0^oC$ = $3K_1$

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $18.0^oC=273+18.0=291K$

$T_2$ = final temperature = $37.0^oC=273+37.0=310K$

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

$\log (\frac{3K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{291K}-\frac{1}{310K}]$

$Ea=43374.66J/mole=43.4KJ$

Therefore, the activation energy for the reaction is, 43.4 KJ