Question

If a temperature increase from 12.0 ∘c to 22.0 ∘c doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction?

Answer 1

Using Arrhenius Equation in Kinematics the answer comes out to be 48.45 kilo joules. I have uploaded photo see it.

Answer 2

Answer : The value of the activation barrier for the reaction is, 48.45 KJ

Solution :

The relation between the rate constant the activation energy is,

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

where,

$K_1$ = initial rate constant = x

$K_1$ = final rate constant = 2x

$T_1$ = initial temperature = $12^oC=273+12=285K$

$T_2$ = final temperature = $22^oC=273+22=295K$

R = gas constant = 8.314 KJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{2x}{x}=\frac{Ea}{2.303\times (8.314KJ/moleK)}\times [\frac{1}{285K}-\frac{1}{295K}]$

$Ea=48.45KJ$

Therefore, the value of the activation barrier for the reaction is, 48.45 KJ