Question

f 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 value of activation barrier for the reaction is, 43.374 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,

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

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

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mol.K

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

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

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

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

$Ea=43,374 J/mol=43.374 KJ/mol$

Therefore, the activation energy for the reaction is, 43.374 kJ/mol.