Question

The rate of most of the reactions double when their temperature is raised from $298{\text{ K}}$to 308${\text{K}}$. Calculate the activation energy of the reaction.

Answer 1

This question can be solved from the equation of activation energy, where activation energy can be defined as the minimum energy required for the reactants to combine with each other and the reaction to take place.

Formula used:
${\text{k = A}}{{\text{e}}^{{\text{ - }}\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}}}$,
where k is the rate constant of the reaction, A is the Arrhenius factor, ${{\text{E}}_{\text{a}}}$is the activation energy..

Complete step by step answer:
Let for the first reaction the rate constant be ${{\text{k}}_{\text{1}}}$and that for the second reaction be ${{\text{k}}_{\text{2}}}$. The Arrhenius Factor is a frequency factor that depends on the frequency of oscillations of the reactant molecules at a standard concentration. Hence we can consider that this factor is the same for both as there is no change in the concentration of the reactants.
Taking log on both sides of the above equation,
${\text{logk = - }}\dfrac{{{\text{Ea}}}}{{{\text{2}}{\text{.303RT}}}}{\text{ + lnA}}$
For two different temperature ${{\text{T}}_{\text{1}}}$and ${{\text{T}}_2}$we have,
${\text{log}}{{\text{k}}_{\text{1}}}{\text{ = - }}\dfrac{{{\text{Ea}}}}{{{\text{2}}{\text{.303R}}{{\text{T}}_{\text{1}}}}}{\text{ + lnA}}$............. (1)
${\text{log}}{{\text{k}}_{\text{2}}}{\text{ = - }}\dfrac{{{\text{Ea}}}}{{{\text{2}}{\text{.303R}}{{\text{T}}_{\text{2}}}}}{\text{ + lnA}}$.................... (2)
Subtracting (2) from (1), we get,
${\text{log}}\dfrac{{{{\text{k}}_{\text{1}}}}}{{{{\text{k}}_2}}}{\text{ = }}\dfrac{1}{{2.303}}\left( {\dfrac{{{\text{Ea}}}}{{{\text{R}}{{\text{T}}_{\text{1}}}}}{\text{ - }}\dfrac{{{\text{Ea}}}}{{{\text{R}}{{\text{T}}_2}}}} \right)$
$\Rightarrow {\text{log}}\dfrac{{{{\text{k}}_{\text{2}}}}}{{{{\text{k}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303R}}}}\left( {\dfrac{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}}}} \right)$
Putting the values of ${{\text{k}}_{\text{1}}}$and ${{\text{k}}_{\text{2}}}$, ${{\text{T}}_{\text{1}}}$and ${{\text{T}}_2}$, we get,
${\text{log}}\dfrac{{\text{2}}}{{\text{1}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303}} \times {\text{8}}{\text{.314}}}}\left( {\dfrac{{{\text{308 - 298}}}}{{308 \times 298}}} \right)$
$\Rightarrow \dfrac{{0.314 \times 2.303 \times 8.314 \times 308 \times 298}}{{\left( {308 - 298} \right)}} = {{\text{E}}_{\text{a}}}$
Solving this, we get:
$\Rightarrow {{\text{E}}_{\text{a}}} = 52.903$ Kilo-Joules.
So the activation energy required for the reaction is $52.903$ Kilo-Joules.

Note:
Before forming the product, the reactants after attaining the activation energy from the activated complex which is the intermediate between the bond breaking stage and bond formation stage.
Some energy is released when the activated complex converts to the products and is known as the Bond Formation Energy.