Question

The rate of a particular reaction doubles when the temperature changes from $300$K to $310$K. calculate the ${{\text{E}}_{\text{a}}}$ of the reaction.

Answer 1

Hint: We will use Arrhenius equation to determine the activation energy. Arrhenius equation gives the relation in activation energy, temperature and rate constant. As the reaction doubles means if the rate constant is one at low temperature and then the rate constant will be two at high temperature.

Formula used: ${\text{K}}\,{\text{ = }}\,{\text{A}}{{\text{e}}^{\dfrac{{ - {{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}}}$

Complete step-by-step solution :
The Arrhenius equation is as follows:
${\text{K}}\,{\text{ = }}\,{\text{A}}{{\text{e}}^{\dfrac{{ - {{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}}}$
Where,
K is the rate constant.
A is the Arrhenius constant.
${{\text{E}}_{\text{a}}}$is the activation energy.
R is the gas constant.
T is the temperature.
At two different temperature, the equation can be written as follows:
$\Rightarrow \operatorname{l} {\text{n}}\,\dfrac{{{{\text{K}}_{\text{2}}}}}{{{{\text{K}}_{\text{1}}}}} = \, - \dfrac{{{{\text{E}}_{\text{a}}}}}{{\text{R}}}\left( {\dfrac{1}{{{{\text{T}}_2}}} - \dfrac{1}{{{{\text{T}}_1}}}} \right)$
It is given that the reaction doubles so, the value of the rate constant at two different temperatures is two.
$\Rightarrow \dfrac{{{{\text{K}}_{\text{2}}}}}{{{{\text{K}}_{\text{1}}}}} = 2$.
Substitute $2$for $\dfrac{{{{\text{K}}_{\text{2}}}}}{{{{\text{K}}_{\text{1}}}}}$,
 $8.314\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}$ for R, $300$K for ${{\text{T}}_1}$ and $310$K for ${{\text{T}}_2}$.
$\Rightarrow \operatorname{l} {\text{n}}\,{\text{2}} = \, - \dfrac{{{{\text{E}}_{\text{a}}}}}{{8.314\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}\left( {\dfrac{1}{{310}} - \dfrac{1}{{300}}} \right)$
$\Rightarrow \operatorname{l} {\text{n}}\,{\text{2}} = \, - \dfrac{{{{\text{E}}_{\text{a}}}}}{{8.314\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}\left( { - \dfrac{{10}}{{93000}}} \right)$
$\Rightarrow {\text{0}}{\text{.69}} = \,\dfrac{{{{\text{E}}_{\text{a}}}}}{{8.314\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}\dfrac{{10}}{{93000}}$
$\Rightarrow {{\text{E}}_{\text{a}}}\, = \,\dfrac{{{\text{0}}{\text{.69}}\, \times 8.314 \times 93000}}{{10}}$
$\Rightarrow {{\text{E}}_{\text{a}}}\, = \,{\text{5}}{\text{.4}}\, \times {10^4}\,{\text{J/mol}}$

So, the ${{\text{E}}_{\text{a}}}$ of the reaction when rate of a particular reaction doubles when the temperature changes from $300$K to $310$K is ${\text{5}}{\text{.4}}\, \times {10^4}$.

Note:Activation energy is the minimum energy required to convert the product into activated complex then product. Activation energy depends upon the temperature. On increasing temperature activation energy decreases. Presence of a catalyst decreases the activation energy hence increases the rate of reaction. Negative catalyst increases the activation energy, so decreases the rate of reaction. Activation energy is measured in units of J/mol or kJ/mol. Activation energy can be negative, positive or zero. It is zero in nuclear reaction. The term ${{\text{E}}_{\text{a}}}{\text{/RT}}$ gives the effective collision.