Question

The rate of a chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate ${{\text{E}}_{\text{a}}}$ .

Answer 1

Hint: Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant ‘k’ is given by the Arrhenius equation which is written as:
${\text{k = A}}{{\text{e}}^{\dfrac{{{\text{ - }}{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}}}$
Here, the pre – exponential factor A represents a constant called the frequency factor, ${{\text{E}}_{\text{a}}}$ represents the energy of activation, R represents the gas constant and T is the absolute temperature.
The value of activation energy can be calculated from the Arrhenius equation by knowing the values of the rate constants of the reaction at two different temperatures.

Complete step by step answer:
Given that the rate of a chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K.
We need to calculate the value of the activation energy.
The energy of activation is an important quantity as it is characteristic of the reaction.
The Arrhenius equation can also be written in another form:
$\log \dfrac{{{{\text{k}}_{\text{2}}}}}{{{{\text{k}}_{\text{1}}}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303R}}}}\left( {\dfrac{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}}}} \right)$
Or $\log \dfrac{{{{\text{k}}_{\text{2}}}}}{{{{\text{k}}_{\text{1}}}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303R}}}}\left( {\dfrac{1}{{{{\text{T}}_{\text{1}}}}} - \dfrac{1}{{{{\text{T}}_2}}}} \right)$
Here, ${{\text{k}}_{\text{1}}}$ represents the value of the rate constant at the temperature ${{\text{T}}_{\text{1}}}$ and ${{\text{k}}_2}$ represents the value of the rate constant at the temperature ${{\text{T}}_2}$ .
Now, by question, ${{\text{T}}_{\text{1}}}$is equal to 298 K. Since temperature increases by 10 K,
 $
  {{\text{T}}_2} = {{\text{T}}_1} + 10{\text{K}} \\
   \Rightarrow {{\text{T}}_2} = 298{\text{K}} + 10{\text{K}} \\
   \Rightarrow {{\text{T}}_2} = 308{\text{K}} \\
 $
Let the value of the rate constant ${{\text{k}}_{\text{1}}}$ at the temperature ${{\text{T}}_{\text{1}}}$ be ‘k’. Since, the rate gets doubled at ${{\text{T}}_2}$ , the value of the rate constant ${{\text{k}}_2}$ will be ‘2k’.
The value of gas constant ${\text{R = 8}}{\text{.314J}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ .
Now, substitute all these values in the Arrhenius equation.
\[
  \log \dfrac{{{\text{2k}}}}{{\text{k}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303}} \times {\text{8}}{\text{.314}}}}\left( {\dfrac{{{\text{308 - 298}}}}{{298 \times 308}}} \right) \\
   \Rightarrow \log 2 = \dfrac{{{{\text{E}}_{\text{a}}}}}{{19.147}}\left( {\dfrac{{10}}{{91784}}} \right) \\
   \Rightarrow 0.30103 = \dfrac{{{{\text{E}}_{\text{a}}}}}{{19.147}}\left( {\dfrac{{10}}{{91784}}} \right) \\
   \Rightarrow {{\text{E}}_{\text{a}}} = \dfrac{{0.30103 \times 19.147 \times 91784}}{{10}} \\
   \Rightarrow {{\text{E}}_{\text{a}}} = 52902.65{\text{Jmo}}{{\text{l}}^{{\text{ - 1}}}} \\
   \Rightarrow {{\text{E}}_{\text{a}}} = 52.9{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}} \\
 \]
So, the value of the activation energy is \[52.9{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}\] .

Note:
The increase in rate of reaction with temperature is not due to the increase in the total number of collisions but due to the increase in the total number of effective collisions.
With increase of temperature, the most probable kinetic energy increases and the fraction of molecules having most probable kinetic energy decreases.