Question

For the reaction of ${{H}_{2}}$ and ${{I}_{2}}$ , the rate constant is $2.5X{{10}^{-4}}d{{m}^{3}}mo{{l}^{-1}}{{s}^{-1}} at {{327}^{o}}C$ and $1.0d{{m}^{3}}mo{{l}^{-1}}{{s}^{-1}}at{{527}^{o}}C$ . the activation energy for the reaction, in $kJmo{{l}^{-1}}is:$ ($R=8.314J{{K}^{-1}}mo{{l}^{-1}}$ ).
(A) 72
(B) 166
(C) 150
(D) 59

Answer 1

Hint: The temperature dependence of the rate of a chemical reaction can be accurate by the Arrhenius equation. Though this equation is applicable under a wide range of circumstances, collision theory. The activation energy of a chemical reaction can be calculated by this Arrhenius equation.


Complete step by step answer:
The given rate of reaction,
${{H}_{2}}(g)+{{I}_{2}}(g)\to 2HI(g)$

According to Arrhenius, this reaction takes place only when a molecule of hydrogen and a molecule of hydrogen and a molecule of iodine collide to form an unstable intermediate. It exists for a very short time and then breaks up to form two molecules of hydrogen iodide.

The energy required to form this intermediate is called activation energy. From the above diagram obtained by plotting potential energy vs reaction coordinate.
According to the Arrhenius equation, the temperature dependence of the rate of a chemical reaction explained by this equation,
$\ln k=-\dfrac{{{E}_{a}}}{RT}+\ln A$
Where k = rate constant, A = constant, ${{E}_{a}}$ = activation energy.
If ${{k}_{1}}\And {{k}_{2}}$ are the values of rate constants at temperature ${{T}_{1}}\And {{T}_{2}}$ respectively,

Then the equation changes as below,
     \[\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}[\dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}}]\] --(1)
Given rate constants of ${{H}_{2}}$ and ${{I}_{2}}$ are ${{k}_{1}}=2.5X{{10}^{-4}}d{{m}^{3}}mo{{l}^{-1}}{{s}^{-1}}$ and ${{k}_{2}}=1.0d{{m}^{3}}mo{{l}^{-1}}{{s}^{-1}}$
And the temperatures given respectively rate constants, ${{T}_{1}}=600K\ And {{T}_{2}}=800K$
Substitute the above values in equation (1),
     \[\begin{align}
  & \log \dfrac{1}{2.5X{{10}^{-4}}}=\dfrac{{{E}_{a}}}{2.303X8.314}[\dfrac{1}{600}-\dfrac{1}{800}] \\
 & \log 4X{{10}^{3}}=\dfrac{{{E}_{a}}}{2.303X8.314}[\dfrac{1}{2400}] \\
 & {{E}_{a}}=165.5kg/mole\cong 166kg/mole \\
\end{align}\]
So, the correct answer is “Option B”.

Note: It is believed that catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy between reactants and products and hence lowering the potential barrier shown in the above diagram. From the Arrhenius equation, that lowers the value of activation energy faster will be the rate of reaction.