Question

Activation energy $({{E}_{a}})$ and rate constant $({{k}_{1}}\,and\,{{k}_{2}})$ of a chemical reaction at two different temperatures $({{T}_{1}}\,and\,{{T}_{2}})$ are related by:
(A)- $\ln \dfrac{{{k}_{2}}}{{{k}_{1}}}=-\dfrac{{{E}_{a}}}{R}\left( \dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}} \right)$
(B)- $\ln \dfrac{{{k}_{2}}}{{{k}_{1}}}=-\dfrac{{{E}_{a}}}{R}\left( \dfrac{1}{{{T}_{2}}}+\dfrac{1}{{{T}_{1}}} \right)$
(C)- $\ln \dfrac{{{k}_{1}}}{{{k}_{2}}}=\dfrac{{{E}_{a}}}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)$
(D)- $\ln \dfrac{{{k}_{2}}}{{{k}_{1}}}=-\dfrac{{{E}_{a}}}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)$

Answer 1

Hint: The relation between the rate constant, temperature and activation energy is given by the Arrhenius equation. The activation energy being independent of temperature, a relation between the rate constants at two different temperatures can be obtained.

Complete step by step answer:
A relation between the rate constant (k) and absolute temperature (T) during the reaction, was given by Arrhenius equation as follows:
$k=A{{e}^{-{{E}_{a}}/RT}}$
where A is the exponential or collision factor, ${{E}_{a}}$ is the activation energy and R is the gas constant.
The activation energy, ${{E}_{a}}$ is the energy required by the reactant in order to overcome the energy level of the transition state, so that the reaction may occur towards the product. This energy is independent of the temperature.
Then, let the rate constant of the reaction at two different temperatures ${{T}_{1}}\,and\,{{T}_{2}}$ be ${{k}_{1}}\,and\,{{k}_{2}}$ respectively. So, the Arrhenius equation will be:
${{k}_{1}}=A{{e}^{-{{E}_{a}}/R{{T}_{1}}}}$ and ${{k}_{2}}=A{{e}^{-{{E}_{a}}/R{{T}_{2}}}}$
Taking logarithm on both the sides of the above equation, we get,
\[\ln \,{{k}_{1}}=\ln A-\dfrac{{{E}_{a}}}{R{{T}_{1}}}\] -----(a)
\[\ln {{k}_{2}}=\ln A-\dfrac{{{E}_{a}}}{R{{T}_{2}}}\] ------(b)
On subtracting equation (a) from (b), we get,
\[\ln \,{{k}_{2}}-\ln \,{{k}_{1}}=-\dfrac{{{E}_{a}}}{R{{T}_{2}}}+\dfrac{{{E}_{a}}}{R{{T}_{1}}}\]

As we know, $\operatorname{lnA}-lnB=\ln \left( \dfrac{A}{B} \right)$ , so we have,
\[\ln \,\left( \dfrac{{{k}_{2}}}{{{k}_{1}}} \right)=-\dfrac{{{E}_{a}}}{R}\left( \dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}} \right)\]
Taking the negative sign inside the bracket, we get,
\[\ln \,\left( \dfrac{{{k}_{2}}}{{{k}_{1}}} \right)=\dfrac{{{E}_{a}}}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)\]

Therefore, the Activation energy $({{E}_{a}})$ and rate constant $({{k}_{1}}\,and\,{{k}_{2}})$ of a chemical reaction at two different temperatures $({{T}_{1}}\,and \,{{T}_{2}})$ are related by using the Arrhenius equation by option (A)- $\ln \dfrac{{{k}_{2}}}{{{k}_{1}}}=-\dfrac{{{E}_{a}}}{R}\left( \dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}} \right)$ or option (C)- $\ln \dfrac{{{k}_{1}}}{{{k}_{2}}}=\dfrac{{{E}_{a}}}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)$.
So, the correct answer is “Option C”.

Note: The activation energy can be lowered by the addition of catalyst, thus increasing the rate of reaction.
The magnitude of the rate constant is dependent on the temperature and the activation energy, as can be seen from the exponential part.
The rate constant is equal to the collision factor, A, when either the activation energy is zero, that is, there is no barrier for the reaction to occur or when the temperature is high enough to allow every collision to be effective and having its kinetic energy more than the activation energy.