Question

Activation energy $\left(E_a\right)$ 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 $\frac{K_2}{K_1}=-\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$
• B. ln $\frac{K_2}{K_1}=-\frac{E_a}{R}(\frac{1}{T_2}+\frac{1}{T_1})$
• C. ln $\frac{K_2}{K_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$
• D. ln $\frac{K_2}{K_1}=-\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

Answer 1

Answer: (A), (C)

The correct options are
A ln $\frac{K_2}{K_1}=-\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$
C ln $\frac{K_2}{K_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

Rate constant of a reaction is given by arrhenius equation

Arrhenius equation :

$K={Ae}^{-Ea/RT}$

Where K $\to$ rate contant.

A $\to$ Collision factor.

$E_a\to$ Activation energy

T $\to$ Temperature on kelvin scale

Let $K_1$ & $K_2$ be rate constant of a reaction of temperatures $T_1$ & $T_2$.

Applying Arrhenius equation.

$K_1={Ae}^{-Ea/R{T}_1}⟶\left(1\right)$

$K_2=A{e}^{-Ea/R{T}_2}⟶\left(2\right)$

$\Rightarrow$ Taking natural log on both sides,

${lnK}_1=lnA-\frac{E_a}{{RT}_1}⟶\left(1\right)$

$in{K}_2=lnA-\frac{E_a}{{RT}_2}⟶\left(2\right)$

Now substracting (1) from (2)

$in{K}_2-in{K}_1=\frac{-E_a}{{RT}_2}+\frac{E_a}{{RT}_1}$

$\Rightarrow ln\left(\frac{K_2}{K_1}\right)=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

OR

$\Rightarrow ln\left(\frac{K_2}{K_1}\right)=\frac{-E_a}{R}[\frac{1}{T_2}-\frac{1}{T_1}]$