d
Rate constant of a reaction is given by arrhenius equation Arrhenius equation

\(K=A e^{-E a / R T}\)

Where \(K \rightarrow\) rate contant.

\(\text { A } \rightarrow \text { Collision factor. }\)

\(E_{a} \rightarrow \text { Activation energy }\)

\(T \rightarrow\) Temperature on kelvin scale

Let \(K_{1} \;and\; K_{2}\) be rate constant of a reaction of temperatures \(T_{1}\;and\; T_{2}\)

Applying Arrhenius equation.

\(K_{1}=A e^{-E a / R T_{1}} \longrightarrow(1)\)

\(K_{2}=A e^{-E a / R T_{2}} \quad \longrightarrow(2)\)

\(\Rightarrow\) Taking natural log on both sides,

\(\ln K_{1}=\ln A-\frac{E_{a}}{R T_{1}} \quad \longrightarrow \dots (1)\)

\(i n K_{2}=\ln A-\frac{E_{a}}{R T_{2}} \quad \longrightarrow \dots (2)\)

Now substracting \((1)\) from \((2)\)

\(i n K_{2}-i n K_{1}=\frac{-E_{a}}{R T_{2}}+\frac{E_{a}}{R T_{1}}\)

\(\Rightarrow \ln \left(\frac{K_{2}}{K_{1}}\right)=\frac{E_{a}}{R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]\)

OR

\(\Rightarrow \ln \left(\frac{K_{2}}{K_{1}}\right)=\frac{-E_{a}}{R}\left[\frac{1}{T_{2}}-\frac{1}{T_{1}}\right]\)