$K=A e^{-E a / R T}$
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
$\Rightarrow \ln \left(\frac{K_{2}}{K_{1}}\right)=\frac{-E_{a}}{R}\left[\frac{1}{T_{2}}-\frac{1}{T_{1}}\right]$
$2 \mathrm{HI}_{(\mathrm{g})} \rightarrow \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})}$
પ્રક્રિયાનો ક્રમ................ છે.
| $1$ | $2$ | $3$ | |
| $\mathrm{HI}\left(\mathrm{mol} \mathrm{L}^{-1}\right)$ | $0.005$ | $0.01$ | $0.02$ |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}-1\right)$ | $7.5 \times 10^{-4}$ | $3.0 \times 10^{-3}$ | $1.2 \times 10^{-2}$ |