$ k=A e^{-\frac{E_a}{R T}} $
$ \therefore \quad \ln k=\ln A-\frac{E_a}{R T}$
In $\mathrm{kv} / \mathrm{s} \frac{1}{\mathrm{~T}}$ gives a straight line graph with slope $=-\frac{E_a}{R}$ and intercept $=\ln \mathrm{A}$
$[$આપેલ છે :${R}=8.31\, {~J} \,{~K}^{-1} \,{~mol}^{-1} ; \log 6.36 \times 10^{-3}=-2.19$ $\left.10^{-4.79}=1.62 \times 10^{-5}\right]$
$\mathrm{R}=8.314\; \mathrm{JK}^{-1} \mathrm{mol}^{-1}$
$2 {~K}_{2} {Cr}_{2} {O}_{7}+8 {H}_{2} {SO}_{4}+3 {C}_{2} {H}_{6} {O} \rightarrow 2 {Cr}_{2}\left({SO}_{4}\right)_{3}+$
$3 {C}_{2} {H}_{4} {O}_{2}+2 {~K}_{2} {SO}_{4}+11 {H}_{2} {O}$
જો ${Cr}_{2}\left({SO}_{4}\right)_{3}$નો દેખાવનો દર $2.67 \,{~mol}$ $\min ^{-1}$ ચોક્કસ સમયે, ${C}_{2} {H}_{6} {O}$નો એક જ સમયે ગાયબ થવાનો દર $....$ ${mol}\, {min}^{-1}$ છે. (નજીકના પૂર્ણાંકમાં)