\(\mathrm{K}^{\prime}=\mathrm{Ae}^{\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}}=10^{6} \mathrm{K}\)
Ae \(\frac{-\mathrm{E}}{\mathrm{RT}}=10^{6} \times \mathrm{Ae}^{\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}}\)
\(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}=\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}+\ln 10^{6}\)
\(\mathrm{E}_{\mathrm{a}}^{\prime}=\mathrm{E}_{\mathrm{a}}-\mathrm{RT} \ln 10^{6}\)
\(\mathrm{E'}_{\mathrm{a}}-\mathrm{E}_{\mathrm{a}}=-\mathrm{RT} \ln 10^{6}\)
\(=-6 \mathrm{RT} \times 2.303\)
$\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$
$S.\ No$ સમય/s કુલ દબાણ/(atm)
$1.$ $0$ $0.1$
$2.$ $115$ $0.28$
પ્રક્રિયાનો વેગ અચળાંક _______________$\times 10^{-2} \mathrm{~s}^{-1}$ (નજીકનાં પૂનાંકમાં)
$\left[\right.$ આપેલ છે $\left.: \log _{10} 2=0.301, \ln 10=2.303\right]$
( $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 4=0.6021$ આપેલ છે.)