\(k=A e^{-E_{\alpha} / R T}\)
\(\ln k=\ln A-\frac{E_{a}}{R T}\)
Hence, if \(\ln k\) is plotted against \(1 / T\) slope of the line will be \(-\frac{E_{a}}{R}\)
$\mathop {2{N_2}{O_5}}\limits_{{\rm{(in}}\,\,{\rm{CC}}{{\rm{l}}_4}{\rm{)}}} \to \mathop {4N{O_2}}\limits_{{\rm{(in}}\,\,{\rm{CC}}{{\rm{l}}_4}{\rm{)}}} + {O_2}$
$1$. $[A]$ $0.1$, $[B]$ $0.1 - $ પ્રારંભિક દર $ \rightarrow 7.5 \times 10^{-3}$
$2$. $[A]$ $0.3$, $[B]$ $0.2 -$ પ્રારંભિક દર $ \rightarrow 9.0 \times 10^{-2}$
$3$. $[A]$ $0.3$, $[B]$ $0.4 -$ પ્રારંભિક દર $ \rightarrow 3.6 \times 10^{-1}$
$4$. $[A]$ $0.4$, $[B]$ $0.1 -$ પ્રારંભિક દર $ \rightarrow 3.0 \times 10^{-2}$
[લો; $R =8.314 \,J\, mol ^{-1}\, K ^{-1}$ In $3.555=1.268$]