$\left( {{\rm{R}} = 8.3\;{\rm{Jmo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}},\ln \left( {\frac{2}{3}} \right) = 0.4,\left. {{e^{ - 3}} = 4.0} \right)} \right.$
\(\ln \left(\frac{60}{40}\right)=\frac{-\operatorname{Ea}}{8.3}\left[\frac{1}{400}-\frac{1}{300}\right]\)
\(\mathrm{E}=0.4 \times 1200 \times 8.3\)
\(=3.984\; \mathrm{kJ} / \mathrm{mole}\)
${A}+{B} \rightarrow {M}+{N}$ $......$ ${kJ} {mol}^{-1}$ બરાબર છે. (નજીકના પૂર્ણાંકમાં)