(આપેલું છે$: \ln 10=2.3, R =8.3 \,J\, K ^{-1} \,mol ^{-1}, \log 2=0.30$ )
$0.3=\frac{ E _{ a }}{2.303 \times 8.3}\left(\frac{9}{300 \times 309}\right)$
$E _{ a }=\frac{0.3 \times 2.303 \times 8.3 \times 300 \times 309}{9}$
$=59065.04\, J$
$E _{ a }=59.06\, kJ$
${A}+{B} \rightarrow {M}+{N}$ $......$ ${kJ} {mol}^{-1}$ બરાબર છે. (નજીકના પૂર્ણાંકમાં)