Question

For a reaction, temperature is increased from 27C to 57C, then the rate of reaction becomes 5 times. Calculate activation energy of reaction in terms of KJ/mol

Answer 1

$k=Ae^{ \frac{-E_a}{RT} }$

$At\ T=27C=300K, k_1=k \\ \\ At\ T=57C=330K, k_2=5k$

$\frac{k_2}{k_1} = \frac{A e^{ \frac{-E_a}{RT_2} } }{A e^{ \frac{-E_a}{RT_1} } } = \frac{e^{ \frac{-E_a}{RT_2} } }{e^{ \frac{-E_a}{RT_1} } } =e^{ \frac{-E_a}{R}( \frac{1}{T_2} - \frac{1}{T_1} ) } \\ \\ \\ 5=e^{ \frac{-E_a}{R}( \frac{1}{T_2} - \frac{1}{T_1} ) } \\ \\ ln\ 5= \frac {-E_a}{R}( \frac{1}{T_2} - \frac{1}{T_1} ) \\ \\ ln\ 5= \frac {-E_a}{R}( \frac{T_1-T_2}{T_1T_2} )$

$-E_a=ln\ 5*R* \frac{T_1T_2}{T_1-T_2} \\ \\ E_a=ln\ 5*R* \frac{T_1T_2}{T_2-T_1}=1.6094*8.314* \frac{330*300}{330-300} \ J = 44155.8\ J/mol$

$Activation\ energy =44155.8\ J/mol=44.15\ kJ/mol$