Question

For a first order reaction $A\to P$, the temperature $\left(T\right)$ dependent rate constant $\left(k\right)$ was found to follow the equation, $\log k=-\left(2000\right)\frac{1}{T}+6.0$. The activation enrgy (in ${\text{kJ mol}}^{-1}$) for the reaction is: (Given, $R=8.3{\text{J mol}}^{-1}{\text{K}}^{-1}$). (Round off your answer upto nearest integer and report your answer upto two decimal only)

Answer 1

By Arrhenius equation, we know that,
$\log k=\log A-\frac{E_a}{2.303RT}$
On comparing with $\log k=-\left(2000\right)\frac{1}{T}+6.0$
$\log A=6,A={10}^6{s}^{-1}$
$\frac{-E_a}{2.303\times 8.3\times T}=-\frac{2000}{T}$
$E_a=2000\times 2.303\times 8.3\text{J}=38.3{\text{kJ mol}}^{-1}$