Chemistry - Chemical Kinetics Question with Solution | TestHub

We know that,
Rate of reaction(K)$=\mathrm{A}{\mathrm{e}}^{-\mathrm{E}/\mathrm{R}\mathrm{T}}$...(i)
And,$10^6.\mathrm{K}=\mathrm{A}{\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{c}}/\mathrm{R}\mathrm{T}}$...(ii)
By dividing equation (ii) by (i)
$\frac{10^6.\mathrm{K}}{\mathrm{K}}=\frac{\mathrm{A}{\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{c}}/\mathrm{R}\mathrm{T}}}{\mathrm{A}{\mathrm{e}}^{-\mathrm{E}/\mathrm{R}\mathrm{T}}}$
${10}^6={\text{e}}^{\left(\text{E}-{\text{E}}_{\text{c}}\right)/\text{RT}}$
$6.\mathrm{l}\mathrm{n}10=(\mathrm{E}-\mathrm{E}\mathrm{c})/\mathrm{R}\mathrm{T}$
$\frac{\mathrm{E}-{\mathrm{E}}_{\mathrm{c}}}{\mathrm{R}\mathrm{T}}=2.303\times 6$
$\mathrm{E}-{\mathrm{E}}_{\mathrm{c}}=6\times 2.303\mathrm{R}\mathrm{T}$
or$\mathrm{\Delta }{\mathrm{E}}_{\mathrm{a}}={\mathrm{E}}_{\mathrm{c}}-\mathrm{E}=-6\times 2.303\mathrm{R}\mathrm{T}$