Question

The rate constant for the decomposition of a certain reaction is described by the equation
$logk\left(s^{-1}\right)=14-\frac{1.25\times {10}^{4}K}{T}$

Energy of activation (in $kcal$) is:

• A. $57.6kcal$
• B. $1.25\times {10}^{4}kcal$
• C. $14.0kcal$
• D. None of these

Answer 1

Answer: (A)

$57.6kcal$

As we know,
$logk=logA-\frac{E_a}{2.303RT}$
$\therefore logA=14\Rightarrow A={10}^{14}{s}^{-1}$
$\frac{E_a}{2.303RT}=\frac{1.25\times {10}^{4}K}{T}$
$\therefore E_a=1.25\times {10}^4\times 2.303\times {10}^{-3}kcal$
$=57.6kcal$