Solve this following
Question:

For the reaction $\mathrm{A} \rightarrow \mathrm{B}$, the rate constant $\mathrm{k}\left(\right.$ in $\left.\mathrm{s}^{-1}\right)$ is given by

$\log _{10} k=20.35-\frac{\left(2.47 \times 10^{3}\right)}{T}$

The energy of activation in $\mathrm{kJ} \mathrm{mol}^{-1}$ is__________________ (Nearest integer)

[Given : $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]

 

Solution:

Given $\log \mathrm{K}=20.35-\frac{2.47 \times 10^{3}}{\mathrm{~T}}$

We know    $\log \mathrm{K}=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}$

$\Rightarrow \quad \frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}=2.47 \times 10^{3}$

$\mathrm{E}_{\mathrm{a}}=2.47 \times 10^{3} \times 2.303 \times \frac{8.314}{1000} \mathrm{~K} \mathrm{~J} / \mathrm{mole}$

$=47.29=47$ (Nearest integer)

 

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