In a study of effect of temperature on reaction rate, the value of $\frac{1}{K}.\frac{dK}{dT}$ is fo

Question

In a study of effect of temperature on reaction rate, the value of $\frac{1}{K}.\frac{dK}{dT}$ is found to be $\frac{1.25 	imes 10^{6}}{T^{3}}K^{- 1}$. Identify the correct statement(s).

Accepted Answer

$\frac{1}{K}.\frac{dK}{dT} = \frac{d\left( \ln K \right)}{dT} = \frac{1.25 \times 10^{6}}{T^{3}} = \frac{E_{a}}{RT^{2}}$

$\therefore E_{a} = \frac{1.25 \times 10^{6}R}{T} = \frac{1.25 \times 10^{6} \times 2}{250} = 10^{4}\text{ cal/mol}$

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