The differential form of the Arrhenius equati | vedclass

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Question

(B) The Arrhenius equation is given by $K = A e^{-\frac{E_a}{RT}}$.Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.

Vedclass · Accepted Answer

$\frac{d \ln K}{dT} = \frac{\Delta E^*}{RT^2}$

Vedclass · Answer

(B) The Arrhenius equation is given by $K = A e^{-\frac{E_a}{RT}}$.Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.Differentiating both sides with respect to temperature $T$,we get $\frac{d \ln K}{dT} = 0 - \frac{E_a}{R} 	imes (-\frac{1}{T^2})$.Therefore,$\frac{d \ln K}{dT} = \frac{E_a}{RT^2}$.Comparing this with the given options,where $\Delta E^*$ represents the activation energy $E_a$,the correct expression is $\frac{d \ln K}{dT} = \frac{\Delta E^*}{RT^2}$.

The differential form of the Arrhenius equation is:

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