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

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$\log K$ versus $1/T$

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(C) The Arrhenius equation is given by $K = A e^{-E_a / RT}$.Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.Converting to base $10$ logarithm,we get $\log K = \log A - \frac{E_a}{2.303 R} 	imes \frac{1}{T}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log K$ and $x = 1/T$,the slope $m$ is equal to $-\frac{E_a}{2.303 R}$.Thus,the plot of $\log K$ versus $1/T$ is a straight line with a slope of $-\frac{E_a}{2.303 R}$,which allows us to calculate the activation energy $E_a$.

Which of the following plots gives the value of activation energy?

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