Which of the following statements is correct | vedclass

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(C) 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

The graph of $\ln k$ vs $\frac{1}{T}$ is a straight line with a slope of $-\frac{E_a}{R}$.

Vedclass · Answer

(C) The Arrhenius equation is given by $k = A e^{-\frac{E_a}{RT}}$.Taking the natural logarithm on both sides,we get $\ln k = \ln A - \frac{E_a}{RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = \frac{1}{T}$,the slope $m = -\frac{E_a}{R}$.Option $C$ is correct because the slope of the plot of $\ln k$ versus $\frac{1}{T}$ is $-\frac{E_a}{R}$.Note: If the plot is $\log_{10} k$ versus $\frac{1}{T}$,the slope is $-\frac{E_a}{2.303R}$.

Which of the following statements is correct regarding the Arrhenius equation and reaction kinetics?

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