Which of the following graphs for ln k versus | vedclass

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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:$\ln k = \ln A - \frac{E_a}{RT}$This eq

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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:$\ln k = \ln A - \frac{E_a}{RT}$This equation is in the form of a straight line $y = mx + c$,where:$y = \ln k$$x = \frac{1}{T}$$m = -\frac{E_a}{R}$ (slope)$c = \ln A$ (intercept)Since the slope is negative $(-\frac{E_a}{R})$,the graph of $\ln k$ versus $\frac{1}{T}$ is a straight line with a negative slope and a positive intercept on the $y$-axis.This corresponds to the graph shown in option $C$.

Which of the following graphs for $\ln k$ versus $\frac{1}{T}$ is correct?

Similar Questions

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