According to the Arrhenius equation, in a gra | vedclass

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(C) The Arrhenius equation is given by $k = A e^{-E_a/RT}$.Taking the logarithm to the base $10$ on both sides, we get:$\log_{10} k = \log_{10} A - \f

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$-\frac{E_a}{2.303R}$

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(C) The Arrhenius equation is given by $k = A e^{-E_a/RT}$.Taking the logarithm to the base $10$ on both sides, we get:$\log_{10} k = \log_{10} A - \frac{E_a}{2.303RT}$.This equation is in the form of a straight line $y = mx + c$, where $y = \log_{10} k$, $x = \frac{1}{T}$, $c = \log_{10} A$, and the slope $m = -\frac{E_a}{2.303R}$.Therefore, the slope of the graph of $\log k$ versus $\frac{1}{T}$ is $-\frac{E_a}{2.303R}$.

According to the Arrhenius equation, in a graph of $\log k$ versus $\frac{1}{T}$, the slope will be . . . . . . .

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