According to the Arrhenius equation,a plot of | vedclass

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

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$1/T$

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(C) The Arrhenius equation is given by $k = A \cdot e^{-E_a / (RT)}$.Taking the logarithm on both sides,we get $\ln \, k = \ln \, A - \frac{E_a}{RT}$.Converting to base $10$ logarithm,we get $\log \, k = \log \, A - \frac{E_a}{2.303 \, R} \cdot \frac{1}{T}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \, k$,$x = 1/T$,$m = -E_a / (2.303 \, R)$,and $c = \log \, A$.Therefore,a plot of $\log \, k$ versus $1/T$ yields a straight line.

According to the Arrhenius equation,a plot of the logarithm of the rate constant $(\log \, k)$ versus which of the following yields a straight line?

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