Consider the following plots of rate constant | vedclass

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(C) The Arrhenius equation is given by $\log K = \frac{-E_{a}}{2.303 R} \left(\frac{1}{T}\right) + \log A$.Comparing this with the equation of a strai

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$E_{c} > E_{a} > E_{d} > E_{b}$

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(C) The Arrhenius equation is given by $\log K = \frac{-E_{a}}{2.303 R} \left(\frac{1}{T}\right) + \log A$.Comparing this with the equation of a straight line $y = mx + c$,the slope of the plot of $\log K$ versus $\frac{1}{T}$ is $m = \frac{-E_{a}}{2.303 R}$.The magnitude of the slope is $|m| = \frac{E_{a}}{2.303 R}$.Since $\frac{1}{2.303 R}$ is a constant,the activation energy $E_{a}$ is directly proportional to the magnitude of the slope $(|m|)$.From the given plot,the order of the magnitude of the slopes is $c > a > d > b$.Therefore,the order of activation energies is $E_{c} > E_{a} > E_{d} > E_{b}$.

Consider the following plots of rate constant versus $\frac{1}{T}$ for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

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