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?

Similar Questions

Which graph shows zero activation energy?

From the given data for the reaction $H_2 + I_2 \rightarrow 2HI$,calculate the activation energy $(E_a)$:
$T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$
$T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$

For a chemical reaction,the rate constant,activation energy,and Arrhenius factor at $25 \, ^\circ C$ are $3.0 \times 10^{-4} \, s^{-1}$,$104.4 \, kJ \, mol^{-1}$,and $6.0 \times 10^{14} \, s^{-1}$ respectively. Find the value of the rate constant as $T \rightarrow \infty$.

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