The rate constant (k) of a reaction is measur | vedclass

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(A) According to the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \

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$2\, R$

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(A) According to the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = \frac{1}{T}$,the slope $(m)$ is equal to $-\frac{E_a}{R}$.From the given graph,the slope is calculated as:$	ext{Slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 10}{(5 	imes 10^{-3}) - 0} = \frac{-10}{5 	imes 10^{-3}} = -2 	imes 10^3$.Equating the slope to $-\frac{E_a}{R}$:$-\frac{E_a}{R} = -2 	imes 10^3$$E_a = 2 	imes 10^3 \, R \, J \, mol^{-1} = 2 \, R \, kJ \, mol^{-1}$.

The rate constant $(k)$ of a reaction is measured at different temperatures $(T),$ and the data are plotted in the given figure. The activation energy of the reaction in $kJ\, mol^{-1}$ is :
($R$ is gas constant)

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The rate constant $(k)$ of a reaction is measured at different temperatures $(T),$ and the data are plotted in the given figure. The activation energy of the reaction in $kJ\, mol^{-1}$ is :($R$ is gas constant)