For a first-order gaseous reaction,a plot of | vedclass

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(A) According to the Arrhenius equation: $k = Ae^{-E_a/RT}$Taking the logarithm on both sides: $\log \, k = \log \, A - \frac{E_a}{2.303R} 	imes \fra

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$36608$

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(A) According to the Arrhenius equation: $k = Ae^{-E_a/RT}$Taking the logarithm on both sides: $\log \, k = \log \, A - \frac{E_a}{2.303R} 	imes \frac{1}{T}$Comparing this with the equation of a straight line $y = mx + c$,the slope $(m)$ is given by: $m = -\frac{E_a}{2.303R}$Given that the slope is $-8000$,we have: $\frac{E_a}{2.303R} = 8000$Using $R = 1.987 \, cal \, K^{-1} \, mol^{-1}$:$E_a = 8000 	imes 2.303 	imes 1.987$$E_a = 36608 \, cal \, mol^{-1}$

For a first-order gaseous reaction,a plot of $\log \, k$ versus $1/T$ gives a straight line with a slope of $-8000$. Calculate the activation energy $(E_a)$ of the reaction in $cal$.

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