The slope of Arrhenius Plot ($\ln k$ v/s $1/T$) of the first-order reaction is $-5 imes 10^3 ext{ K}$. The

Question

The slope of Arrhenius Plot ($\ln k$ v/s $1/T$) of the first-order reaction is $-5 	imes 10^3 	ext{ K}$. The value of $E_a$ of the reaction is: [Given $R = 8.314 	ext{ J K}^{–1} 	ext{ mol}^{–1}$]

Accepted Answer

According to the Arrhenius equation,
$k = A e^{-E_a/RT}$
Taking the natural logarithm on both sides yields:
$\ln k = -\frac{E_a}{R} \left(\frac{1}{T}ight) + \ln A$
Comparing this with the equation of a straight line, $y = mx + c$, the plot of $\ln k$ versus $1/T$ gives a straight line with slope $m = -\frac{E_a}{R}$.
Given slope $= -5 	imes 10^3 	ext{ K}$
$-\frac{E_a}{R} = -5 	imes 10^3 	ext{ K}$
$E_a = 5 	imes 10^3 	imes R$
$E_a = 5 	imes 10^3 	ext{ K} 	imes 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1} = 41570 	ext{ J mol}^{-1} = 41.57 	ext{ kJ mol}^{-1} \approx 41.5 	ext{ kJ mol}^{-1}$.

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