The slope of Arrhenius plot $\left( \ln k ext{ v/s } \frac{1}{T} ight)$ of first order reaction is $-5 i

Question

The slope of Arrhenius plot $\left( \ln k 	ext{ v/s } \frac{1}{T} ight)$ of first order reaction is $-5 	imes 10^3 	ext{K}$. The value of $E_a$ of the reaction is. Choose the correct option for your answer. [Given $R = 8.314 	ext{ JK}^{-1}	ext{mol}^{-1}$]

Accepted Answer

Arrhenius equation: $k = Ae^{-E_a/RT}$. Taking natural log: $\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} ight)$. The slope $m = -\frac{E_a}{R}$. Given $m = -5 	imes 10^3 	ext{K}$. So, $-5 	imes 10^3 = -\frac{E_a}{8.314}$. $E_a = 5 	imes 10^3 	imes 8.314 = 41570 	ext{ J/mol} = 41.57 	ext{ kJ/mol} \approx 41.5 	ext{ kJ/mol}$.

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