The plot of $\ln k$ vs $1/T$ for the following reaction, $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$ gives a st

Question

The plot of $\ln k$ vs $1/T$ for the following reaction, $2N_2O_5(g) ightarrow 4NO_2(g) + O_2(g)$ gives a straight line with the slope of the line equal to $-1.0 	imes 10^4 	ext{ K}$. What is the activation energy for the reaction in $	ext{J mol}^{–1}$? (Given: $R = 8.3 	ext{ J K}^{–1} 	ext{ mol}^{–1}$)

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 $= -1.0 	imes 10^4 	ext{ K}$
$-\frac{E_a}{R} = -1.0 	imes 10^4 	ext{ K}$
$E_a = 1.0 	imes 10^4 	imes R$
$E_a = 1.0 	imes 10^4 	ext{ K} 	imes 8.3 	ext{ J K}^{-1} 	ext{ mol}^{-1} = 8.3 	imes 10^4 	ext{ J mol}^{-1}$.

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