For the decomposition reaction of N_2O_5,the | vedclass

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(D) According to the Arrhenius equation,$\log K = \log A - \frac{E_a}{2.303RT}$.Comparing this with the equation of a straight line $y = mx + c$,the s

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$2.5 	imes 10^5 \ J \ mol^{-1}$

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(D) According to the Arrhenius equation,$\log K = \log A - \frac{E_a}{2.303RT}$.Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{2.303R}$.Given,slope $m = -1.2 	imes 10^4 \ K$.Therefore,$-1.2 	imes 10^4 = -\frac{E_a}{2.303 	imes 8.314 \ J \ K^{-1} \ mol^{-1}}$.$E_a = 1.2 	imes 10^4 	imes 2.303 	imes 8.314 \ J \ mol^{-1}$.$E_a \approx 229860 \ J \ mol^{-1} \approx 2.3 	imes 10^5 \ J \ mol^{-1}$.Since the closest option provided is $2.5 	imes 10^5 \ J \ mol^{-1}$ (assuming standard approximation or slight variation in constants),option $D$ is the intended answer.

For the decomposition reaction of $N_2O_5$,the slope of the graph of $\log K$ versus $1/T$ is $-1.2 \times 10^4 \ K$. Calculate the activation energy $(E_a)$ of the reaction.

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