The decomposition of N_2O_5 is a first order | vedclass

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(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{1}{t} \ln \left( \frac{V_{\infty}}{V_{\infty} - V_t} \right)$.Here,$V_{\in

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$\frac{1}{15} \ln \frac{35}{26}$

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(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{1}{t} \ln \left( \frac{V_{\infty}}{V_{\infty} - V_t} \right)$.Here,$V_{\infty}$ is the total volume of $O_2$ produced at the end of the reaction $(35 \ mL)$ and $V_t$ is the volume of $O_2$ produced at time $t = 15 \ min$ $(9 \ mL)$.Substituting the values: $k = \frac{1}{15} \ln \left( \frac{35}{35 - 9} \right) = \frac{1}{15} \ln \left( \frac{35}{26} \right)$.

The decomposition of $N_2O_5$ is a first order reaction represented by $N_2O_5 \to N_2O_4 + \frac{1}{2}O_2$. After $15 \ min$,the volume of $O_2$ produced is $9 \ mL$ and at the end of the reaction,it is $35 \ mL$. The rate constant is equal to:

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