For the reaction shown in the image,the half- | vedclass

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(A) The reaction is a first-order reaction because the half-life is independent of the initial concentration.For a first-order reaction,the rate const

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$\frac{2.303}{10} \log \frac{5}{4} \, 	ext{min}^{-1}$

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(A) The reaction is a first-order reaction because the half-life is independent of the initial concentration.For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{V_{\infty}}{V_{\infty} - V_t}$,where $V_{\infty}$ is the total volume of $N_2$ at completion and $V_t$ is the volume at time $t$.Given: $V_{\infty} = 100 \, 	ext{L}$,$V_t = 20 \, 	ext{L}$,and $t = 10 \, 	ext{min}$.Substituting the values:$k = \frac{2.303}{10} \log \frac{100}{100 - 20}$$k = \frac{2.303}{10} \log \frac{100}{80}$$k = \frac{2.303}{10} \log \frac{5}{4} \, 	ext{min}^{-1}$

For the reaction shown in the image,the half-life does not depend on the concentration of the reactant. After $10 \, \text{min}$,the volume of $N_2$ gas evolved is $20 \, \text{L}$ and after the completion of the reaction,it is $100 \, \text{L}$. Hence,the rate constant is:

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