An organic compound undergoes first order dec | vedclass

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Question

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{1}{t} \ln \frac{a_0}{a_t}$.For $60 \%$ decomposition,$a_t = a_0 - 0.6 a_0

Vedclass · Accepted Answer

$1350$

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(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{1}{t} \ln \frac{a_0}{a_t}$.For $60 \%$ decomposition,$a_t = a_0 - 0.6 a_0 = 0.4 a_0$ and $t_1 = 540 \ s$.$k = \frac{1}{540} \ln \frac{a_0}{0.4 a_0} = \frac{1}{540} \ln \frac{10}{4} = \frac{1}{540} (\ln 10 - \ln 4) = \frac{1}{540} (2.3 - 2 	imes 0.3 	imes 2.3 / 2.3) = \frac{1}{540} (2.3 - 1.38) = \frac{0.92}{540}$.For $90 \%$ decomposition,$a_t = a_0 - 0.9 a_0 = 0.1 a_0$.$t_2 = \frac{1}{k} \ln \frac{a_0}{0.1 a_0} = \frac{1}{k} \ln 10 = \frac{1}{k} 	imes 2.3$.Substituting $k$: $t_2 = \frac{540}{0.92} 	imes 2.3 = \frac{540 	imes 2.3}{0.92} = \frac{1242}{0.92} = 1350 \ s$.

An organic compound undergoes first order decomposition. If the time taken for the $60 \%$ decomposition is $540 \ s$,then the time required for $90 \%$ decomposition will be $.......... \ s$. (Nearest integer). Given : $\ln 10 = 2.3 ; \log 2 = 0.3$

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