For a reaction,r = k[A]. If 50 % of A is deco | vedclass

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(B) The rate law $r = k[A]$ indicates that the reaction is of the first order.For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.

Vedclass · Accepted Answer

$399$

Vedclass · Answer

(B) The rate law $r = k[A]$ indicates that the reaction is of the first order.For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k} = 120 \ 	ext{min}$.Thus,$k = \frac{0.693}{120} \ 	ext{min}^{-1}$.For $90 \%$ decomposition,the amount remaining is $100 - 90 = 10 \%$.The time $t$ is given by the formula $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} ight)$.Substituting the values: $t = \frac{2.303}{(0.693 / 120)} \log \left( \frac{100}{10} ight)$.$t = \frac{2.303 	imes 120}{0.693} 	imes \log(10)$.Since $\frac{2.303}{0.693} \approx \frac{1}{0.301} \approx 3.32$,we get $t = 120 	imes 3.32 \approx 399 \ 	ext{minutes}$.

For a reaction,$r = k[A]$. If $50 \%$ of $A$ is decomposed in $120 \ \text{minutes}$,the time taken for $90 \%$ decomposition of $A$ is $............ \ \text{minutes}$.

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