At 30^{circ}C,the half-life for the decomposi | vedclass

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(C) Since the half-life is independent of the initial concentration,it is a $1^{st}$ order reaction.$k = \frac{0.693}{T_{1/2}} = \frac{0.693}{200} = 3

Vedclass · Accepted Answer

$467$

Vedclass · Answer

(C) Since the half-life is independent of the initial concentration,it is a $1^{st}$ order reaction.$k = \frac{0.693}{T_{1/2}} = \frac{0.693}{200} = 3.465 	imes 10^{-3} \, s^{-1}$.For $80\%$ decomposition,the remaining concentration is $20\%$ of the initial concentration $(A = 0.2 A_{0})$.The rate equation is $t = \frac{2.303}{k} \log \frac{A_{0}}{A}$.$t = \frac{2.303}{3.465 	imes 10^{-3}} \log \frac{A_{0}}{0.2 A_{0}} = \frac{2.303}{3.465 	imes 10^{-3}} \log 5$.Using $\log 5 = 0.70$ and $\frac{2.303}{k} = \frac{200}{0.693} \approx 288.66$.$t = 288.66 	imes 0.70 \approx 202.06 	imes 2.303 \approx 466.67 \, s \approx 467 \, s$.

At $30^{\circ}C$,the half-life for the decomposition of $AB_{2}$ is $200\,s$ and is independent of the initial concentration of $AB_{2}$. The time required for $80\%$ of the $AB_{2}$ to decompose is $....s$ (Given: $\log 2 = 0.30; \log 5 = 0.70$)

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