Two first order reactions have half-lives in | vedclass

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(A) For a first order reaction,$k = \frac{0.693}{t_{1/2}}$. Given $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{3}{2}$,so $\frac{k_2}{k_1} = \frac{3}{2}$.F

Vedclass · Accepted Answer

$0.3$

Vedclass · Answer

(A) For a first order reaction,$k = \frac{0.693}{t_{1/2}}$. Given $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{3}{2}$,so $\frac{k_2}{k_1} = \frac{3}{2}$.For the first reaction,$t_1 = \frac{2.303}{k_1} \log \left( \frac{100}{100 - 25} ight) = \frac{2.303}{k_1} \log \left( \frac{4}{3} ight)$.For the second reaction,$t_2 = \frac{2.303}{k_2} \log \left( \frac{100}{100 - 75} ight) = \frac{2.303}{k_2} \log (4)$.Taking the ratio: $\frac{t_1}{t_2} = \frac{k_2}{k_1} 	imes \frac{\log(4/3)}{\log(4)} = \frac{3}{2} 	imes \frac{0.602 - 0.477}{0.602} = \frac{3}{2} 	imes \frac{0.125}{0.602} \approx 1.5 	imes 0.2076 \approx 0.311$.Thus,the ratio is approximately $0.3 : 1$.

Two first order reactions have half-lives in the ratio $3 : 2$. Calculate the ratio of time intervals $t_1 : t_2$ if $t_1$ is the time period for $25\%$ completion of the first reaction and $t_2$ for $75\%$ completion of the second reaction. (in $: 1$)

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