A reactant undergoes 90% decomposition in 366 | vedclass

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(B) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given $t = 366 	ext{ min}$ and $[A]_

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$366 \left( \frac{\ln 2}{\ln 10} \right)$

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(B) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given $t = 366 	ext{ min}$ and $[A]_t = 10\% 	ext{ of } [A]_0$,so $\frac{[A]_0}{[A]_t} = \frac{100}{10} = 10$. $K = \frac{2.303}{366} \log 10 = \frac{2.303}{366}$. The half-life $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{K} = \frac{\ln 2}{K}$. Substituting $K$: $t_{1/2} = \frac{\ln 2}{(2.303 / 366)} = \frac{366 	imes \ln 2}{2.303}$. Since $2.303 = \ln 10$,we get $t_{1/2} = 366 \left( \frac{\ln 2}{\ln 10} ight)$.

Time	$t$	$\infty$
$P_{\text{system}}$	$P_t$	$P_{\infty}$

$A$ reactant undergoes $90\%$ decomposition in $366 \text{ minutes}$. What is the half-life of this first-order reaction?

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