The rate for a first order reaction is 0.6932 | vedclass

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(B) For a first order reaction,the rate law is given by $r = k[A]$. Given the rate $r = 0.6932 	imes 10^{-2} \, mol \, L^{-1} \, min^{-1}$ and initia

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) For a first order reaction,the rate law is given by $r = k[A]$. Given the rate $r = 0.6932 	imes 10^{-2} \, mol \, L^{-1} \, min^{-1}$ and initial concentration $[A] = 1 \, M$. Calculating the rate constant $k$: $k = \frac{r}{[A]} = \frac{0.6932 	imes 10^{-2} \, mol \, L^{-1} \, min^{-1}}{1 \, mol \, L^{-1}} = 0.6932 	imes 10^{-2} \, min^{-1}$. For a first order reaction,the half-life $T_{1/2}$ is calculated as: $T_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.6932 	imes 10^{-2}} \approx 100 \, min$.

The rate for a first order reaction is $0.6932 \times 10^{-2} \, mol \, L^{-1} \, min^{-1}$ and the initial concentration of the reactants is $1 \, M$. The half-life $T_{1/2}$ is equal to ........ $min$.

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