The first order reaction $\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+2…
The first order reaction $\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+2 \mathrm{C}(\mathrm{g})$ occurs at $25^{\circ} \mathrm{C}$. After 24 minutes the ratio of the concentration of products to the concentration of the reactant is $1: 3$. What is the half life of the reaction $($ in $\min ) ?(\log 1.11=0.046)$
$150.5$
$142.2$
$157.8$
$15.78$
Solution
$\begin{array}{llll} & \mathrm{A}(\mathrm{g}) & \rightarrow \mathrm{B}(\mathrm{g})+2 \mathrm{C}(\mathrm{g}) \\ \mathrm{t}=0 \text { min } & 100 & & \\ \mathrm{t}=24 \min & 100-\mathrm{x} & \qquad \mathrm{x} \qquad \quad 2 \mathrm{x}\end{array}$
$\begin{aligned}
& \frac{\text { Amount of product }}{\text { Amount of reactant }}=\frac{1}{3}=\frac{x+2 x}{100-x} \\
& \therefore x=10
\end{aligned}$
Hence at $\mathrm{t}=24 \mathrm{~min}$, amount of A left is 90
$\begin{aligned}
& K=\frac{2.303}{24} \log \frac{100}{90} \\
& \therefore K=0.00441 \\
& t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{0.00441}=157.14 \mathrm{~min}
\end{aligned}$