The gaseous reaction $\mathrm{A}(\mathrm{g}) ightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$ is…

The gaseous reaction $\mathrm{A}(\mathrm{g}) ightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$ is found to be first order. If the reaction is started with $p_{\mathrm{A}}=90$ $\mathrm{mm} \mathrm{Hg}$, the total pressure after $10 \mathrm{~min}$ is found to be $180 \mathrm{mmHg}$. The rate constant of the reaction is
  1. $1.15 \times 10^{-3} \mathrm{~s}^{-1}$
  2. $2.30 \times 10^{-3} \mathrm{~s}^{-1}$
  3. $3.45 \times 10^{-3} \mathrm{~s}^{-1}$
  4. $4.60 \times 10^{-3} \mathrm{~s}^{-1}$

Solution

$\mathrm{A}(\mathrm{g}) ightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$
$p_{0}-p \quad 2 p \quad p$
Total pressure $=\left(p_{0}-pight)+(2 p)+p=p_{0}+2 p$
i.e. $\quad 180 \mathrm{mmHg}=90 \mathrm{mmHg}+2 p \quad$ or $\quad p=45 \mathrm{mmHg}$
Now $\quad \log \frac{p_{0}-p}{p_{0}}=-\frac{k}{2.303} t$
$$
\log \frac{45}{90}=-\frac{k}{2.303}(10 \times 60 \mathrm{~s}) . \text { Thus } \quad k=\frac{0.301 \times 2.303}{10 \times 60 \mathrm{~s}}=1.155 \times 10^{-3} \mathrm{~s}^{-1}
$$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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