Reaction $\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$ is a first…

Reaction $\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$ is a first order reaction. It was started with pure A

Which of the following option is incorrect?
  1. Initial pressure of A is 80 mm Hg
  2. The reaction never goes to completion
  3. Rate constant of the reaction is $1.693 \mathrm{~min}^{-1}$
  4. Partial pressure of A after 10 minute is 40 mm Hg

Solution

$\mathrm{A}(\mathrm{g}) \longrightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$
$\begin{array}{llll}\mathrm{t}=0 & \mathrm{P}_0 & & \\ \mathrm{t} \rightarrow \infty & 0 & 2 \mathrm{P}_0 & \mathrm{P}_0\end{array}$
$\begin{aligned} \mathrm{P}_{\infty}=3 \mathrm{P}_0 & =240 \\ \mathrm{P}_0 & =80 \mathrm{~mm} \text { of } \mathrm{Hg}\end{aligned}$
$\mathrm{Kt}=\ln \left(\frac{\mathrm{P}_{\infty}-\mathrm{P}_0}{\mathrm{P}_{\infty}-\mathrm{Pt}}\right)$
$K \times 10=\ln \left(\frac{240-80}{240-160}\right)$
$\mathrm{K}=\frac{\ell \mathrm{n} 2}{10}=0.0693 \mathrm{~min}^{-1}$
Option (3) is incorrect

Asked in: JEE Main 2025 (07 Apr Shift 1)

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