Q6 / 75 · JEE Main 2025 (7 Apr Shift 2) · Chemistry
$\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ is a first order reaction.
$\begin{array}{|l|l|l|}\hline \text{Time} & T & \infty \\\hline \mathbf{P}_{\text {system }} & \mathrm{P}_{\mathrm{t}} & \mathrm{P}_{\infty} \\\hline\end{array}$
The reaction was started with reactant A only. Which of the following expression is correct for rate constant k ?
$\begin{array}{|l|l|l|}\hline \text{Time} & T & \infty \\\hline \mathbf{P}_{\text {system }} & \mathrm{P}_{\mathrm{t}} & \mathrm{P}_{\infty} \\\hline\end{array}$
The reaction was started with reactant A only. Which of the following expression is correct for rate constant k ?
- A$\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{2\left(\mathrm{p}_{\infty}-\mathrm{P}_{\mathrm{t}}\right)}{\mathrm{P}_{\mathrm{t}}}$
- B$\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{p}_{\infty}}{\mathrm{P}_{\mathrm{t}}}$
- Correct answerC$\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{p}_{\infty}}{2\left(\mathrm{p}_{\infty}-\mathrm{P}_{\mathrm{t}}\right)}$
- D$\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{p}_{\infty}}{\left(\mathrm{p}_{\infty}-\mathrm{P}_{\mathrm{t}}\right)}$
Solution
$\begin{aligned}<br />& \begin{array}{lllll} <br />& A_{(g)} & \rightarrow & B_{(g)} & + <br />C_{(g)} \\<br>t=0 & P^o & & 0 & 0 \\<br>t=t & P^o-x & & x & x \\<br>t=\infty & 0 & & P^o & P^o<br />\end{array} \\<br>& \mathrm{P}_{\mathrm{t}}=\mathrm{P}^{\mathrm{o}}+\mathrm{x} \Rightarrow \mathrm{x}=\mathrm{P}_{\mathrm{t}}-\mathrm{P}^{\mathrm{o}}=\mathrm{P}_{\mathrm{t}}-\frac{\mathrm{P}_{\infty}}{2} \\<br>& \mathrm{P}_{\infty}=2 \mathrm{P}^{\mathrm{o}} \Rightarrow \mathrm{P}^0=\frac{\mathrm{P} \infty}{2} \\<br>& \mathrm{k}=\frac{1}{\mathrm{t}} \ell \ln \frac{\mathrm{P}^{\mathrm{o}}}{\mathrm{P}^{\mathrm{o}}-\mathrm{x}} \\<br>& k=\frac{1}{t} \ell n \frac{P_{\infty}}{2\left(P_{\infty}-P_t\right)}<br />\end{aligned}$
Hence, the correct answer is option: 'C'.