JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
\(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) 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 ?
- A \(\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{2\left(\mathrm{p}_{\infty}-\mathrm{P}_{\mathrm{t}}\right)}{\mathrm{P}_{\mathrm{t}}}\)
- B \(\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{p}_{\infty}}{\mathrm{P}_{\mathrm{t}}}\)
- C \(\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{p}_{\infty}}{2\left(\mathrm{p}_{\infty}-\mathrm{P}_{\mathrm{t}}\right)}\)
- D \(\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{p}_{\infty}}{\left(\mathrm{p}_{\infty}-\mathrm{P}_{\mathrm{t}}\right)}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{p}_{\infty}}{2\left(\mathrm{p}_{\infty}-\mathrm{P}_{\mathrm{t}}\right)}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \begin{array}{lllll} & A_{(g)} & \rightarrow & B_{(g)} & + C_{(g)} \\ t=0 & P^o & & 0 & 0 \\ t=t & P^o-x & & x & x \\ t=\infty & 0 & & P^o & P^o \end{array} \\ & \mathrm{P}_{\mathrm{t}}=\mathrm{P}^{\mathrm{o}}+\mathrm{x} \Rightarrow…
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