Chemistry · Chemical Kinetics

JEE Advanced 2018 — Paper 2 — Question 20

For a first order reaction $\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$ at constant volume and 300 K , the total pressure at the beginning $(t=0)$ and at time $t$ are $P_{0}$ and $P_{t}$ , respectively. Initially, only $A$ is present with concentration $[A]_{0}$ , and $t_{1 / 3}$ is the time required for the partial pressure of $A$ to reach $1 / 3^{\text {rd }}$ of its initial value. The correct option(s) is (are) (Assume that all these gases behave as ideal gases)

Option A:
Correct
Option B:
Option C:
Option D:
Correct

Answer: A, D

Step-by-step solution

\text{A(g)} \longrightarrow \text{2B(g)} + \text{C(g)}

\text{at t = 0 p}_0 \quad 0 \quad 0

\text{(p}_0 - \text{p}) \quad 2\text{p} \quad \text{p}

\text{p}_t = \text{p}_0 - \text{p} + 2\text{p} + \text{p} = \text{p}_0 + 2\text{p}

\text{p} = \frac{(\text{p}_t - \text{p}_0)}{2}

\text{t} = \frac{1}{\text{k}} \ln \frac{2\text{p}_0}{3\text{p}_0 - \text{p}_t}

\text{kt} = \ln 2\text{p}_0 - \ln (3\text{p}_0 - \text{p}_t)

\ln (3\text{p}_0 - \text{p}_t) = \ln 2\text{p}_0 - \text{kt}

\text{y} = \text{mx} + \text{c}

\text{So, m} = -\text{k}

\text{c} = \ln(2\text{p}_0)

Solution figure

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws

For a first order reaction A ( g ) rightarrow 2 B ( g )+ C ( g ) at… | JEE Advanced 2018 PYQ with Solution · DhiX AI