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Chemistry · Chemical Kinetics

JEE Main 2024 — 9 April, Shift 2 — Question 77

Consider the following first order gas phase reaction at constant temperature A(g)→2 B( g)+C(g)\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) If the total pressure of the gases is found to be 200 torr after 23 sec . and 300 torr upon the complete decomposition of A after a very long time, then the rate constant of the given reaction is \qquad ×10−2 s−1\times 10^{-2} \mathrm{~s}^{-1} (nearest integer) [Given : log⁡10(2)=0.301]\left.\log _{10}(2)=0.301\right]

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

A(g)→2 B( g)+C(g)\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) P23=P0+2x=200P_{23}=P_{0}+2 x=200 P∞=3P0=300\mathrm{P}_{\infty}=3 \mathrm{P}_{0}=300 P0=100\mathrm{P}_{0}=100 K=1tln⁡P∞−P0P∞−PtK=\frac{1}{t} \ln \frac{P_{\infty}-P_{0}}{P_{\infty}-P_{t}}

K=2.323log⁡300−100300−200\mathrm{K}=\frac{2.3}{23} \log \frac{300-100}{300-200} =2.3×0.30123=0.0301=3.01×10−2sec−1=\frac{2.3 \times 0.301}{23}=0.0301=3.01 \times 10^{-2} \mathrm{sec}^{-1}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Integrated Rate Laws