Consider the following first order gas phase reaction at constant temperature ( A(g) arrow 2 ~B(~g)+C(g) ) If the total pressure of the gases is found to be 200 torr after (23 sec ). and 300 torr upon the complete

Correct answer: 0 aligned & A(g) arrow 2 ~B(~g)+C(g) & P_ 23 =P_0+2 x=200 & P_ ∞ =3 P_0=300 & P_0=100 & ~K= 1 t P_ ∞ -P_0 P_ ∞ -P_ t & K=(2.3)/(23) (300-100)/(300-200) & =(2.3 × 0.301)/(23)=0.0301=3.01 × 10^ -2 sec^ -1 & aligned

JEE Main 20249 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2024 · 9 April · Evening Shift · Chemistry · Chemical Kinetics

Consider the following first order gas phase reaction at constant...

Consider the following first order gas phase reaction at constant temperature (

\mathrm{A}(\mathrm{g}) \rightarrow

\mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})

) If the total pressure of the gases is found to be 200 torr after (23

\mathrm{sec}

). and 300 torr upon the complete decomposition of A after a very long time, then the rate constant of the given reaction is ______ (

\times

10^{-2}

\mathrm{~s}^{-1}

) (nearest integer) [Given : (\log _{10}(2)=0.301)]

Correct answer

Step-by-step solution

\begin{aligned} & \mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \\ & \mathrm{P}_{23}=\mathrm{P}_0+2 \mathrm{x}=200 \\ & \mathrm{P}_{\infty}=3 \mathrm{P}_0=300 \\ & \mathrm{P}_0=100 \\ & \mathrm{~K}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{P}_{\infty}-\mathrm{P}_0}{\mathrm{P}_{\infty}-\mathrm{P}_{\mathrm{t}}} \\ & \mathrm{K}=\frac{2.3}{23} \log \frac{300-100}{300-200} \\ & =\frac{2.3 \times 0.301}{23}=0.0301=3.01 \times 10^{-2} \mathrm{sec}^{-1} \\ & \end{aligned}

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