Consider the following gas phase reaction being carried out in a closed vessel at 25 ° C . 2 A ( g ) rightarrow 4 B ( g )+ C ( g ) The pressure of C ( g ) at 30 minutes time interval would be \ \ \ \ mm Hg. (Nearest integer).

Chemistry · Chemical Kinetics

JEE Main 2026 — 2 April, Evening Shift — Question 64

Consider the following gas phase reaction being carried out in a closed vessel at $25^{\circ} \mathrm{C}$ . $2 \mathrm{~A}(\mathrm{~g}) \rightarrow 4 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{~g})$

The pressure of $\mathrm{C}(\mathrm{g})$ at 30 minutes time interval would be $\_\_\_\_$ mm Hg. (Nearest integer).

Answer: 20

Numerical answer — enter this value.

Step-by-step solution

\ce{2A(g) -> 4B(g) + C(g)}

\begin{array}{c c c c} & p^\circ & & \\ t = 30\ \text{min} & p^\circ - 2x & 4x & x \\ t = \infty & - & 2p^\circ & \dfrac{p^\circ}{2} \end{array}

p_\infty = 2p^\circ + \frac{p^\circ}{2} = 600

\frac{5p^\circ}{2} = 600

p^\circ = 240\ \text{mm Hg}

\text{At } t = 30\ \text{min},\quad p = 300 = p^\circ - 2x + 4x + x

300 = 240 + 3x

3x = 60

x = 20

\therefore\ \text{Pressure of } C(g) = 20\ \text{mm Hg}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Rate Laws and Rate Constant