Q22 / 75 · JEE Main 2026 (02 April Shift 2) · Chemistry
Consider the following gas phase reaction being carried out in a closed vessel at $25°C$.
$2A(g) \rightarrow 4B(g) + C(g)$
time (min) total pressure of the system (mm Hg) $30$ $300$ $\infty$ $600$
The pressure of $C(g)$ at $30$ minutes time interval would be _______ mm Hg. (nearest integer)
$2A(g) \rightarrow 4B(g) + C(g)$
| time (min) | total pressure of the system (mm Hg) |
|---|---|
| $30$ | $300$ |
| $\infty$ | $600$ |
Correct answer: 20
Solution
For the given gas phase reaction:
$2A(g) \rightarrow 4B(g) + C(g)$
Let the initial pressure of $A(g)$ be $P_0$.
At $t = 0$: $P_A = P_0$, $P_B = 0$, $P_C = 0$
At $t = t$: $P_A = P_0 - 2p$, $P_B = 4p$, $P_C = p$
Total pressure at time $t$, $P_t = P_0 - 2p + 4p + p = P_0 + 3p$
At $t = \infty$, the reaction is complete, so $P_A = 0 \Rightarrow 2p = P_0 \Rightarrow p = \dfrac{P_0}{2}$
Total pressure at $t = \infty$, $P_{\infty} = 4\left(\dfrac{P_0}{2}\right) + \dfrac{P_0}{2} = \dfrac{5P_0}{2}$
Given $P_{\infty} = 600$ mm Hg:
$\dfrac{5P_0}{2} = 600 \Rightarrow P_0 = 240$ mm Hg
At $t = 30$ min, $P_t = 300$ mm Hg:
$P_0 + 3p = 300$
$240 + 3p = 300 \Rightarrow 3p = 60 \Rightarrow p = 20$ mm Hg
The pressure of $C(g)$ at $30$ minutes is $p = 20$ mm Hg.
Answer: $20$
$2A(g) \rightarrow 4B(g) + C(g)$
Let the initial pressure of $A(g)$ be $P_0$.
At $t = 0$: $P_A = P_0$, $P_B = 0$, $P_C = 0$
At $t = t$: $P_A = P_0 - 2p$, $P_B = 4p$, $P_C = p$
Total pressure at time $t$, $P_t = P_0 - 2p + 4p + p = P_0 + 3p$
At $t = \infty$, the reaction is complete, so $P_A = 0 \Rightarrow 2p = P_0 \Rightarrow p = \dfrac{P_0}{2}$
Total pressure at $t = \infty$, $P_{\infty} = 4\left(\dfrac{P_0}{2}\right) + \dfrac{P_0}{2} = \dfrac{5P_0}{2}$
Given $P_{\infty} = 600$ mm Hg:
$\dfrac{5P_0}{2} = 600 \Rightarrow P_0 = 240$ mm Hg
At $t = 30$ min, $P_t = 300$ mm Hg:
$P_0 + 3p = 300$
$240 + 3p = 300 \Rightarrow 3p = 60 \Rightarrow p = 20$ mm Hg
The pressure of $C(g)$ at $30$ minutes is $p = 20$ mm Hg.
Answer: $20$
Hence, the correct answer is 20.