JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

Consider the following gas phase reaction being carried out in a closed vessel at $2 5^{\circ}$ C.

$2 A (g) \to 4 B (g) + C (g)$

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)

Your answer

Solution

Let initial pressure of A be $p_{0}$ . After consumption of $2 x$ partial pressure of A:

$P_{A} = p_{0} - 2 x$ , $P_{B} = 4 x$ , $P_{C} = x$ , total $= p_{0} + 3 x$ .

At $t = \infty$ (reaction complete, $2 x = p_{0}$ ): total $= 0 + 2 p_{0} + p_{0} /2 = \frac{5 p _{0}}{2} = 600$ , giving $p_{0} = 240$ mm Hg.

At $t = 30$ min: total $= p_{0} + 3 x = 240 + 3 x = 300$ , so $3 x = 60$ and $x = 20$ .

Therefore $P_{C} = x = 20$ mm Hg.

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