JEE Main 2025 Apr 7 Shift 2, Chemistry Q4: Reactions of Various Orders

JEE Main2025Apr 7, Shift 2Chemistry

A(g) $\to$ B(g) + C(g) is a first order reaction.

Time	T	$\infty$
$P_{sy s t e m}$	$P_{t}$	$P_{\infty}$

The reaction was started with reactant A only. Which of the following expression is correct for rate constant k ?

A
$k = \frac{1}{t} ln \frac{2 ( p _{\infty} - P _{t} )}{P _{t}}$
B
$k = \frac{1}{t} ln \frac{p _{\infty}}{P _{t}}$
C
$k = \frac{1}{t} ln \frac{p _{\infty}}{2 ( p _{\infty} - P _{t} )}$
D
$k = \frac{1}{t} ln \frac{p _{\infty}}{( p _{\infty} - P _{t} )}$

Solution

Let the initial pressure of A be $P^{\circ}$ . Since one mole of A produces two moles of gaseous products, if $x$ is the pressure of A consumed at time $t$ :

$P_{t} = (P^{\circ} - x) + x + x = P^{\circ} + x \Rightarrow x = P_{t} - P^{\circ}$

At $t = \infty$ , all of A is consumed, so $P_{\infty} = 2 P^{\circ}$ , giving $P^{\circ} = \frac{P _{\infty}}{2}$ .

The pressure of A remaining is $P^{\circ} - x = P^{\circ} - (P_{t} - P^{\circ}) = 2 P^{\circ} - P_{t} = P_{\infty} - P_{t}$ .

For a first order reaction, $k = \frac{1}{t} ln \frac{P ^{\circ}}{P ^{\circ} - x} = \frac{1}{t} ln \frac{P _{\infty} /2}{P _{\infty} - P _{t}} = \frac{1}{t} ln \frac{P _{\infty}}{2 ( P _{\infty} - P _{t} )}$ .

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JEE Main 2025 Apr 7 Shift 2, Chemistry Q4: Reactions of Various Orders

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