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Chemistry Question 57 – JEE-MAIN 2025

A(g)→B(g)+C(g) is a first order reaction. Timet∞PsystemPtP∞ The reaction was started with reactant A only. Which of the following expression is correct for rate constant k?

Relate the partial pressures of reactants and products to the total system pressure at different times.

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Ninja Strategy•Initial Condition Check

For the integrated rate law k=1tln⁡[A]0[A]t, at t=0, the argument of the logarithm must be 1 (since [A]t=[A]0), making ln⁡(1)=0. Only option 2 satisfies this condition when Pt=PA0 and P∞=2PA0 at t=0.

Step 1: Relate Pressures to Reaction Progress✦ Active

For the reaction A(g)→B(g)+C(g), if PA0 is the initial partial pressure of A, then at time t, the partial pressures are PA=PA0−x, PB=x, and PC=x. The total pressure at time t is Pt=PA0+x. At t=∞, all A is consumed, so PA=0, PB=PA0, PC=PA0. The total pressure at t=∞ is P∞=2PA0.

PA0=P∞2 x=Pt−PA0=Pt−P∞2 PA=PA0−x=P∞2−(Pt−P∞2)=P∞−Pt
Step 2: Apply First-Order Integrated Rate Law○ Expand

The integrated rate law for a first-order reaction in terms of partial pressures is k=1tln⁡PA0PA.

Step 3: Substitute and Simplify○ Expand

Substitute the expressions for PA0 and PA derived in Step 1 into the rate law.

k=1tln⁡P∞2P∞−Pt k=1tln⁡P∞2(P∞−Pt)

This matches option 2.

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation, especially with fractions and signs.

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