JEE Main 2023 — Chemical Equilibrium Question with Solution

Given the reaction:

$${\mathrm{A}}_{(\mathrm{g})}\rightleftharpoons 2{\mathrm{B}}_{(\mathrm{g})}+{\mathrm{C}}_{(\mathrm{g})}$$

At the beginning of the reaction (at time = 0), the total pressure is solely due to A, hence it is 450 mmHg.

As the reaction progresses, let's denote 'x' as the pressure decrease due to the decomposition of A. Correspondingly, the pressure increases by '2x' and 'x' for B and C respectively, following the stoichiometry of the reaction.

At time 't', the total pressure (P(T)) is the sum of the pressures of A, B, and C, which is given to be 720 mmHg.

This gives us the equation:

$$450\text{ mmHg}-x\text{ mmHg (decrease in A’s pressure)}+2x\text{ mmHg (increase in B’s pressure)}+x\text{ mmHg (increase in C’s pressure)}=720\text{ mmHg (total pressure at time t)}$$

Solving for 'x' gives:

$$x=135\text{ mmHg}$$

The fraction of A decomposed would then be this change in pressure divided by the initial pressure:

$$\text{Fraction decomposed}=\frac{x}{\text{Initial pressure}}=\frac{135\text{ mmHg}}{450\text{ mmHg}}=0.3=3\times 10^{-1}$$