Chemistry Q56 – JEE-MAIN 2026

The reaction

A_{(g)} ⇌ B_{(g)} + C_{(g)}

was initiated with the amount 'a' of

A_{(g)}

. At equilibrium it is found that the amount of

A_{(g)}

remaining is

(a - x)

at a total pressure of

p

. The equilibrium constant

K_{p}

of the reaction can be calculated from the expression :

Determine the moles of each species at equilibrium based on the initial amount and the amount reacted.

Step 1: Determine Equilibrium Moles and Total Moles✦ Active

For the reaction $A_{(g)} ⇌ B_{(g)} + C_{(g)}$ :

\begin{array}{lccc} Initial moles: & a & 0 & 0 \\ Equilibrium moles: & (a - x) & x & x \end{array}

The total moles at equilibrium, $n_{t o t a l}$ , are the sum of moles of all species:

n_{t o t a l} = (a - x) + x + x = a + x

Step 2: Express Partial Pressures○ Expand

The partial pressure of each gas ( $P_{i}$ ) is given by its mole fraction multiplied by the total pressure ( $p$ ):

P_{A} = \frac{n_{A}}{n_{t o t a l}} \times p = \frac{a - x}{a + x} \times p

P_{B} = \frac{n_{B}}{n_{t o t a l}} \times p = \frac{x}{a + x} \times p

P_{C} = \frac{n_{C}}{n_{t o t a l}} \times p = \frac{x}{a + x} \times p

💡 Teacher's Secret Hint

Remember that partial pressure is directly proportional to the mole fraction.

Step 3: Calculate

K_{p}

○ Expand

For the given reaction, the equilibrium constant $K_{p}$ is expressed as:

K_{p} = \frac{P_{B} \times P_{C}}{P_{A}}

Substitute the partial pressure expressions into the $K_{p}$ equation:

K_{p} = \frac{(\frac{x}{a + x} \times p) \times (\frac{x}{a + x} \times p)}{(\frac{a - x}{a + x} \times p)}

K_{p} = \frac{\frac{x^{2}}{(a + x)^{2}} \times p^{2}}{\frac{a - x}{a + x} \times p} = \frac{x^{2} \times p^{2}}{(a + x)^{2}} \times \frac{a + x}{(a - x) p}

K_{p} = \frac{x^{2} \times p}{(a + x) (a - x)} = \frac{x^{2} \times p}{a^{2} - x^{2}}

This expression matches option 2.

💡 Teacher's Secret Hint

Simplify the algebraic expression carefully to avoid errors.

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Chemistry Question 56 – JEE-MAIN 2026