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Chemistry Question 56 – JEE-MAIN 2026

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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 Kp 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.

● Video Walkthrough
Step 1: Determine Equilibrium Moles and Total Moles✦ Active

For the reaction A(g)⇌B(g)+C(g):

Initial moles:a00Equilibrium moles:(a−x)xx

The total moles at equilibrium, ntotal, are the sum of moles of all species:

ntotal=(a−x)+x+x=a+x
Step 2: Express Partial Pressures○ Expand

The partial pressure of each gas (Pi) is given by its mole fraction multiplied by the total pressure (p):

PA=nAntotal×p=a−xa+x×p
PB=nBntotal×p=xa+x×p
PC=nCntotal×p=xa+x×p
💡 Teacher's Secret Hint

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

Step 3: Calculate Kp○ Expand

For the given reaction, the equilibrium constant Kp is expressed as:

Kp=PB×PCPA

Substitute the partial pressure expressions into the Kp equation:

Kp=(xa+x×p)×(xa+x×p)(a−xa+x×p)
Kp=x2(a+x)2×p2a−xa+x×p=x2×p2(a+x)2×a+x(a−x)p
Kp=x2×p(a+x)(a−x)=x2×pa2−x2

This expression matches option 2.

💡 Teacher's Secret Hint

Simplify the algebraic expression carefully to avoid errors.

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