JEE Main 2026 Apr 5 Shift 2, Chemistry Q6: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2026Apr 5, Shift 2Chemistry

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:

A
$\frac{x ^{2}}{a ^{2} + x ^{2}} \cdot p$
B
$\frac{x ^{2}}{a ^{2} - x ^{2}} \cdot p$
C
$\frac{a + x ^{2}}{x ^{2}} \cdot p$
D
$\frac{a - x ^{2}}{x ^{2}} \cdot p$

Solution

At equilibrium, moles of $A = a - x$ , $B = x$ , $C = x$ . Total moles $n_{T} = a + x$ .

Mole fractions: $χ_{A} = \frac{a - x}{a + x}$ , $χ_{B} = χ_{C} = \frac{x}{a + x}$ .

Partial pressures: $p_{A} = χ_{A} p$ , etc.

$K_{p} = \frac{p _{B} \cdot p _{C}}{p _{A}} = \frac{x ^{2}}{( a + x ) ( a - x )} \cdot p = \frac{x ^{2}}{a ^{2} - x ^{2}} \cdot p$ .

Concept behind this question

Law of Chemical Equilibrium And Equilibrium Constant Notes, formulas and examples →

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Apr 5 Shift 2, Chemistry Q6: Law of Chemical Equilibrium And Equilibrium Constant

Concept behind this question

More previous year questions on Law of Chemical Equilibrium And Equilibrium Constant

Practice more Chemistry