The reaction A(g) \rightleftharpoons 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 :

\frac{x^{2}}{a^{2} - x^{2}} \times p

\text{ }\frac{x^{2}}{a^{2} + x^{2}} \times p

\frac{a + x^{2}}{x^{2}} \times p

\frac{a^{2} - x^{2}}{x^{2}} \times p

JEE Main · Chemistry · Chapter 6

Equilibrium JEE Main PYQs (2020–2026)

85 questions from Equilibrium have appeared in JEE Main between 2020 and 2026, including 59 in the last five papers (2022–2026). The most-asked topic is Applications of Equilibrium Constants & Degree of Dissociation (40 questions). Practise them all with instant answers and step-by-step solutions.

Practise all 85 questions

Check answers, read solutions, filter by year & topic, get a report.

Start free practice

Free · sign in with Google

Questions

Years covered

Topics

Numerical

Solved Equilibrium PYQs

3 recent questions with answers. The full set of 85 is free to practise.

JEE Main 2026EasyEquilibrium Constants (Kc, Kp) and their Relationship
Q1
Consider the following reactions in which all the reactants and products are present in gaseous state

$2 x y ⇌ x_{2} + y_{2} K_{1} = 2.5 \times 1 0^{5} x y + \frac{1}{2} z_{2} ⇌ x yz K_{2} = 5 \times 1 0^{- 3}$

The value of $K_{3}$ for the equilibrium $\frac{1}{2} x_{2} + \frac{1}{2} y_{2} + \frac{1}{2} z_{2} ⇌ x yz$ is :
- A $2.5 \times 1 0^{- 3}$
- B $2.5 \times 1 0^{3}$
- C $1.0 \times 1 0^{- 5}$
- D $5 \times 1 0^{- 3}$
Show answer & solution
Correct answer: (C)
$x_{2} + y_{2} ⇌ 2 x y, K^{'} = \frac{1}{K _{1}} = \frac{1}{2.5 \times 1 0 ^{5}} = \frac{1}{25 \times 1 0 ^{4}}$

$\frac{1}{2} x_{2} + \frac{1}{2} y_{2} ⇌ x y, K^{^{''}} = (\frac{1}{25 \times 1 0 ^{4}})^{1/2} = \frac{1}{5 \times 1 0 ^{2}} \dots\dots (1)$

$x y + \frac{1}{2} z_{2} ⇌ x yz, K_{2} = 5 \times 1 0^{- 3}$ …….(2)

On adding equation (1) & (2)

$\frac{1}{2} x_{2} + \frac{1}{2} y_{2} + \frac{1}{2} z_{2} ⇌ x yz K_{3} = \frac{1}{5 \times 1 0 ^{2}} \times 5 \times 1 0^{- 3} = 1 \times 1 0^{- 5}$
JEE Main 2026MediumApplications of Equilibrium Constants & Degree of Dissociation
Q2
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 :
- A $\frac{x ^{2}}{a ^{2} + x ^{2}} \times p$
- B $\frac{x ^{2}}{a ^{2} - x ^{2}} \times p$
- C $\frac{a + x ^{2}}{x ^{2}} \times p$
- D $\frac{a ^{2} - x ^{2}}{x ^{2}} \times p$
Show answer & solution
Correct answer: (B)
For the reaction

$A (g) ⇌ B (g) + C (g)$

initial amount of $A = a$ .

At equilibrium, amount of $A$ left $= a - x$ .

So, amount of $A$ dissociated $= x$ .

Hence at equilibrium:

$A = a - x$

$B = x$

$C = x$

Now total moles at equilibrium:

$(a - x) + x + x = a + x$

Since total pressure is $p$ , the partial pressures are:

$P_{A} = \frac{a - x}{a + x} p$

$P_{B} = \frac{x}{a + x} p$

$P_{C} = \frac{x}{a + x} p$

Now,

$K_{p} = \frac{P _{B} \cdot P _{C}}{P _{A}}$

Substitute the values:

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

Simplifying,

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

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

Using

$(a + x) (a - x) = a^{2} - x^{2}$

we get

$K_{p} = \frac{x ^{2}}{a ^{2} - x ^{2}} p$

So the correct expression is:

$K_{p} = \frac{x ^{2}}{a ^{2} - x ^{2}} \times p$

Therefore, the correct option is:

$Option B$
JEE Main 2026MediumReaction Quotient (Qc, Qp) and Direction of Reaction
Q3
Observe the following equilibrium in a 1 L flask.

A(g) ⇌ B(g)

At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively
- A0.742, 0.557.
- B0.367, 0.275.
- C0.53, 0.4.
- D0.557, 0.418.
Show answer & solution
Correct answer: (D)
$A ⇌ B 0.5 M 0.375 M (At equilibrium) K_{e q} = \frac{[ B ] _{e q}}{[ A ] _{e q}} = \frac{0.375}{0.5} = 0.75$

Now 0.1 mole of A is added so reaction will move in forward direction.

$A ⇌ B 0.6 - x 0.375 + x K_{e q} = 0.75 = \frac{0.375 + x}{0.6 - x} 0.45 - 0.75 x = 0.375 + x 1.75 x = 0.075 X = \frac{0.075}{1.75} = \frac{3}{70} = 0.043$

Moles of $A = 0.043 = 0.557$

Moles of $B = 0.418$