Chemical Equilibrium — JEE Main practice

15 questions

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Sample questions with solutions

Q1 · 2026

Consider the general reaction given below at 400 K

$xA(g)\rightleftharpoons yB(g).$

The values of $K_p$ and $K_c$ are studied under the same condition of temperature but variation in $x$ and $y$ .

(i) $K_p=85.87$ and $K_c=2.586$ appropriate units

(ii) $K_p=0.862$ and $K_c=28.62$ appropriate units

The values of $x$ and $y$ in (i) and (ii) respectively are :

A.
(i): 1,2 (ii): 2,1
B.
(i): 1,3 (ii): 2,1
C.
(i): 3,1 (ii): 3,1
D.
(i): 4,1 (ii): 4,1

Answer: A

Use the relation between $K_p$ and $K_c$ , which depends on the change in gas moles $\Delta n_g = y-x$ , so first determine the sign of $\Delta n_g$ for case (i). Given $K_p>K_c$ in case (i), and $K_p=K_c(RT)^{\Delta n_g}$ , this requires $\Delta n_g>0$ (positive exponent increases the value when $RT>1$ ).
Solve for $\Delta n_g$ numerically using the given values and $RT=0.0821\times400\approx32.84$ . Given $85.87=2.586\times(32.84)^{\Delta n_g}$ , solving gives $\Delta n_g\approx1$ , so $y-x=1$ .
Choose the simplest whole numbers satisfying $y=x+1$ : take $x=1,y=2$ .
For case (ii), $K_p<K_c$ implies $\Delta n_g<0$ . Given $0.862=28.62\times(32.84)^{\Delta n_g}$ , solving gives $\Delta n_g\approx-1$ , so $x-y=1$ .
Choose the simplest whole numbers satisfying $x=y+1$ : take $x=2,y=1$ .

Hence, the values are (i) 1,2 and (ii) 2,1, so the answer is option A.

Q2 · 2026

Consider the following gaseous equilibrium in a closed container of volume 'V' at $\mathrm{T}(\mathrm{K})$ .

$\mathrm{P}_2(\mathrm{g})+\mathrm{Q}_2(\mathrm{g})\rightleftharpoons2\mathrm{PQ}(\mathrm{g})$

2 moles each of $\mathrm{P}_2(\mathrm{g}), \mathrm{Q}_2(\mathrm{g})$ and $\mathrm{PQ}(\mathrm{g})$ are present at equilibrium. Now one mole each of 'P₂' and 'Q₂' are added to the equilibrium keeping the temperature at $\mathrm{T}(\mathrm{K})$ . The number of moles of $\mathrm{P}_2, \mathrm{Q}_2$ and PQ at the new equilibrium, respectively, are

A.
2.56, 1.62, 2.24
B.
2.67, 2.67, 2.67
C.
1.21, 2.24, 1.56
D.
1.66, 1.66, 1.66

Answer: B

Since concentration is moles divided by the fixed volume V, and V appears equally in numerator and denominator of the $K_c$ expression, it cancels out, allowing us to work directly with mole numbers. Given the initial equilibrium has 2 mol each of P₂, Q₂, PQ,

K_c = \frac{n_{PQ}^2}{n_{P_2}n_{Q_2}} = \frac{2^2}{2\times2} = 1

After adding 1 mole each of P₂ and Q₂, the new starting point is 3 mol P₂, 3 mol Q₂, 2 mol PQ. Let the reaction proceed forward by extent $x$ . Given this,

n_{P_2}=3-x, \quad n_{Q_2}=3-x, \quad n_{PQ}=2+2x

Apply $K_c=1$ to this new equilibrium and solve for $x$ . Given

1 = \frac{(2+2x)^2}{(3-x)^2} \quad\Rightarrow\quad 2+2x = 3-x \quad\Rightarrow\quad x=\frac{1}{3}

Substitute $x=1/3$ back to find each mole amount at the new equilibrium. Hence, $n_{P_2}=n_{Q_2}=3-\frac{1}{3}=\frac{8}{3}=2.67, \qquad n_{PQ}=2+\frac{2}{3}=\frac{8}{3}=2.67$

Hence, the new equilibrium has 2.67 mol of each species, so the answer is option B.

Q3 · 2026

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

A.
0.742, 0.557.
B.
0.557, 0.418.
C.
0.53, 0.4.
D.
0.367, 0.275.

Answer: B

Calculate the equilibrium constant from the initial equilibrium data, since this remains constant when more A is added (temperature unchanged). Given $[A]=0.5$ M, $[B]=0.375$ M,

K_{eq} = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75

After adding 0.1 mol of A (in a 1 L flask, so concentration increases by 0.1 M), the new starting point before re-equilibrating is $[A]=0.6$ M, $[B]=0.375$ M. Let $x$ be the amount converted to re-reach equilibrium. Given this, at new equilibrium: A = $0.6-x$ , B = $0.375+x$ .
Set up the equilibrium expression using the same $K_{eq}=0.75$ and solve for $x$ . Given

0.75 = \frac{0.375+x}{0.6-x}

$0.45-0.75x = 0.375+x \quad\Rightarrow\quad 1.75x = 0.075 \quad\Rightarrow\quad x = 0.043$

Compute the new equilibrium concentrations using this value of $x$ . Hence, $[A] = 0.6-0.043 = 0.557\ \text{M}, \qquad [B] = 0.375+0.043 = 0.418\ \text{M}$

Hence, the new equilibrium concentrations are 0.557 M and 0.418 M, so the answer is option B.

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