Equilibrium — Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Chemical & Ionic Equilibrium — Kc/Kp manipulation, buffers, pH, solubility product, degree of dissociation and ΔG° — with step-by-step solutions.
Solved JEE Main 2026 questions from the Equilibrium chapter, covering equilibrium constant manipulation, Kp/Kc calculations, buffers and pH, solubility product, degree of dissociation, and standard free energy change, each with a concise step-by-step solution.
Solutions are AI-generated and pending review.
Solution
The target reaction is the original reaction multiplied by $\frac{1}{3}$. When a balanced equation is scaled by a factor $n$, the new equilibrium constant is $K^{n}$.
$$K' = K^{1/3} = \left(2.7 \times 10^{-5}\right)^{1/3}$$Write $2.7 \times 10^{-5} = 27 \times 10^{-6}$:
$$K' = \left(27 \times 10^{-6}\right)^{1/3} = 27^{1/3} \times \left(10^{-6}\right)^{1/3} = 3 \times 10^{-2}$$Answer: D ($3 \times 10^{-2}$)
Solution
This is a basic buffer (weak base $\text{NH}_4\text{OH}$ + its salt $\text{NH}_4\text{Cl}$).
Millimoles:
$$\text{base} = 5 \times 0.1 = 0.5\ \text{mmol}, \qquad \text{salt} = 250 \times 0.1 = 25\ \text{mmol}$$Henderson equation (basic form):
$$\text{pOH} = \text{p}K_b + \log\frac{[\text{salt}]}{[\text{base}]} = 4.74 + \log\frac{25}{0.5} = 4.74 + \log 50$$$$\log 50 = \log\frac{100}{2} = 2 - 0.30 = 1.70$$$$\text{pOH} = 4.74 + 1.70 = 6.44 \implies \text{pH} = 14 - 6.44 = 7.56$$$$\text{pH} = 7.56 = 756 \times 10^{-2}$$Answer: 756
Solution
Let $x$ mol of A convert to B. In the 10 L flask:
$$[A] = \frac{1-x}{10}, \qquad [B] = \frac{x}{10}$$$$K_c = \frac{[B]}{[A]} = \frac{x}{1-x} = 4 \implies x = 0.8\ \text{mol}$$Moles at equilibrium: He $= 1$, B $= 0.8$.
Partial pressures from $p = \dfrac{nRT}{V}$ with $\dfrac{RT}{V} = \dfrac{0.082 \times 400}{10} = 3.28$ atm/mol:
$$p_{\text{He}} = 1 \times 3.28 = 3.28\ \text{atm}$$$$p_{B} = 0.8 \times 3.28 = 2.624\ \text{atm}$$Answer: A (3.28, 2.624)
Solution
Ag$_2$CrO$_4$ ($\text{A}_2\text{B}$ type): $K_{sp} = 4s^3$.
$$32x = 4s_1^3 \implies s_1^3 = 8x \implies s_1 = (8x)^{1/3} = 2\,x^{1/3}$$AgBr ($\text{AB}$ type): $K_{sp} = s^2$.
$$4y = s_2^2 \implies s_2 = \sqrt{4y} = 2\sqrt{y}$$Ratio:
$$\frac{s_1}{s_2} = \frac{2\,x^{1/3}}{2\sqrt{y}} = \frac{\sqrt[3]{x}}{\sqrt{y}}$$Answer: D $\left(\dfrac{\sqrt[3]{x}}{\sqrt{y}}\right)$
Solution
Find moles of acid from the full neutralization:
$$n_{\text{acid}} = 28.4 \times 0.1 = 2.84\ \text{mmol}$$In solution (X): NaOH added $= 14.2 \times 0.1 = 1.42\ \text{mmol}$.
This is exactly half of the acid ($1.42 = \tfrac{1}{2}\times 2.84$), converting half the acid to acetate:
$$\text{salt} = 1.42\ \text{mmol}, \qquad \text{acid left} = 2.84 - 1.42 = 1.42\ \text{mmol}$$Henderson equation — at the half-equivalence point $[\text{salt}] = [\text{acid}]$:
$$\text{pH} = \text{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]} = 4.75 + \log 1 = 4.75$$Answer: B (4.75)
Solution
Take 2 mol N$_2$O$_5$ initially; 50% dissociates $\Rightarrow$ 1 mol reacts.
| Species | N$_2$O$_5$ | N$_2$O$_4$ | O$_2$ |
|---|---|---|---|
| mol at eqm | $1$ | $1$ | $0.5$ |
Total moles $= 2.5$. With $P = 2$ atm, partial pressures $p_i = x_i P$:
$$p_{\text{N}_2\text{O}_5} = \frac{1}{2.5}(2) = 0.8, \quad p_{\text{N}_2\text{O}_4} = 0.8, \quad p_{\text{O}_2} = \frac{0.5}{2.5}(2) = 0.4$$$$K_p = \frac{p_{\text{N}_2\text{O}_4}^2\, p_{\text{O}_2}}{p_{\text{N}_2\text{O}_5}^2} = \frac{(0.8)^2 (0.4)}{(0.8)^2} = 0.4$$$$\log K_p = \log\frac{4}{10} = 2\log 2 - 1 = 0.60 - 1 = -0.40$$$$\Delta G^\circ = -2.303\,RT \log K_p = -2.303 \times 8.314 \times 323 \times (-0.40) \approx 2474\ \text{J mol}^{-1}$$$$T = 50 + 273 = 323\ \text{K}$$Answer: 2474
Solution
The strong acid HCl fixes $[\text{H}^+] \approx 0.1\ \text{M}$, which suppresses the (already very weak) ionization of $H_2A$, so $[H_2A] \approx 0.1\ \text{M}$.
First ionization: $H_2A \rightleftharpoons H^+ + HA^-$
$$K_{a1} = \frac{[H^+][HA^-]}{[H_2A]}$$Solve for $[HA^-]$:
$$[HA^-] = \frac{K_{a1}\,[H_2A]}{[H^+]} = \frac{(8.1 \times 10^{-8})(0.1)}{0.1} = 8.1 \times 10^{-8}\ \text{M}$$(The second ionization, $K_{a2} = 10^{-13}$, is negligible.)
Answer: C ($8.1 \times 10^{-8}\,\text{M}$)
Solution
(a) Before adding NaOH — pure 0.2 M weak acid:
$$[\text{H}^+] = \sqrt{K_a\,C} = \sqrt{5 \times 10^{-4} \times 0.2} = \sqrt{1 \times 10^{-4}} = 10^{-2}\ \text{M}$$$$\text{pH} = 2.0$$(b) After 10 mL of 0.2 M NaOH:
$$n_{\text{acid}} = 20 \times 0.2 = 4\ \text{mmol}, \qquad n_{\text{NaOH}} = 10 \times 0.2 = 2\ \text{mmol}$$Exactly half the acid is neutralized $\Rightarrow$ half-equivalence point, where $[\text{salt}] = [\text{acid}]$:
$$\text{pH} = \text{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]} = 3.3 + \log 1 = 3.3$$Answer: B ((a) 2.0 (b) 3.3)
Solution
Dissociation: $\text{A}_x\text{B}_y \rightleftharpoons x\,\text{A}^{y+} + y\,\text{B}^{x-}$.
With degree of dissociation $\alpha$ (and $\alpha \ll 1$, so $[\text{A}_x\text{B}_y] \approx c$):
$$[\text{A}^{y+}] = xc\alpha, \qquad [\text{B}^{x-}] = yc\alpha$$$$K = \frac{[\text{A}^{y+}]^x\,[\text{B}^{x-}]^y}{[\text{A}_x\text{B}_y]} = \frac{(xc\alpha)^x (yc\alpha)^y}{c} = x^x y^y\, c^{x+y-1}\, \alpha^{x+y}$$Solve for $\alpha$:
$$\alpha^{x+y} = \frac{K}{c^{x+y-1}\,x^x y^y} \implies \alpha = \left(\frac{K}{c^{x+y-1}\,x^x y^y}\right)^{\frac{1}{x+y}}$$Answer: B $\left(\left(\dfrac{K}{c^{x+y-1}x^x y^y}\right)^{\frac{1}{x+y}}\right)$
Solution
Start with 1 mol X$_2$Y$_4$; $\alpha = 0.75$.
| Species | X$_2$Y$_4$ | XY$_2$ |
|---|---|---|
| mol at eqm | $1-\alpha = 0.25$ | $2\alpha = 1.5$ |
Total moles $= 1 + \alpha = 1.75$. With $P = 1$ atm:
$$p_{X_2Y_4} = \frac{0.25}{1.75} = \frac{1}{7}, \qquad p_{XY_2} = \frac{1.5}{1.75} = \frac{6}{7}$$$$K_p = \frac{p_{XY_2}^2}{p_{X_2Y_4}} = \frac{(6/7)^2}{1/7} = \frac{36/49}{1/7} = \frac{36}{7}$$$$\log K_p = \log 36 - \log 7 = (2\log 2 + 2\log 3) - \log 7 = (0.60 + 0.96) - 0.84 = 0.72$$$$\Delta_rG^\ominus = -2.3\,RT\log K_p = -2.3 \times 8.3 \times 600 \times 0.72 \approx -8247\ \text{J mol}^{-1} \approx -8.25\ \text{kJ mol}^{-1}$$Magnitude $\approx 8\ \text{kJ mol}^{-1}$.
Answer: 8
Solution
Molar solubility $s = \dfrac{\text{solubility (g/L)}}{\text{molar mass}} = \dfrac{x}{y}\ \text{mol L}^{-1}$.
Dissociation: $M_3A_2 \rightleftharpoons 3M^{2+} + 2A^{3-}$, so
$$[A^{3-}] = 2s, \qquad [M^{2+}] = 3s$$$$K_{sp} = [M^{2+}]^3[A^{3-}]^2 = (3s)^3 (2s)^2 = 27s^3 \cdot 4s^2 = 108\,s^5$$Required ratio:
$$\frac{[A^{3-}]}{K_{sp}} = \frac{2s}{108\,s^5} = \frac{1}{54\,s^4} = \frac{1}{54}\cdot\frac{1}{(x/y)^4} = \frac{1}{54}\cdot\frac{y^4}{x^4}$$Answer: A $\left(\frac{1}{54} \cdot \frac{y^4}{x^4}\right)$
Solution
Compute milliequivalents (meq) of H$^+$ and OH$^-$ in each.
A. OH$^-$ $= 10 \times 0.2 \times 2 = 4$ meq; H$^+$ $= 25 \times 0.1 = 2.5$ meq. Excess OH$^-$ $= 1.5$ meq in 35 mL:
$$[\text{OH}^-] = \frac{1.5}{35} \approx 0.043 \implies \text{pOH} \approx 1.37,\ \text{pH} \approx 12.6\ (\text{strongly basic})$$B. H$^+$ $= 10 \times 0.01 \times 2 = 0.2$ meq; OH$^-$ $= 10 \times 0.01 \times 2 = 0.2$ meq. Exactly neutralized $\Rightarrow \text{pH} = 7$.
C. H$^+$ $= 10 \times 0.1 \times 2 = 2$ meq; OH$^-$ $= 10 \times 0.1 = 1$ meq. Excess H$^+$ $= 1$ meq in 20 mL:
$$[\text{H}^+] = \frac{1}{20} = 0.05 \implies \text{pH} \approx 1.3\ (\text{acidic})$$Increasing pH: $\text{C}\ (1.3) < \text{B}\ (7) < \text{A}\ (12.6)$.
Answer: C ($C < B < A$)
Solution
The van’t Hoff equation in log form:
$$\log_{10} K_p = -\frac{\Delta H^\circ}{2.303\,R}\cdot\frac{1}{T} + \text{constant}$$So the plot of $\log_{10}K_p$ vs $\frac{1}{T}$ is a straight line of slope $-\dfrac{\Delta H^\circ}{2.303\,R}$.
Slope from any two points:
$$\text{slope} = \frac{1.5 - 3.5}{0.07 - 0.05} = \frac{-2}{0.02} = -100$$Therefore:
$$-\frac{\Delta H^\circ}{2.303\,R} = -100 \implies \frac{\Delta H^\circ}{R} = 100 \times 2.303 = 230.3$$Magnitude (nearest integer) $= 230$.
Answer: 230
Solution
Moles at equilibrium:
| $A$ | $B$ | $C$ | total |
|---|---|---|---|
| $a-x$ | $x$ | $x$ | $a+x$ |
Partial pressures ($p_i = \text{mole fraction} \times p$):
$$p_A = \frac{a-x}{a+x}\,p, \qquad p_B = p_C = \frac{x}{a+x}\,p$$$$K_p = \frac{p_B\,p_C}{p_A} = \frac{\left(\dfrac{x}{a+x}\,p\right)^2}{\dfrac{a-x}{a+x}\,p} = \frac{x^2\,p}{(a+x)(a-x)} = \frac{x^2}{a^2 - x^2}\times p$$Answer: B $\left(\dfrac{x^2}{a^2 - x^2} \times p\right)$
Solution
Build the target reaction from the given ones.
Reverse and halve reaction 1: $\ \frac{1}{2}x_2 + \frac{1}{2}y_2 \rightleftharpoons xy$, with $K = \left(\dfrac{1}{K_1}\right)^{1/2}$.
Add reaction 2: $\ xy + \frac{1}{2}z_2 \rightleftharpoons xyz$, with $K_2$.
Adding these gives the target, so multiply the constants:
$$K_3 = \left(\frac{1}{K_1}\right)^{1/2}\times K_2 = \left(\frac{1}{2.5 \times 10^5}\right)^{1/2}\times (5 \times 10^{-3})$$$$\left(\frac{1}{2.5 \times 10^5}\right)^{1/2} = \left(4 \times 10^{-6}\right)^{1/2} = 2 \times 10^{-3}$$$$K_3 = (2 \times 10^{-3})(5 \times 10^{-3}) = 1.0 \times 10^{-5}$$Answer: C ($1.0 \times 10^{-5}$)
Solution
For the first equilibrium (only solids besides CO$_2$), $K_{p_1}$ fixes the CO$_2$ pressure:
$$p_{\text{CO}_2} = K_{p_1} = 0.08\ \text{atm}$$For the second equilibrium:
$$K_{p_2} = \frac{p_{\text{CO}}^2}{p_{\text{CO}_2}} = 2$$$$p_{\text{CO}}^2 = K_{p_2}\times p_{\text{CO}_2} = 2 \times 0.08 = 0.16$$$$p_{\text{CO}} = \sqrt{0.16} = 0.4\ \text{atm} = 4 \times 10^{-1}\ \text{atm}$$Answer: 4