Chemical Equilibrium: JEE Main 2026 PYQs

Answer: Option (4)

Key idea: Solubility Product (Ksp) and Solubility Calculations

1. For Ag$_2$CrO$_4$: Write dissolution as Ag$_2$CrO$_4$(s) $\rightleftharpoons$ 2Ag$^+$ + CrO$_4^{2-}$. If solubility = $s_1$, then K$_{sp}$ = $(2s_1)^2(s_1)$ = $4s_1^3$
2. From K$_{sp}$ = 32$x$ = $4s_1^3$, we get $s_1^3$ = 8$x$, so $s_1$ = 2$\sqrt[3]{x}$
3. For AgBr: Write dissolution as AgBr(s) $\rightleftharpoons$ Ag$^+$ + Br$^-$. If solubility = $s_2$, then K$_{sp}$ = $s_2^2$
4. From K$_{sp}$ = 4$y$ = $s_2^2$, we get $s_2$ = 2$\sqrt{y}$
5. Calculate the ratio: $\frac{s_1}{s_2}$ = $\frac{2\sqrt[3]{x}}{2\sqrt{y}}$ = $\frac{\sqrt[3]{x}}{\sqrt{y}}$

Answer: Option (3)

Key idea: Common Ion Effect and Weak Acid Equilibrium

1. Given: $H_2A$ with $K_1 = 8.1 \times 10^{-8}$, $K_2 = 1.0 \times 10^{-13}$, dissolved in 0.1 M HCl
2. Initial concentrations: $[H_2A] = 0.1$ M, $[H^+] = 0.1$ M (from HCl)
3. For first ionization: $H_2A \rightleftharpoons H^+ + HA^-$ with $K_1 = \frac{[H^+][HA^-]}{[H_2A]}$
4. Due to common ion effect (high $[H^+]$ from HCl), ionization is suppressed. Let $[HA^-] = x$
5. At equilibrium: $[H_2A] \approx 0.1$ M, $[H^+] \approx 0.1$ M (since $x \ll 0.1$)
6. Substitute in $K_1$: $8.1 \times 10^{-8} = \frac{(0.1)(x)}{(0.1)} = x$
7. Therefore, $[HA^-] = 8.1 \times 10^{-8}$ M

Answer: Option (2)

Key idea: Weak Acid Titration and Buffer Solution pH Calculation

1. **Part (a):** For weak acid HX at start, use $[H^+] = \sqrt{K_a \times C}$ where $C = 0.2$ M
2. Calculate: $[H^+] = \sqrt{5 \times 10^{-4} \times 0.2} = \sqrt{10^{-4}} = 10^{-2}$ M
3. Therefore, $pH = -\log(10^{-2}) = 2.0$
4. **Part (b):** Initial moles HX = $0.2 \times 20 = 4$ mmol; NaOH added = $0.2 \times 10 = 2$ mmol
5. After neutralization: HX remaining = 2 mmol, salt NaX formed = 2 mmol (buffer at half-equivalence point)
6. Using Henderson-Hasselbalch: $pH = pK_a + \log\frac{[A^-]}{[HA]} = 3.3 + \log(1) = 3.3$
7. Answer: pH values are **(2.0, 3.3)**

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Key idea: Relationship between equilibrium constants when reaction equations are scaled

Answer: Option (1)

Key idea: Solubility Product and Ionic Equilibrium

1. Convert solubility to molar concentration: $s = \frac{x}{y}$ mol/L
2. Write dissolution equilibrium: $M_3A_2(s) \rightleftharpoons 3M^{2+}(aq) + 2A^{3-}(aq)$
3. Find ion concentrations: $[M^{2+}] = 3s$ and $[A^{3-}] = 2s = \frac{2x}{y}$
4. Calculate $K_{sp} = [M^{2+}]^3[A^{3-}]^2 = (3s)^3(2s)^2 = 108s^5 = \frac{108x^5}{y^5}$
5. Find ratio: $\frac{[A^{3-}]}{K_{sp}} = \frac{2x/y}{108x^5/y^5} = \frac{2x}{y} \times \frac{y^5}{108x^5} = \frac{y^4}{54x^4}$

Answer: Option (3)

Key idea: Acid-Base Neutralization and pH Calculation

1. Calculate moles of H⁺ and OH⁻ for each mixture. Remember: $Ca(OH)_2$ gives 2 OH⁻, $H_2SO_4$ gives 2 H⁺
2. **Mixture A:** OH⁻ = 4 mmol, H⁺ = 2.5 mmol. Excess OH⁻ = 1.5 mmol in 35 mL. [OH⁻] = 0.0429 M, pH ≈ 12.63
3. **Mixture B:** H⁺ = 0.2 mmol, OH⁻ = 0.2 mmol. Complete neutralization occurs, pH = 7
4. **Mixture C:** H⁺ = 2 mmol, OH⁻ = 1 mmol. Excess H⁺ = 1 mmol in 20 mL. [H⁺] = 0.05 M, pH ≈ 1.30
5. Arrange in increasing order: C (pH 1.30) < B (pH 7) < A (pH 12.63)

Answer: 230

Key idea: Van't Hoff equation and determination of enthalpy change from equilibrium constant-temperature data

1. Use the van't Hoff equation: $\log_{10}K_p = -\frac{\Delta H^\circ}{2.303R} \times \frac{1}{T} + \text{constant}$
2. This is a linear equation with slope $m = -\frac{\Delta H^\circ}{2.303R}$
3. Calculate slope using two points: $m = \frac{2.5-3.5}{0.06-0.05} = \frac{-1.0}{0.01} = -100$
4. Verify with another pair: $m = \frac{1.5-2.5}{0.07-0.06} = \frac{-1.0}{0.01} = -100$ ✓
5. From slope: $-\frac{\Delta H^\circ}{2.303R} = -100$, so $\frac{\Delta H^\circ}{R} = 100 \times 2.303 = 230.3$
6. Magnitude of $\frac{\Delta H^\circ}{R}$ = 230 (nearest integer)

Answer: Option (1)

Key idea: Chemical Equilibrium with Ideal Gas Law - calculating partial pressures using Kc and PV=nRT

1. Set up equilibrium: Initial moles are He = 1, A = 1, B = 0. At equilibrium: He = 1, A = (1-x), B = x
2. Apply $K_c = \frac{[B]}{[A]} = \frac{x/10}{(1-x)/10} = 4.0$
3. Solve for x: $\frac{x}{1-x} = 4 \Rightarrow x = 4 - 4x \Rightarrow 5x = 4 \Rightarrow x = 0.8$ moles
4. At equilibrium: He = 1 mole, A = 0.2 moles, B = 0.8 moles
5. Calculate partial pressure of He: $P_{He} = \frac{nRT}{V} = \frac{1 \times 0.082 \times 400}{10} = 3.28$ atm
6. Calculate partial pressure of B: $P_B = \frac{nRT}{V} = \frac{0.8 \times 0.082 \times 400}{10} = 2.624$ atm

Answer: Option (2)

Key idea: Degree of dissociation of weak electrolyte using equilibrium constant with small alpha approximation

1. Write dissociation: $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$
2. Set up equilibrium concentrations: $[A_xB_y] = c(1-\alpha)$, $[A^{y+}] = xc\alpha$, $[B^{x-}] = yc\alpha$
3. Write $K = \frac{(xc\alpha)^x(yc\alpha)^y}{c(1-\alpha)} = \frac{x^x y^y c^{x+y} \alpha^{x+y}}{c(1-\alpha)}$
4. For concentrated solution, $\alpha \ll 1$, so $(1-\alpha) \approx 1$
5. Simplify: $K = x^x y^y c^{x+y-1} \alpha^{x+y}$
6. Solve for $\alpha$: $\alpha = \left(\frac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}$

Answer: 5

Key idea: Coordination Chemistry - Ionizable vs Non-ionizable Ligands

1. Write the formula of tetraaquadichloridochromium(III) chloride: $[Cr(H_2O)_4Cl_2]Cl$
2. Identify ionizable chlorides: Only the $Cl^-$ outside the coordination sphere (counter ion) precipitates with $AgNO_3$. The 2 $Cl^-$ inside are non-ionizable.
3. Calculate moles of complex: $n = M \times V = 0.05 \times 0.1 = 5 \times 10^{-3}$ mol
4. Since 1 mole of complex gives 1 mole of ionizable $Cl^-$: Moles of $AgCl$ = $5 \times 10^{-3}$ mol
5. Express in required format: $5 \times 10^{-3}$ mol, so answer = 5

Answer: 8

Key idea: Chemical Equilibrium, Gibbs Free Energy, Equilibrium Constant

1. Set up ICE table: Initial 1 mol $X_2Y_4$, at equilibrium with 75% dissociation gives 0.25 mol $X_2Y_4$ and 1.5 mol $XY_2$, total = 1.75 mol
2. Calculate partial pressures: $P_{X_2Y_4} = \frac{0.25}{1.75} \times 1 = \frac{1}{7}$ atm, $P_{XY_2} = \frac{1.5}{1.75} \times 1 = \frac{6}{7}$ atm
3. Calculate $K_p = \frac{(P_{XY_2})^2}{P_{X_2Y_4}} = \frac{(6/7)^2}{1/7} = \frac{36}{7}$
4. Find $\ln K_p = \ln 36 - \ln 7 = (2\ln 2 + 2\ln 3) - \ln 7 = 2(0.69) + 2(1.104) - 1.932 = 1.656$
5. Calculate $\Delta_rG^\circ = -RT \ln K_p = -8.3 \times 600 \times 1.656 = -8247$ J/mol $= -8.247$ kJ/mol
6. Magnitude of $\Delta_rG^\circ = 8$ kJ/mol (nearest integer)

Answer: Option (1.0 × 10⁻⁵)

Key idea: Equilibrium constant manipulation: reversing reactions, multiplying by coefficients, and adding reactions

1. Reverse Reaction 1: $X_2 + Y_2 \rightleftharpoons 2XY$, so $K_1' = \frac{1}{K_1} = \frac{1}{2.5 \times 10^5} = 4 \times 10^{-6}$
2. Multiply reversed Reaction 1 by $\frac{1}{2}$: $\frac{1}{2}X_2 + \frac{1}{2}Y_2 \rightleftharpoons XY$, so $K_1'' = (K_1')^{1/2} = (4 \times 10^{-6})^{1/2} = 2 \times 10^{-3}$
3. Keep Reaction 2 as given: $XY + \frac{1}{2}Z_2 \rightleftharpoons XYZ$, $K_2 = 5 \times 10^{-3}$
4. Add the modified reactions to get target: $\frac{1}{2}X_2 + \frac{1}{2}Y_2 + \frac{1}{2}Z_2 \rightleftharpoons XYZ$
5. Multiply equilibrium constants: $K_3 = K_1'' \times K_2 = (2 \times 10^{-3}) \times (5 \times 10^{-3}) = 1.0 \times 10^{-5}$

Answer: 4

Key idea: Simultaneous equilibria with heterogeneous reactions

1. Given two equilibrium reactions with $Kp_1 = 0.08$ atm and $Kp_2 = 2$ atm
2. For Reaction 1: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$, we have $Kp_1 = P_{CO_2} = 0.08$ atm
3. For Reaction 2: $C(s) + CO_2(g) \rightleftharpoons 2CO(g)$, we have $Kp_2 = \frac{(P_{CO})^2}{P_{CO_2}} = 2$ atm
4. Substitute $P_{CO_2} = 0.08$ atm into the second equation: $(P_{CO})^2 = 2 \times 0.08 = 0.16$
5. Solve for $P_{CO}$: $P_{CO} = \sqrt{0.16} = 0.4$ atm $= 4 \times 10^{-1}$ atm