NEET Equilibrium Previous Year Questions with Solutions

Silver bromide dissociates into one silver cation and one bromide anion.

• • S₁ = sqrt(5.0 × 10⁻¹³) = sqrt(50 × 10⁻¹⁴) ≈ 7.07 × 10⁻⁷ mol L⁻¹.

Zinc hydroxide dissociates into one Zn²⁺ cation and two OH⁻ anions.

• • S₂³ = K_sp / 4 = (1.0 × 10⁻¹⁵) / 4 = 2.5 × 10⁻¹⁶ = 250 × 10⁻¹⁸.
• • S₂ = (250)^(1/3) × 10⁻⁶ ≈ 6.30 × 10⁻⁶ mol L⁻¹.

Mercurous chloride contains the dimeric mercurous ion Hg₂²⁺ and two chloride ions Cl⁻.

• • S₃³ = K_sp / 4 = (1.3 × 10⁻¹⁸) / 4 = 0.325 × 10⁻¹⁸.
• • S₃ = (0.325)^(1/3) × 10⁻⁶ ≈ 0.687 × 10⁻⁶ mol L⁻¹ = 6.87 × 10⁻⁷ mol L⁻¹.

Arranging the three solubilities: S[Zn(OH)₂] = 63.0 × 10⁻⁷ M S[AgBr] = 7.07 × 10⁻⁷ M S[Hg₂Cl₂] = 6.87 × 10⁻⁷ M Therefore: Zn(OH)₂ > AgBr > Hg₂Cl₂.

• • Matches Option (4).

Phosphoric acid (H₃PO₄) is a tribasic mineral acid ionizing in three consecutive stages:

• • Taking logarithms of both sides converts the multiplication into addition:
• • \log K = \log K_{a1} + \log K_{a2} + \log K_{a3}.
• • Thus Statement A is rigorously TRUE.

Electrostatic principles dictate the relative difficulty of successive proton removals:

• • It is far easier to remove a positively charged proton from a neutral molecule (H₃PO₄) than from a negatively charged ion (H₂PO₄⁻), and even harder from a doubly negative ion (HPO₄²⁻).
• • Consequently, H₃PO₄ is a significantly stronger acid than H₂PO₄⁻ and HPO₄²⁻ (Statement B is TRUE).
• • The hierarchy K_{a1} > K_{a2} > K_{a3} holds universally for all polybasic acids (Statement C is TRUE).

Statement D asserts that K_{a1} = (K_{a3} + K_{a2}) / 2. Because these equilibrium constants differ by orders of magnitude (10⁻³ vs 10⁻⁸ vs 10⁻¹³), an arithmetic mean is physically and mathematically impossible. Statement D is FALSE.

• • Matches Option (4) exactly.

At dynamic chemical equilibrium, the rate of the forward reaction equals the rate of the backward reaction:

• • Given k_b = 2500 k_f, so:
• • K_c = \frac{k_f}{2500 k_f} = \frac{1}{2500} = 4.0 \times 10^{-4}\text{ mol L}^{-1}.

Calculate the difference between the number of moles of gaseous products and gaseous reactants:

• • The thermodynamic relationship linking K_p and K_c is:
• • K_p = K_c (R T)^{\Delta n_g} = K_c (R T)^1.

Substitute R = 0.0831 L atm mol⁻¹ K⁻¹, T = 1000 K, and K_c = 1/2500:

• • Matches Option (3) 0.033.

For an endothermic reversible reaction (ΔH > 0), absorption of heat accompanies the forward reaction:

• • An increase in temperature supplies thermal energy, driving the system to absorb heat by shifting to the right (forward).
• • Consequently, higher temperature (A) increases the equilibrium constant K_p and enhances the yield of NO.
• • Condition A is correct; condition B is incorrect.

Examine the reaction quotient Q_c when reactant concentrations are increased:

• • Increasing [N₂] or [O₂] makes Q_c < K_c, driving the reaction in the forward direction until Q_c = K_c again.
• • Therefore, higher concentration of N₂ (C) and higher concentration of O₂ (D) both increase the yield of NO.

Combine all favorable operational parameters:

• • Matches Option (4) A, C, D only.

For the generalized reversible reaction 2A ⇌ B + C, the reaction quotient Qc is defined using non-equilibrium instantaneous concentrations in the same functional form as the equilibrium constant:

Substituting the instantaneous concentration values [A] = [B] = [C] = 2 × 10⁻³ M into the quotient expression:

Comparing the calculated reaction quotient with the given equilibrium constant: - Qc = 1.0 - Kc = 4 × 10⁻³ = 0.004 Because Qc > Kc, the current concentration ratio of products to reactants is significantly greater than that at equilibrium. To establish dynamic equilibrium, the reaction must proceed from right to left (backward direction) to consume B and C while producing A until Qc decreases to equal Kc.

For a general gas-phase equilibrium aA + bB ⇌ cC + dD, partial pressures relate to molar concentrations by the ideal gas law: p_i = [i]RT. Substituting this into the Kp expression yields:

For Kp to equal Kc identically (independent of temperature): (RT)^Δng = 1 ⟹ Δng = 0. Thus, Kp = Kc if and only if the total stoichiometric sum of gaseous product coefficients equals that of the gaseous reactants.

Calculating Δng for all four reactions: 1. CO(g) + H2O(g) ⇌ CO2(g) + H2(g): Δng = (1 + 1) - (1 + 1) = 0 ⟹ Kp = Kc. 2. 2 BrCl(g) ⇌ Br2(g) + Cl2(g): Δng = (1 + 1) - 2 = 0 ⟹ Kp = Kc. 3. PCl5(g) ⇌ PCl3(g) + Cl2(g): Δng = (1 + 1) - 1 = +1 ⟹ Kp = Kc(RT) ≠ Kc! 4. H2(g) + I2(g) ⇌ 2 HI(g): Δng = 2 - (1 + 1) = 0 ⟹ Kp = Kc.

For the gas-phase reaction 2 NO(g) ⇌ N2(g) + O2(g), the equilibrium constant expression is:

Let initial concentration of NO be C = 0.1 mol/L and degree of dissociation be α: - Initial: [NO] = C, [N2] = 0, [O2] = 0 - Change: [NO] = -Cα, [N2] = +Cα/2, [O2] = +Cα/2 - Equilibrium: [NO] = C(1 - α), [N2] = Cα/2, [O2] = Cα/2

Substituting equilibrium expressions into Kc:

Taking square roots on both sides:

Expanding and solving the linear equation in α:

According to G.N. Lewis (1923), an acid is defined as an electron-pair acceptor, and a base is defined as an electron-pair donor. A coordinate covalent (dative) bond is formed when a Lewis base donates its electron pair into a vacant orbital of a Lewis acid:

Boron (Z = 5) has ground-state configuration 1s² 2s² 2p¹. In BF₃, boron undergoes sp² hybridization, forming three B–F σ-bonds. The central boron atom is surrounded by only 6 valence electrons, leaving one unhybridized 2p_z orbital completely empty:

Because boron possesses an empty 2p orbital, BF₃ readily accepts an electron pair from Lewis bases (such as NH₃ or F⁻) to complete its stable octet (8 e⁻):

All other three species possess available lone pairs of electrons and act as electron-pair donors (Lewis bases):

Dimethylammonium acetate ($CH_3COONH_2(CH_3)_2$) is formed by the neutralization of acetic acid ($CH_3COOH$, a weak acid with $pK_a = 4.77$) and dimethylamine ($(CH_3)_2NH$, a weak base with $pK_b = 3.27$). Both the cation and the anion undergo simultaneous hydrolysis in aqueous solution.

The hydrolysis equilibrium is: $$CH_3COO^- + (CH_3)_2NH_2^+ + H_2O \rightleftharpoons CH_3COOH + (CH_3)_2NH$$ The hydronium ion concentration in such a solution is independent of salt concentration and given by: $$[H_3O^+] = \sqrt{\frac{K_w \cdot K_a}{K_b}}$$ Taking negative logarithms on both sides:

Substitute $pK_a = 4.77$ and $pK_b = 3.27$:

Sodium hydroxide (NaOH) is a strong base that dissociates completely in aqueous solution: $$\text{NaOH (aq)} \to \text{Na}^+ \text{(aq)} + \text{OH}^- \text{(aq)}$$ Given $$\text{[NaOH]} = 0.1\text{ M}$$, the contribution of hydroxide ions from NaOH is $$[\text{OH}^-]_{\text{NaOH}} = 0.1\text{ M} = 10^{-1}\text{ M}$$.

Nickel hydroxide is a sparingly soluble salt that establishes the following dissolution equilibrium: $$\text{Ni(OH)}_2 \text{ (s)} \rightleftharpoons \text{Ni}^{2+} \text{(aq)} + 2\text{OH}^- \text{(aq)}$$ If $$s$$ is the molar solubility of $$\text{Ni(OH)}_2$$ in this solution: $$[\text{Ni}^{2+}] = s$$ $$[\text{OH}^-] = 0.1 + 2s$$

Because $$K_{sp} = 2 \times 10^{-15}$$ is extremely small, $$s \ll 0.1$$, and hence $$2s$$ is negligible compared to $$0.1$$: $$[\text{OH}^-] \approx 0.1\text{ M} = 10^{-1}\text{ M}$$

Substitute into the solubility product expression: $$K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2$$ $$2 \times 10^{-15} = s \times (0.1)^2 = s \times 10^{-2}$$ $$s = \frac{2 \times 10^{-15}}{10^{-2}} = 2 \times 10^{-13}\text{ M}$$ This confirms the approximation ($$2 \times (2 \times 10^{-13}) = 4 \times 10^{-13} \ll 0.1$$).