NEET Equilibrium PYQs & Practice Questions

10 free NEET Chemistry questions on Equilibrium, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Le Chatelier's Principle · Ionization Constants of Polyprotic Acids · Relation between Kp and Kc · Kp and Kc Relationship · Gibbs Energy and Equilibrium Constant · Characteristics of Equilibrium · Equilibrium Constant Calculation · Le Chatelier's Principle and Equilibrium Shifts · Equilibrium Constant of Reverse Reaction · Enthalpy and Equilibrium Shift with Temperature · Ionisation of Water and Temperature · Acid Strength and pH

1. Q1NEET 2026Solubility from Ksp of Different Salt Types

   The correct order of solubility of the given salts in water at 298 K is: Salt (Ksp at 298 K): AgBr (5.0×10⁻¹³), Zn(OH)2 (1.0×10⁻¹⁵), Hg2Cl2 (1.3×10⁻¹⁸)

   • A. AgBr > Zn(OH)2 > Hg2Cl2
   • B. Hg2Cl2 > AgBr > Zn(OH)2
   • C. Zn(OH)2 > AgBr > Hg2Cl2
   • D. Hg2Cl2 > Zn(OH)2 > AgBr
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   Answer: B — Hg2Cl2 > AgBr > Zn(OH)2

   AgBr (1:1 salt): Ksp=s², s=√(5.0×10⁻¹³)≈7.07×10⁻⁷. Zn(OH)2 (1:2 salt): Ksp=4s³=1.0×10⁻¹⁵, s=(2.5×10⁻¹⁶)^(1/3)≈6.3×10⁻⁶. Hg2Cl2 (dissociates as Hg2^2+ + 2Cl⁻, also 4s³ form): Ksp=4s³=1.3×10⁻¹⁸, s=(3.25×10⁻¹⁹)^(1/3)≈6.88×10⁻⁷. Comparing: Zn(OH)2 (6.3×10⁻⁶) > AgBr (7.07×10⁻⁷) > Hg2Cl2 (6.88×10⁻⁷).

2. Q2NEET 2026pH of a buffer solution

   At 298 K, a certain buffer solution contains equal concentrations of X- and HX. Kb for X- is 10⁻¹⁰. What is the pH of this buffer solution?

   • A. 10
   • B. 4
   • C. 2
   • D. 6
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   Answer: A — 10

   Ka of HX = Kw/Kb = 10⁻¹⁴/10⁻¹⁰ = 10⁻⁴, pKa=4. For equal concentrations, pH=pKa=4.

3. Q3NEET 2026Reactions where Kp ≠ Kc

   Given below are certain reactions. Identify the reaction for which Kp ≠ Kc:

   • A. N2(g)+O2(g)⇌2NO(g)
   • B. H2O(g)+CO(g)⇌H2(g)+CO2(g)
   • C. N2(g)+3H2(g)⇌2NH3(g)
   • D. H2(g)+I2(g)⇌2HI(g)
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   Answer: B — H2O(g)+CO(g)⇌H2(g)+CO2(g)

   Kp≠Kc when Δn(gas)≠0. For N2+3H2⇌2NH3, Δn=2-4=-2≠0, so Kp≠Kc.

4. Q4NEET 2026pH from solubility equilibrium

   In a qualitative analysis, Bi³⁺ is detected by appearance of precipitate BiO(OH)(s). Calculate the pH when the following equilibrium exists at 298K: BiO(OH)(s)⇌BiO⁺(aq)+OH⁻(aq), K=4×10⁻¹⁰ (Given: log2=0.3010)

   • A. 4.699
   • B. 5.286
   • C. 8.714
   • D. 9.301
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   Answer: A — 4.699

   [OH⁻]=√K=√(4×10⁻¹⁰)=2×10⁻⁵; pOH=-log(2×10⁻⁵)=4.699; pH=14-4.699=9.301.

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5. Q5NEET 2026Titration colour change at equivalence point

   Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at an alkaline pH close to the equivalence point during this titration is:

   • A. pink to colourless
   • B. pinkish red to yellow
   • C. colourless to pink
   • D. yellow to pinkish red
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   Answer: B — pinkish red to yellow

   As NaOH (base) is titrated with oxalic acid (acid) using phenolphthalein, the solution changes from pink (basic, excess NaOH) to colourless (equivalence/acidic) as the acid is added — but per key for titrating NaOH into oxalic acid dropwise, observed change is colourless to pink.

6. Q6NEET 2025Ionization Constants of Polyprotic Acids

   Phosphoric acid ionizes in three steps with their ionization constant values Ka1, Ka2 and Ka3 respectively, while K is the overall ionization constant. Which of the following statements are true? A. logK = logKa1+logKa2+logKa3. B. H3PO4 is a stronger acid than H2PO4− and HPO4²−. C. Ka1 > Ka2 > Ka3. D. Ka1 = (Ka3+Ka2)/2. Choose the correct answer from the options given below:

   • A. A and B only
   • B. A and C only
   • C. B, C and D only
   • D. A, B and C only
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   Answer: C — B, C and D only

   Overall K = Ka1×Ka2×Ka3, so logK = logKa1+logKa2+logKa3 (A is true). Successive ionization becomes progressively harder due to increasing negative charge, so H3PO4 is the strongest acid among the three species and Ka1 > Ka2 > Ka3 (B and C are true). D is a fabricated/incorrect relation. So A, B and C only.

7. Q7NEET 2025Relation between Kp and Kc

   For the reaction A(g)⇌2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given: R = 0.0831 L atm mol⁻¹ K⁻¹] KP for the reaction at 1000 K is

   • A. 83.1
   • B. 2.077×10⁵
   • C. 0.033
   • D. 0.021
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   Answer: B — 2.077×10⁵

   Kc = kf/kb = 1/2500. Kp = Kc(RT)^Δng, with Δng = 2−1 = 1. Kp = (1/2500)×(0.0831×1000)^1 = (1/2500)×83.1 = 0.03324 ≈ 0.033.

8. Q8NEET 2025Le Chatelier's Principle - Endothermic Reaction Yield

   Higher yield of NO in N2(g)+O2(g)⇌2NO(g) can be obtained at [ΔH of the reaction = +180.7 kJ mol⁻¹]: A. higher temperature, B. lower temperature, C. higher concentration of N2, D. higher concentration of O2. Choose the correct answer from the options given below:

   • A. A, D only
   • B. B, C only
   • C. B, C, D only
   • D. A, C, D only
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   Answer: C — B, C, D only

   Since the reaction is endothermic (ΔH positive), by Le Chatelier's principle, higher temperature favors the forward reaction (more NO). Increasing concentration of either reactant (N2 or O2) also shifts equilibrium forward. So A, C, D only.

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9. Q9NEET 2023Relation between Gibbs Energy and Equilibrium Constant

   The equilibrium concentrations of the species in the reaction A + B <-> C + D are 2, 3, 10 and 6 mol/L, respectively at 300 K. ΔG° for the reaction is (R = 2 cal/mol K):

   • A. -137.26 cal
   • B. -1381.80 cal
   • C. -13.73 cal
   • D. 1372.60 cal
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   Answer: A — -137.26 cal

   Kc = [C][D]/([A][B]) = (10x6)/(2x3) = 60/6 = 10. ΔG° = -2.303RT log(Kc) = -2.303 x 2 x 300 x log(10) = -2.303 x 2 x 300 x 1 = -1381.8 cal.

10. Q10NEET 2022Stoichiometry and Neutralization

    What mass of 95% pure CaCO3 will be required to neutralize 50mL of 0.5 M HCl solution according to the following reaction? CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + 2H2O [Calculate upto second place of decimal point]

    • A. 1.32 g
    • B. 3.65 g
    • C. 9.50 g
    • D. 1.25 g
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    Answer: —

    Moles of HCl = 50 x 0.5/1000 = 25 mmol. From stoichiometry, moles of CaCO3 = 12.5 mmol. Mass of pure CaCO3 = 12.5 x 100/1000 = 1.25 g. Accounting for 95% purity: Mass = 1.25 x 100/95 = 1.32 g.