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Chemical Equilibrium NEET Practice MCQs & Notes

Chapter-wise equilibrium MCQs for NEET with detailed solutions, short revision notes and related ChemNexa mock tests to build problem-solving skills.

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Equilibrium Revision Guide for NEET Chemistry

Equilibrium covers reversible chemical processes and the quantitative relationship between reactants and products at equilibrium. It also includes acid-base, ionic and solubility equilibria in many syllabi.

For NEET Chemistry, build confidence with equilibrium constants, reaction quotient, Le Chatelier principle, pH calculations, buffer concepts and solubility product.

Important Topics

  • Dynamic chemical equilibrium
  • Kc, Kp and reaction quotient
  • Le Chatelier principle
  • Acid-base equilibrium and pH
  • Buffer solutions
  • Solubility product and common-ion effect

Important Formulae & Relationships

  • Kp = Kc(RT)^Δn
  • pH = −log[H+]
  • pOH = −log[OH−]
  • pH + pOH = 14 at 25°C
  • Ka × Kb = Kw for a conjugate pair

Use formulas only after checking the required units, sign convention and assumptions for the question.

What to practise in Equilibrium

Concept Revision

Revise definitions, principles, equations, trends and core ideas from Equilibrium before attempting MCQs.

Exam-style MCQs

Use chapter-focused multiple-choice practice to improve accuracy, recall and application for NEET Chemistry.

Performance Practice

Attempt ChemNexa tests, review results and return to weak areas for another round of targeted revision.

Top 20 Equilibrium MCQs with Answers

These public questions are selected from the exact Equilibrium chapter, screened for topic relevance and deduplicated so repeated versions of the same MCQ are not shown publicly. Use them for quick revision, then sign in to attempt the complete test experience with more questions, results and performance review.

NEET Chemistry · Equilibrium · Very Hard

1. If the reaction quotient Qc is smaller than Kc, the reaction will proceed:

  • A. Until Kc becomes zero
  • B. In the forward direction
  • C. In the reverse direction
  • D. Not at all
Answer: B
Explanation: Qc < Kc means the product-to-reactant ratio is below equilibrium, so the reaction proceeds forward.

NEET Chemistry · Equilibrium · Very Hard

2. For an acidic buffer, the Henderson-Hasselbalch equation is:

  • A. pH = pKa − log([salt]/[acid])
  • B. pOH = pKa + log([acid]/[salt])
  • C. pH = 14 − pKa
  • D. pH = pKa + log([salt]/[acid])
Answer: D
Explanation: For a weak acid and its salt, pH = pKa + log([A−]/[HA]).

NEET Chemistry · Equilibrium · Very Hard

3. Increasing pressure shifts which equilibrium toward products?

  • A. 2SO3 ⇌ 2SO2 + O2
  • B. N2 + 3H2 ⇌ 2NH3
  • C. H2 + I2 ⇌ 2HI
  • D. PCl5 ⇌ PCl3 + Cl2
Answer: B
Explanation: Higher pressure favours the side with fewer moles of gas. The ammonia side has two moles versus four on the reactant side.

NEET Chemistry · Equilibrium · Very Hard

4. For N2(g) + 3H2(g) ⇌ 2NH3(g), the relation between Kp and Kc is:

  • A. Kp = Kc
  • B. Kp = Kc(RT)−2
  • C. Kp = Kc(RT)2
  • D. Kp = KcRT
Answer: B
Explanation: Kp = Kc(RT)^Δng and Δng = 2 − 4 = −2. Therefore Kp = Kc(RT)−2.

NEET Chemistry · Equilibrium · Very Hard

5. For H2(g) + I2(g) ⇌ 2HI(g), Kp is related to Kc by:

  • A. Kp = KcRT
  • B. Kp = Kc/(RT)
  • C. Kp = Kc(RT)2
  • D. Kp = Kc
Answer: D
Explanation: Δng = 2 − 2 = 0, so Kp = Kc(RT)^0 = Kc.

NEET Chemistry · Equilibrium · Very Hard

6. For CaCO3(s) ⇌ CaO(s) + CO2(g), the equilibrium expression Kp is:

  • A. PCaO × PCO2/PCaCO3
  • B. 1/PCO2
  • C. PCO2²
  • D. PCO2
Answer: D
Explanation: Activities of pure solids are unity, so only the partial pressure of CO2 appears.

NEET Chemistry · Equilibrium · Very Hard

7. At 25 °C, if pH = 5, the pOH is:

  • A. 14
  • B. 9
  • C. 5
  • D. 7
Answer: B
Explanation: At 25 °C, pH + pOH = 14. Therefore pOH = 9.

NEET Chemistry · Equilibrium · Very Hard

8. For a weak monoprotic acid HA of concentration C and dissociation constant Ka, when ionisation is small,

  • A. KaC
  • B. √(KaC)
  • C. Ka/C
  • D. C/Ka
Answer: B
Explanation: For small dissociation, Ka ≈ x²/C, so x = [H+] ≈ √(KaC).

NEET Chemistry · Equilibrium · Very Hard

9. A buffer solution resists change in pH because it contains:

  • A. A neutral salt only
  • B. A weak acid and its conjugate base or a weak base and its conjugate acid
  • C. Only a strong acid
  • D. Only a strong base
Answer: B
Explanation: The conjugate pair consumes added H+ or OH−, limiting pH change.

NEET Chemistry · Equilibrium · Very Hard

10. The solubility product expression for Ag2CrO4 is:

  • A. Ksp = [Ag+]²/[CrO4²−]
  • B. Ksp = [Ag+]²[CrO4²−]
  • C. Ksp = [Ag+][CrO4²−]
  • D. Ksp = 2[Ag+][CrO4²−]
Answer: B
Explanation: Ag2CrO4(s) ⇌ 2Ag+ + CrO4²−, so Ksp = [Ag+]²[CrO4²−].

NEET Chemistry · Equilibrium · Very Hard

11. If the molar solubility of AgCl is s, its Ksp equals:

  • A. 2s²
  • B. 4s³
  • C. s
  • D. s²
Answer: D
Explanation: AgCl ⇌ Ag+ + Cl− gives [Ag+] = [Cl−] = s, so Ksp = s².

NEET Chemistry · Equilibrium · Very Hard

12. The solubility of AgCl decreases when NaCl is added because of:

  • A. Increased temperature necessarily
  • B. Common-ion effect
  • C. Salt hydrolysis
  • D. Buffer action
Answer: B
Explanation: Added Cl− shifts AgCl dissolution equilibrium to the left, lowering solubility.

NEET Chemistry · Equilibrium · Very Hard

13. A catalyst added to a system at equilibrium:

  • A. Decreases Kc
  • B. Does not change Kc but accelerates attainment of equilibrium
  • C. Shifts equilibrium toward products
  • D. Increases Kc
Answer: B
Explanation: A catalyst lowers activation energies of forward and reverse reactions equally; it does not alter the equilibrium constant or position.

NEET Chemistry · Equilibrium · Very Hard

14. For an exothermic reaction at equilibrium, increasing temperature generally:

  • A. Decreases K
  • B. Does not affect K
  • C. Makes K equal to one
  • D. Increases K
Answer: A
Explanation: Heat behaves as a product for an exothermic reaction. Raising temperature favours the reverse reaction and decreases K.

NEET Chemistry · Equilibrium · Very Hard

15. If all stoichiometric coefficients in a balanced equilibrium equation are doubled, the new equilibrium constant is:

  • A. √K
  • B. K²
  • C. 2K
  • D. K/2
Answer: B
Explanation: Multiplying a reaction by a factor of two raises its equilibrium constant to the second power.

NEET Chemistry · Equilibrium · Very Hard

16. When a reaction is reversed, its equilibrium constant becomes:

  • A. −K
  • B. K²
  • C. Unchanged
  • D. 1/K
Answer: D
Explanation: Reversing the reaction interchanges numerator and denominator in the equilibrium expression, giving 1/K.

NEET Chemistry · Equilibrium · Very Hard

17. The pH of 1.0 × 10−3 M HCl at 25 °C is approximately:

  • A. 7
  • B. 3
  • C. 2
  • D. 11
Answer: B
Explanation: HCl is a strong acid and [H+] ≈ 10−3 M, so pH = 3.

NEET Chemistry · Equilibrium · Very Hard

18. The pOH of 1.0 × 10−4 M NaOH at 25 °C is:

  • A. 10
  • B. 3
  • C. 7
  • D. 4
Answer: D
Explanation: NaOH is a strong base, so [OH−] = 10−4 M and pOH = 4.

NEET Chemistry · Equilibrium · Very Hard

19. The conjugate base of H2PO4− is:

  • A. H3PO4
  • B. PO4³−
  • C. H3O+
  • D. HPO4²−
Answer: D
Explanation: Removal of one proton from H2PO4− gives HPO4²−.

NEET Chemistry · Equilibrium · Very Hard

20. Which species is amphiprotic?

  • A. Na+
  • B. HCO3−
  • C. NH4+
  • D. Cl−
Answer: B
Explanation: HCO3− can donate H+ to form CO3²− or accept H+ to form H2CO3.

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Equilibrium Practice Tests

Equilibrium Frequently Asked Questions

What is the difference between Q and K?

The reaction quotient Q uses the current composition; K describes the equilibrium composition at a fixed temperature. Comparing Q with K predicts the direction of shift.

Does a catalyst change the equilibrium constant?

No. A catalyst changes the rate at which equilibrium is reached but does not change the equilibrium constant at a given temperature.

How should I practise ionic equilibrium?

Work systematically with species, approximations, charge or mass balance when needed, and verify that the final concentration or pH is chemically reasonable.

Continue with another chapter to build connected concepts and strengthen your overall Chemistry preparation.

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