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Equilibrium JEE Main MCQ & Practice Questions

Practice Equilibrium MCQs for JEE Main Chemistry with answers, explanations, chapter revision and related ChemNexa mock tests.

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Equilibrium Revision Guide for JEE Main 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 JEE Main 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 JEE Main 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.

JEE Main Chemistry · Equilibrium · Very Very Hard

1. Henderson-Hasselbalch equation for an acidic buffer is:

  • A. pH=pKa+log([salt]/[acid])
  • B. pH=pKa-log([salt]/[acid])
  • C. pH=pKb+log([acid]/[salt])
  • D. pH=14-pKa always
Answer: A
Explanation: This relates buffer pH to pKa and conjugate-base/acid ratio.

JEE Main Chemistry · Equilibrium · Very Very Hard

2. JEE Main advanced: Henderson-Hasselbalch equation for an acidic buffer is:

  • A. pH=pKb+log([acid]/[salt])
  • B. pH=14-pKa always
  • C. pH=pKa+log([salt]/[acid])
  • D. pH=pKa-log([salt]/[acid])
Answer: C
Explanation: This relates buffer pH to pKa and conjugate-base/acid ratio.

JEE Main Chemistry · Equilibrium · Very Very Hard

3. Consider: (I) catalyst does not change K; (II) higher pressure favours fewer gas moles; (III) pH+pOH=14 at 25°C; (IV) common-ion effect suppresses weak-electrolyte ionisation. Which are correct?

  • A. I, III and IV only
  • B. I, II, III and IV
  • C. I and II only
  • D. II and III only
Answer: B
Explanation: All four statements are correct.

JEE Main Chemistry · Equilibrium · Very Very Hard

4. JEE Main advanced: Consider: (I) catalyst does not change K; (II) higher pressure favours fewer gas moles; (III) pH+pOH=14 at 25°C; (IV) common-ion effect suppresses weak-electrolyte ionisation. Which are correct?

  • A. I and II only
  • B. II and III only
  • C. I, III and IV only
  • D. I, II, III and IV
Answer: D
Explanation: All four statements are correct.

JEE Main Chemistry · Equilibrium · Very Very Hard

5. For gaseous equilibria, Kp and Kc are related by:

  • A. Kp=Kc+RT
  • B. Kp=Kc(RT)^-1 always
  • C. Kp=Kc(RT)^Δn
  • D. Kp=Kc/RT
Answer: C
Explanation: Kp=Kc(RT)^Δn, where Δn is gaseous product moles minus reactant moles.

JEE Main Chemistry · Equilibrium · Very Very Hard

6. Le Chatelier's principle predicts that an equilibrium shifts to:

  • A. Oppose an imposed change
  • B. Always favour products
  • C. Always favour reactants
  • D. Stop the reaction
Answer: A
Explanation: The equilibrium responds in a direction that counteracts the disturbance.

JEE Main Chemistry · Equilibrium · Very Very Hard

7. At 25°C, pH+pOH equals:

  • A. 14
  • B. 7
  • C. 1
  • D. 0
Answer: A
Explanation: pH+pOH=pKw=14.

JEE Main Chemistry · Equilibrium · Very Very Hard

8. For a conjugate acid-base pair at 25°C:

  • A. Ka=Kb always
  • B. Ka×Kb=Kw
  • C. Ka+Kb=Kw
  • D. Ka/Kb=Kw
Answer: B
Explanation: The product of conjugate acid/base constants equals Kw.

JEE Main Chemistry · Equilibrium · Very Very Hard

9. Common-ion effect generally:

  • A. Suppresses ionisation of a weak electrolyte
  • B. Increases ionisation infinitely
  • C. Changes temperature
  • D. Destroys equilibrium
Answer: A
Explanation: Adding a common ion shifts the ionisation equilibrium backward.

JEE Main Chemistry · Equilibrium · Very Very Hard

10. A buffer solution resists changes in:

  • A. Temperature
  • B. Volume
  • C. Density
  • D. pH
Answer: D
Explanation: Buffers minimise pH changes after small additions of acid or base.

JEE Main Chemistry · Equilibrium · Very Very Hard

11. Solubility product Ksp applies to:

  • A. Sparingly soluble salts
  • B. Only highly soluble salts
  • C. Gases only
  • D. Strong acids only
Answer: A
Explanation: Ksp describes dissolution equilibrium of a sparingly soluble ionic solid.

JEE Main Chemistry · Equilibrium · Very Very Hard

12. JEE Main advanced: For gaseous equilibria, Kp and Kc are related by:

  • A. Kp=Kc(RT)^Δn
  • B. Kp=Kc/RT
  • C. Kp=Kc+RT
  • D. Kp=Kc(RT)^-1 always
Answer: A
Explanation: Kp=Kc(RT)^Δn, where Δn is gaseous product moles minus reactant moles.

JEE Main Chemistry · Equilibrium · Very Very Hard

13. JEE Main advanced: Le Chatelier's principle predicts that an equilibrium shifts to:

  • A. Always favour reactants
  • B. Stop the reaction
  • C. Oppose an imposed change
  • D. Always favour products
Answer: C
Explanation: The equilibrium responds in a direction that counteracts the disturbance.

JEE Main Chemistry · Equilibrium · Very Very Hard

14. JEE Main advanced: At 25°C, pH+pOH equals:

  • A. 1
  • B. 0
  • C. 14
  • D. 7
Answer: C
Explanation: pH+pOH=pKw=14.

JEE Main Chemistry · Equilibrium · Very Very Hard

15. JEE Main advanced: For a conjugate acid-base pair at 25°C:

  • A. Ka+Kb=Kw
  • B. Ka/Kb=Kw
  • C. Ka=Kb always
  • D. Ka×Kb=Kw
Answer: D
Explanation: The product of conjugate acid/base constants equals Kw.

JEE Main Chemistry · Equilibrium · Very Very Hard

16. JEE Main advanced: Common-ion effect generally:

  • A. Changes temperature
  • B. Destroys equilibrium
  • C. Suppresses ionisation of a weak electrolyte
  • D. Increases ionisation infinitely
Answer: C
Explanation: Adding a common ion shifts the ionisation equilibrium backward.

JEE Main Chemistry · Equilibrium · Very Very Hard

17. JEE Main advanced: A buffer solution resists changes in:

  • A. Density
  • B. pH
  • C. Temperature
  • D. Volume
Answer: B
Explanation: Buffers minimise pH changes after small additions of acid or base.

JEE Main Chemistry · Equilibrium · Very Very Hard

18. JEE Main advanced: Solubility product Ksp applies to:

  • A. Gases only
  • B. Strong acids only
  • C. Sparingly soluble salts
  • D. Only highly soluble salts
Answer: C
Explanation: Ksp describes dissolution equilibrium of a sparingly soluble ionic solid.

JEE Main Chemistry · Equilibrium · Very Very Hard

19. Equilibrium reasoning: For gaseous equilibria, Kp and Kc are related by:

  • A. Kp=Kc+RT
  • B. Kp=Kc(RT)^-1 always
  • C. Kp=Kc(RT)^Δn
  • D. Kp=Kc/RT
Answer: C
Explanation: Kp=Kc(RT)^Δn, where Δn is gaseous product moles minus reactant moles.

JEE Main Chemistry · Equilibrium · Very Very Hard

20. Equilibrium reasoning: Le Chatelier's principle predicts that an equilibrium shifts to:

  • A. Oppose an imposed change
  • B. Always favour products
  • C. Always favour reactants
  • D. Stop the reaction
Answer: A
Explanation: The equilibrium responds in a direction that counteracts the disturbance.

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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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