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Equilibrium JENPAS MCQ & Practice Questions

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

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Equilibrium Revision Guide for JENPAS 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 JENPAS 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 JENPAS 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.

JENPAS Chemistry · Equilibrium · Very Very Hard

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

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

JENPAS Chemistry · Equilibrium · Very Very Hard

2. JENPAS advanced: The Henderson-Hasselbalch equation for an acidic buffer is:

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

JENPAS Chemistry · Equilibrium · Very Very Hard

3. For gaseous equilibrium, Kp and Kc are related by:

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

JENPAS Chemistry · Equilibrium · Very Very Hard

4. Le Chatelier's principle predicts that an equilibrium responds to disturbance by:

  • A. Always shifting to products
  • B. Always shifting to reactants
  • C. Stopping the reaction
  • D. Shifting to oppose the disturbance
Answer: D
Explanation: An equilibrium shifts to counteract imposed change.

JENPAS Chemistry · Equilibrium · Very Very Hard

5. Increasing concentration of a reactant generally shifts equilibrium toward:

  • A. No direction always
  • B. Only gases
  • C. Products
  • D. Reactants
Answer: C
Explanation: The system consumes part of the added reactant.

JENPAS Chemistry · Equilibrium · Very Very Hard

6. Removing a product generally shifts equilibrium toward:

  • A. Catalyst formation
  • B. Products
  • C. Reactants
  • D. No change
Answer: B
Explanation: The system forms more product to oppose removal.

JENPAS Chemistry · Equilibrium · Very Very Hard

7. Increasing pressure favours the side of a gaseous equilibrium with:

  • A. Fewer moles of gas
  • B. More moles of gas
  • C. More solids
  • D. More liquids only
Answer: A
Explanation: Higher pressure is opposed by shifting toward fewer gaseous particles.

JENPAS Chemistry · Equilibrium · Very Very Hard

8. The conjugate base of H2CO3 is:

  • A. H3O+
  • B. OH-
  • C. HCO3-
  • D. CO3^2-
Answer: C
Explanation: Loss of one proton from H2CO3 gives HCO3-.

JENPAS Chemistry · Equilibrium · Very Very Hard

9. Common-ion effect generally:

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

JENPAS Chemistry · Equilibrium · Very Very Hard

10. Solubility product Ksp applies to:

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

JENPAS Chemistry · Equilibrium · Very Very Hard

11. Consider: (I) catalyst does not change K; (II) increasing 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.

JENPAS Chemistry · Equilibrium · Very Very Hard

12. JENPAS advanced: For gaseous equilibrium, Kp and Kc are related by:

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

JENPAS Chemistry · Equilibrium · Very Very Hard

13. JENPAS advanced: Le Chatelier's principle predicts that an equilibrium responds to disturbance by:

  • A. Stopping the reaction
  • B. Shifting to oppose the disturbance
  • C. Always shifting to products
  • D. Always shifting to reactants
Answer: B
Explanation: An equilibrium shifts to counteract imposed change.

JENPAS Chemistry · Equilibrium · Very Very Hard

14. JENPAS advanced: Increasing concentration of a reactant generally shifts equilibrium toward:

  • A. Products
  • B. Reactants
  • C. No direction always
  • D. Only gases
Answer: A
Explanation: The system consumes part of the added reactant.

JENPAS Chemistry · Equilibrium · Very Very Hard

15. JENPAS advanced: Removing a product generally shifts equilibrium toward:

  • A. Reactants
  • B. No change
  • C. Catalyst formation
  • D. Products
Answer: D
Explanation: The system forms more product to oppose removal.

JENPAS Chemistry · Equilibrium · Very Very Hard

16. JENPAS advanced: Increasing pressure favours the side of a gaseous equilibrium with:

  • A. More solids
  • B. More liquids only
  • C. Fewer moles of gas
  • D. More moles of gas
Answer: C
Explanation: Higher pressure is opposed by shifting toward fewer gaseous particles.

JENPAS Chemistry · Equilibrium · Very Very Hard

17. JENPAS advanced: The conjugate base of H2CO3 is:

  • A. HCO3-
  • B. CO3^2-
  • C. H3O+
  • D. OH-
Answer: A
Explanation: Loss of one proton from H2CO3 gives HCO3-.

JENPAS Chemistry · Equilibrium · Very Very Hard

18. JENPAS advanced: Common-ion effect generally:

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

JENPAS Chemistry · Equilibrium · Very Very Hard

19. JENPAS advanced: Solubility product Ksp applies to:

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

JENPAS Chemistry · Equilibrium · Very Very Hard

20. JENPAS advanced: Consider: (I) catalyst does not change K; (II) increasing 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.

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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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