Chemical Equilibrium MCQs 2026

19 questions with detailed answers · 6 from past papers · 2 quiz batches available

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Page 1 of 1 Questions 110 of 19
  1. Q1 medium

    Increasing pressure has no effect on gaseous equilibria where

    1. A Reactant moles exceed product moles
    2. B Product moles exceed reactant moles
    3. C The reaction is exothermic
    4. D The number of gas moles is equal on both sides
    💡 Explanation:

    When gas moles are equal on both sides, changing pressure does not shift the equilibrium position.

  2. Q2 Past Paper · PPSC/FPSC/NTS medium

    The degree of dissociation of a weak electrolyte at equilibrium is affected by

    1. A Nothing at all
    2. B Only the container's shape
    3. C Catalysts only
    4. D Dilution (per Ostwald's dilution law)
    💡 Explanation:

    Ostwald's dilution law shows that dissociation of a weak electrolyte increases upon dilution.

  3. Q3 medium

    Which of the following equilibria is heterogeneous

    1. A N2(g) + 3H2(g) ⇌ 2NH3(g)
    2. B H2(g) + I2(g) ⇌ 2HI(g)
    3. C 2SO2(g) + O2(g) ⇌ 2SO3(g)
    4. D CaCO3(s) ⇌ CaO(s) + CO2(g)
    💡 Explanation:

    This equilibrium involves solids and a gas in different phases, making it heterogeneous.

  4. Q4 Past Paper · PPSC/FPSC/NTS medium

    For an equilibrium at constant temperature, changing the volume of the container affects

    1. A K only, not the position
    2. B Nothing at all
    3. C Only solid-state equilibria
    4. D The position of equilibrium, but not K, for gaseous reactions with unequal moles
    💡 Explanation:

    Volume changes shift equilibrium position for unequal-mole gas reactions, but K itself stays constant at fixed temperature.

  5. Q5 medium

    The equilibrium constant expression is written using the

    1. A Coefficients of reactants only
    2. B Rate law of the reaction
    3. C Activation energy
    4. D Balanced chemical equation's stoichiometric coefficients as exponents
    💡 Explanation:

    Equilibrium expressions raise each species' concentration to the power of its coefficient in the balanced equation.

  6. Q6 medium

    Which condition favors maximum yield of ammonia in the Haber process according to Le Chatelier's principle

    1. A Low pressure and high temperature
    2. B High pressure and moderately low temperature
    3. C Low pressure and low temperature
    4. D High temperature and low pressure only
    💡 Explanation:

    High pressure favors fewer gas moles (NH3), while a moderate temperature balances yield with reasonable reaction rate.

  7. Q7 Past Paper · PPSC/FPSC/NTS medium

    In practice, industrial processes like the Haber process use a compromise temperature to balance

    1. A Yield (favored by low T) and reaction rate (favored by high T)
    2. B Cost of catalyst and volume of reactor
    3. C Color of product and purity
    4. D Pressure and humidity only
    💡 Explanation:

    A moderate temperature is chosen so the reaction proceeds fast enough while still giving a reasonable yield.

  8. Q8 medium

    Which statement about reversible reactions is correct

    1. A They can proceed in both forward and backward directions
    2. B They go to completion in one direction only
    3. C They never reach equilibrium
    4. D They require a catalyst to be reversible
    💡 Explanation:

    Reversible reactions can proceed in both directions, eventually reaching a state of equilibrium.

  9. Q9 medium

    The equilibrium position of a weak acid dissociation in water is described by its

    1. A Acid dissociation constant, Ka
    2. B Solubility product, Ksp
    3. C Rate constant, k
    4. D Molar mass
    💡 Explanation:

    Ka quantifies the equilibrium extent of ionization for a weak acid in water.

  10. Q10 Past Paper · PPSC/FPSC/NTS hard

    A high value of Ka for an acid indicates that the acid is

    1. A Weak
    2. B Strong (highly dissociated)
    3. C Neutral
    4. D Not soluble in water
    💡 Explanation:

    A large Ka value shows the acid dissociates extensively, indicating greater acid strength.

  11. Q11 hard

    The equilibrium constant for the reverse of a reaction is related to the equilibrium constant of the forward reaction (K) by

    1. A Kreverse = K
    2. B Kreverse = -K
    3. C Kreverse = K^2
    4. D Kreverse = 1/K
    💡 Explanation:

    Reversing a reaction inverts its equilibrium constant, giving Kreverse = 1/K.

  12. Q12 hard

    If the equilibrium constant of reaction 1 is K1 and reaction 2 is K2, and reaction 3 is the sum of reactions 1 and 2, then K3 equals

    1. A K1 + K2
    2. B K1 × K2
    3. C K1 − K2
    4. D K1 / K2
    💡 Explanation:

    When reactions are added together, their equilibrium constants multiply to give the overall constant.

  13. Q13 Past Paper · PPSC/FPSC/NTS hard

    In a saturated solution of a sparingly soluble salt, the equilibrium exists between the

    1. A Gas and liquid phases only
    2. B Two different solutes
    3. C Undissolved solid and its dissolved ions
    4. D Solvent and container walls
    💡 Explanation:

    Saturated solution equilibrium is between the solid precipitate and its ions in solution.

  14. Q14 hard

    The value of the equilibrium constant for a reaction at a fixed temperature is independent of

    1. A The initial concentrations of reactants and products
    2. B Temperature
    3. C The nature of the reaction
    4. D The stoichiometry of the reaction
    💡 Explanation:

    K remains the same value regardless of the starting concentrations used, as long as temperature is fixed.

  15. Q15 hard

    Which of the following would increase the rate of both forward and reverse reactions equally without shifting equilibrium

    1. A Increasing reactant concentration
    2. B Removing product
    3. C Adding a catalyst
    4. D Increasing temperature
    💡 Explanation:

    A catalyst lowers activation energy for both directions equally, speeding equilibrium attainment without shifting it.

  16. Q16 Past Paper · PPSC/FPSC/NTS hard

    Equilibrium involving a solid and its saturated solution, such as sugar dissolving in water, is an example of

    1. A Chemical equilibrium only
    2. B Redox equilibrium
    3. C Physical equilibrium
    4. D Nuclear equilibrium
    💡 Explanation:

    Dissolution equilibria without a chemical reaction are classified as physical equilibria.

  17. Q17 hard

    For the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g), decreasing the pressure will shift equilibrium towards

    1. A The products (more moles of gas)
    2. B The reactants (fewer moles of gas)
    3. C No shift occurs
    4. D Only PCl5 remains
    💡 Explanation:

    Lower pressure favors the side with more gas moles, shifting equilibrium toward the products.

  18. Q18 hard

    Le Chatelier's principle can be applied to predict changes in equilibrium due to all of the following EXCEPT

    1. A Concentration changes
    2. B Pressure/volume changes
    3. C Changes in the amount of catalyst used
    4. D Temperature changes
    💡 Explanation:

    Catalysts affect reaction rate, not equilibrium position, so they fall outside Le Chatelier's predictions.

  19. Q19 hard

    At chemical equilibrium, the Gibbs free energy of the system is

    1. A Maximum
    2. B At a minimum
    3. C Increasing continuously
    4. D Equal to enthalpy
    💡 Explanation:

    The system reaches equilibrium at the point of minimum Gibbs free energy for the given conditions.