Chemical Equilibrium

What to remember

• At equilibrium the forward and reverse rates are equal, so the concentrations stay constant. The state is dynamic and can be reached from either side.
• For aA + bB ⇌ cC + dD: Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ. Pure solids and pure liquids are left out. Kp = Kc(RT)^Δn, where Δn is the change in moles of gas.
• Compare Q with K. If Q < K the reaction goes forward, if Q > K it goes backward, and if Q = K it is at equilibrium. K depends only on temperature.
• Le Chatelier's principle: a system at equilibrium opposes any change in concentration, pressure, volume or temperature. A catalyst does not shift the equilibrium.

Equilibrium in physical processes

Equilibrium is the state in which the measurable properties of a system no longer change with time. In a physical equilibrium only the physical state changes, not the chemical composition.

• At the melting point of ice at 1 atm, the rate of melting equals the rate of freezing.
• In a closed vessel, water evaporates until the rate of evaporation equals the rate of condensation. The pressure of the vapour then is the equilibrium vapour pressure. It depends only on the temperature.
• In a saturated solution the rate of dissolution equals the rate of crystallisation.
• Henry's law (qualitative): at constant temperature the mass of a gas dissolved in a given volume of liquid is directly proportional to the pressure of the gas above the liquid. A soda bottle is opened and the pressure falls, so dissolved CO₂ escapes as bubbles. Solubility of a gas in a liquid decreases when the temperature rises.
• Equilibrium in a physical process is possible only in a closed system at constant temperature.

Dynamic nature, law of mass action and the equilibrium constant

At chemical equilibrium the forward and reverse reactions continue at the same rate. Nothing seems to change, but the reactions have not stopped. This is called dynamic equilibrium. It is shown by using a labelled substance: if D₂ is added to an equilibrium mixture of H₂, I₂ and HI, deuterium soon appears in HI molecules and in H₂.

Features of chemical equilibrium:

• It can be approached from either side: starting from reactants only, or from products only.
• At constant temperature, the concentrations of all species stay constant.
• It is reached only in a closed system.
• A catalyst makes the equilibrium be reached sooner but does not change the equilibrium composition.
• At a given temperature the equilibrium constant has one value.

Law of mass action: the rate of a chemical reaction is proportional to the product of the active masses (molar concentrations) of the reactants, each raised to its stoichiometric coefficient.

Law of chemical equilibrium: for the reaction aA + bB ⇌ cC + dD at a given temperature,

Kc = [C]ᶜ [D]ᵈ / ([A]ᵃ [B]ᵇ)

Concentrations are in mol L⁻¹. For gaseous reactions, partial pressures can be used:

Kp = (pC)ᶜ (pD)ᵈ / ((pA)ᵃ (pB)ᵇ)

The unit of Kc is (mol L⁻¹)^Δn and that of Kp is (bar)^Δn (or atm^Δn). K has no unit when Δn = 0.

Relation between Kp and Kc

For a gas, pV = nRT, so p = (n/V)RT = CRT. Substituting for each gas gives

Kp = Kc (RT)^Δn

Here Δn = (moles of gaseous products) − (moles of gaseous reactants), and R must be consistent with the unit of pressure: R = 0.0831 L bar K⁻¹ mol⁻¹ or 0.0821 L atm K⁻¹ mol⁻¹. If Δn = 0, then Kp = Kc.

Worked example 1. For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 0.1 mol L⁻¹ at 1000 K. Find Kp (R = 0.082 L atm K⁻¹ mol⁻¹).

Δn = 2 − 1 = +1. Kp = 0.1 × (0.082 × 1000) = 0.1 × 82 = 8.2 atm.

Worked example 2. For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 36. One mole each of H₂ and I₂ are placed in a 1 L vessel. Find the equilibrium concentrations.

Let x mol of H₂ react. Then [H₂] = [I₂] = 1 − x and [HI] = 2x.

Kc = (2x)² / (1 − x)² = 36, so 2x / (1 − x) = 6, giving 2x = 6 − 6x and x = 0.75.

[HI] = 1.5 mol L⁻¹ and [H₂] = [I₂] = 0.25 mol L⁻¹. Check: (1.5)²/(0.25)² = 36.

Homogeneous and heterogeneous equilibria

• Homogeneous equilibrium: all species are in the same phase. Examples: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l); Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq).
• Heterogeneous equilibrium: species are in more than one phase. Examples: CaCO₃(s) ⇌ CaO(s) + CO₂(g); Ni(s) + 4CO(g) ⇌ Ni(CO)₄(g).

The concentration of a pure solid or pure liquid is constant, so it is taken as 1 and omitted from the expression.

For CaCO₃, the amounts of solid CaCO₃ and CaO do not change the value of pCO₂ at a given temperature, provided some of each solid is present.

Applications of the equilibrium constant

1. Extent of reaction. A large K means mostly products at equilibrium.

K tells how far a reaction goes, not how fast it goes.

2. Direction of reaction: reaction quotient Q. Q has the same form as K, but it is written with the concentrations (or pressures) at any instant.

Worked example 3. For H₂ + I₂ ⇌ 2HI, Kc = 36. In a mixture, [H₂] = [I₂] = 0.5 mol L⁻¹ and [HI] = 4 mol L⁻¹. Q = 4² / (0.5 × 0.5) = 16/0.25 = 64. Since Q > K, the reaction shifts backward.

3. Equilibrium concentrations. Write the initial concentrations, let x change, write the equilibrium expression, and solve for x (as in worked example 2).

Relation between K, Q and Gibbs energy

ΔG = ΔG° + RT ln Q. At equilibrium ΔG = 0 and Q = K, hence

ΔG° = −RT ln K = −2.303 RT log K

Also, ΔG = RT ln (Q/K). When Q < K, ΔG is negative and the forward reaction is spontaneous. When Q > K, ΔG is positive and the backward reaction is spontaneous.

Worked example 4. For a reaction, K = 1000 at 300 K. ΔG° = −2.303 × 8.314 × 300 × log 1000 = −2.303 × 8.314 × 300 × 3 = −17232 J, about −17.2 kJ mol⁻¹.

Le Chatelier's principle

If a system at equilibrium is disturbed by a change of concentration, pressure, volume or temperature, it shifts so as to partly cancel the effect of the change.

• If Δn = 0 (for example H₂ + I₂ ⇌ 2HI), pressure has no effect on the position of equilibrium.
• For an exothermic reaction, K decreases when the temperature rises. For an endothermic reaction, K increases when the temperature rises.
• Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.4 kJ mol⁻¹. High pressure (about 200 atm) favours NH₃ because 4 mol of gas become 2 mol. Low temperature favours NH₃ because the reaction is exothermic, but it would be too slow, so a moderate temperature (about 700 K) with an iron catalyst (molybdenum as promoter) is used. Removing NH₃ as it forms also shifts the equilibrium forward.
• N₂(g) + O₂(g) ⇌ 2NO(g) is endothermic (ΔH = +180.7 kJ mol⁻¹) and has Δn = 0. High temperature increases the yield of NO, and pressure has no effect.
• PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is endothermic with Δn = +1. High temperature and low pressure favour dissociation.
• Ice ⇌ water: ice is less dense than water, so water has a smaller volume. Increasing pressure favours melting.
• In Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq), adding Fe³⁺ or SCN⁻ deepens the red colour, because the equilibrium shifts forward.

Manipulation of the equilibrium constant

Worked example 5. For A ⇌ B, K = 4. Then B ⇌ A has K = 1/4 = 0.25; 2A ⇌ 2B has K = 4² = 16; ½A ⇌ ½B has K = 4^½ = 2.

Worked example 6. If N₂ + O₂ ⇌ 2NO has K₁ and 2NO + O₂ ⇌ 2NO₂ has K₂, then N₂ + 2O₂ ⇌ 2NO₂ has K = K₁ × K₂.

The same rules link ΔG° values: adding reactions adds their ΔG° values, which multiplies the K values.

NEET traps

• 1. Equilibrium is dynamic. The reactions do not stop; their rates are equal.
• 2. Equal rates do not mean equal concentrations. At equilibrium the concentrations are constant, not equal.
• 3. A catalyst changes the rate of both directions equally. It does not change K or the equilibrium composition.
• 4. Only temperature changes the value of K. Concentration, pressure, volume and inert gas change Q or the position, not K.
• 5. Pure solids and liquids do not appear in K. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂].
• 6. Kp = Kc only if Δn = 0. R must match the unit of pressure.
• 7. An inert gas added at constant volume does not shift the equilibrium, but at constant pressure it does when Δn ≠ 0.
• 8. K tells the extent, not the speed, of a reaction.
• 9. Q is not K. Q can be calculated at any moment; K only at equilibrium.
• 10. For a reversed reaction K′ = 1/K; for doubled coefficients K′ = K², not 2K.

One-liners

• 1. At equilibrium, rate of forward reaction equals rate of reverse reaction.
• 2. Ice and water are in equilibrium at 273 K and 1 atm.
• 3. Equilibrium vapour pressure depends only on temperature.
• 4. Henry's law: gas solubility is proportional to its pressure at constant temperature.
• 5. Kc for H₂ + I₂ ⇌ 2HI is [HI]² / ([H₂][I₂]).
• 6. Kp = Kc(RT)^Δn.
• 7. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = pCO₂.
• 8. K > 10³ means the reaction goes nearly to completion.
• 9. Q < K: reaction goes forward. Q > K: it goes backward.
• 10. ΔG° = −2.303 RT log K.
• 11. Reverse reaction has K′ = 1/K, and adding reactions multiplies the K values.
• 12. Higher temperature favours the endothermic direction.
• 13. High pressure favours the side with fewer gas moles.
• 14. A catalyst does not change K.
• 15. Haber process uses about 200 atm, about 700 K and an iron catalyst.

Practice questions

1. At chemical equilibrium,

   1. the concentrations of reactants and products are equal
   2. the forward reaction has stopped
   3. the rate of the reverse reaction is zero
   4. the rates of the forward and reverse reactions are equal
   Answer

   D. the rates of the forward and reverse reactions are equal

   Equilibrium is dynamic: both reactions go on at equal rates, so concentrations stay constant.

2. The equilibrium constant Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is

   1. [N₂][H₂]³ / [NH₃]²
   2. [NH₃]² / ([N₂][H₂]³)
   3. [NH₃]² / ([N₂][H₂]²)
   4. [NH₃] / ([N₂][H₂])
   Answer

   B. [NH₃]² / ([N₂][H₂]³)

   Concentrations of products over reactants, each raised to its coefficient.

3. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the expression for Kc is

   1. [CO₂]
   2. [CaO][CO₂] / [CaCO₃]
   3. [CaO] / [CaCO₃]
   4. 1 / [CO₂]
   Answer

   A. [CO₂]

   Pure solids are omitted from the equilibrium constant expression.

4. For COCl₂(g) ⇌ CO(g) + Cl₂(g), Kc = 0.4 mol L⁻¹ at a temperature where RT = 25 L bar mol⁻¹. The value of Kp is

   1. 0.016 bar
   2. 100 bar
   3. 10 bar
   4. 0.4 bar
   Answer

   C. 10 bar

   Δn = +1, so Kp = Kc(RT) = 0.4 × 25 = 10 bar.

5. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Kc = 50 L mol⁻¹ at a temperature where RT = 25 L bar mol⁻¹. The value of Kp is

   1. 2 bar⁻¹
   2. 1250 bar⁻¹
   3. 50 bar⁻¹
   4. 25 bar⁻¹
   Answer

   A. 2 bar⁻¹

   Δn = 2 − 3 = −1, so Kp = Kc/(RT) = 50/25 = 2 bar⁻¹.

6. For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 49. The value of Kc for 2HI(g) ⇌ H₂(g) + I₂(g) at the same temperature is

   1. 7
   2. 49
   3. 1/7
   4. 1/49
   Answer

   D. 1/49

   For the reverse reaction K′ = 1/K = 1/49.

7. For H₂ + I₂ ⇌ 2HI, Kc = 50. A mixture has [H₂] = 0.2, [I₂] = 0.1 and [HI] = 1.0 mol L⁻¹. This mixture is

   1. shifting forward
   2. shifting backward
   3. at equilibrium
   4. at equilibrium only after a catalyst is added
   Answer

   C. at equilibrium

   Q = (1.0)²/(0.2 × 0.1) = 50 = Kc, so there is no net change.

8. A reaction has K = 10 at 300 K. Taking R = 8.314 J K⁻¹ mol⁻¹ and log 10 = 1, ΔG° is nearest to

   1. −57.4 kJ mol⁻¹
   2. −5.74 kJ mol⁻¹
   3. −2.50 kJ mol⁻¹
   4. +5.74 kJ mol⁻¹
   Answer

   B. −5.74 kJ mol⁻¹

   ΔG° = −2.303 RT log K = −2.303 × 8.314 × 300 × 1 = −5744 J.

9. The value of the equilibrium constant of a reaction changes with

   1. addition of a catalyst
   2. the initial concentration of a reactant
   3. temperature
   4. pressure
   Answer

   C. temperature

   K depends only on temperature.

10. An inert gas is added at constant volume to the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The equilibrium

    1. shifts backward
    2. does not shift
    3. shifts and Kc increases
    4. shifts forward
    Answer

    B. does not shift

    At constant volume the concentrations (partial pressures) of reacting gases are unchanged.

11. The yield of ammonia in the Haber process is favoured by

    1. high pressure and low temperature
    2. low pressure and high temperature
    3. high pressure and high temperature
    4. low pressure and low temperature
    Answer

    A. high pressure and low temperature

    4 mol of gas become 2 mol, and the forward reaction is exothermic.

12. At equilibrium in a 1 L vessel, [NO₂] = 0.1 mol L⁻¹ and [N₂O₄] = 0.4 mol L⁻¹ for 2NO₂ ⇌ N₂O₄. The value of Kc is

    1. 0.25 L mol⁻¹
    2. 4 L mol⁻¹
    3. 0.025 L mol⁻¹
    4. 40 L mol⁻¹
    Answer

    D. 40 L mol⁻¹

    Kc = [N₂O₄]/[NO₂]² = 0.4/(0.1)² = 40.

13. For A ⇌ B, K₁ = 2, and for B ⇌ C, K₂ = 5. The equilibrium constant for A ⇌ C is

    1. 7
    2. 2.5
    3. 10
    4. 0.4
    Answer

    C. 10

    Adding reactions multiplies the constants: 2 × 5 = 10.

14. Which of the following is a heterogeneous equilibrium?

    1. CaCO₃(s) ⇌ CaO(s) + CO₂(g)
    2. CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
    3. H₂(g) + I₂(g) ⇌ 2HI(g)
    4. N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
    Answer

    A. CaCO₃(s) ⇌ CaO(s) + CO₂(g)

    A solid and a gas are present, so there is more than one phase.

15. Bubbles of gas appear when a soda bottle is opened because

    1. CO₂ reacts with water to give a gas
    2. the pressure above the liquid rises and the solubility of CO₂ increases
    3. the temperature of the liquid falls and CO₂ is released
    4. the pressure above the liquid falls and the solubility of CO₂ decreases
    Answer

    D. the pressure above the liquid falls and the solubility of CO₂ decreases

    Henry's law: the solubility of a gas is proportional to its pressure.

16. For a reaction at 300 K, K = 100 and Q = 1. Taking R = 8.314 J K⁻¹ mol⁻¹ and log 100 = 2, ΔG is nearest to

    1. +11.5 kJ mol⁻¹
    2. −11.5 kJ mol⁻¹
    3. 0 kJ mol⁻¹
    4. −23.0 kJ mol⁻¹
    Answer

    B. −11.5 kJ mol⁻¹

    ΔG = 2.303 RT log (Q/K) = 2.303 × 8.314 × 300 × (−2) = −11488 J.

17. Given below are two statements: Statement I: Chemical equilibrium is dynamic in nature. Statement II: At equilibrium the forward and reverse reactions stop.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Both reactions continue at equal rates, so Statement II is incorrect.

18. Given below are two statements: Statement I: Kp = Kc for H₂(g) + I₂(g) ⇌ 2HI(g). Statement II: Δn = 0 for this reaction.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Kp = Kc(RT)^Δn, and with Δn = 0 the two constants are equal.

19. Given below are two statements: Statement I: Addition of a catalyst increases the equilibrium constant. Statement II: A catalyst increases the rates of forward and reverse reactions equally.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    A catalyst does not change K; Statement II is true.

20. Given below are two statements: Statement I: When an inert gas is added at constant pressure to N₂ + 3H₂ ⇌ 2NH₃ at equilibrium, the equilibrium shifts backward. Statement II: Addition of inert gas at constant pressure increases the volume, which favours the side with more moles of gas.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    The volume increases, partial pressures fall, and the side with 4 mol (reactants) is favoured.

21. Which of the following statements are correct? 1. For a reaction with coefficients doubled, K′ = K². 2. For the reverse reaction, K′ = 1/K. 3. On adding two reactions, K = K₁ + K₂.

    1. 1, 2 and 3
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1 and 2 only
    Answer

    D. 1 and 2 only

    Adding reactions multiplies their constants, so statement 3 is incorrect.

22. Given below are two statements: Statement I: The equilibrium constant of an endothermic reaction decreases when the temperature is raised. Statement II: The equilibrium constant of an exothermic reaction increases when the temperature is raised.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    K of an endothermic reaction rises and K of an exothermic reaction falls with temperature.

23. Which of the following statements are correct? 1. If Q < K, the reaction proceeds in the forward direction. 2. If Q > K, the reaction proceeds in the forward direction. 3. If Q = K, the system is at equilibrium.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    C. 1 and 3 only

    If Q > K the reaction goes backward, so statement 2 is wrong.

24. Given below are two statements: Statement I: For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = pCO₂. Statement II: Pure solids are not included in the equilibrium constant expression.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Solids have constant concentration and are taken as 1.

25. Given below are two statements: Statement I: The equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction. Statement II: The equilibrium constants of the forward and reverse reactions are equal.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    K′ = 1/K, so the two are not equal unless K = 1.

26. Given below are two statements: Statement I: Increase in pressure increases the yield of NO in N₂(g) + O₂(g) ⇌ 2NO(g). Statement II: When Δn = 0, pressure has no effect on the position of equilibrium.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Δn = 0 here, so pressure has no effect; Statement I is incorrect.

27. Assertion (A): Pressure has no effect on the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g). Reason (R): The number of moles of gas is the same on both sides.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Equal gas moles on both sides means pressure change does not favour either side.

28. Assertion (A): Dissolved CO₂ escapes when a soda bottle is opened. Reason (R): The solubility of a gas in a liquid increases with rise of temperature.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    A is true (pressure falls), but gas solubility generally decreases when temperature rises, so R is false.

29. Assertion (A): A catalyst does not change the value of the equilibrium constant. Reason (R): The Haber process uses an iron catalyst.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both are true, but the use of iron in the Haber process does not explain why K is unchanged.

30. Assertion (A): The equilibrium constant of an exothermic reaction increases when the temperature is raised. Reason (R): High temperature favours the endothermic direction.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    A is false: K falls for an exothermic reaction. R is true.

31. Assertion (A): Addition of an inert gas at constant volume does not disturb the equilibrium. Reason (R): An inert gas changes the partial pressures of the reacting gases.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    A is true, but the inert gas does not change the partial pressures at constant volume, so R is false.

32. Assertion (A): ΔG = 0 for a system at equilibrium. Reason (R): At equilibrium Q = K, and ΔG = RT ln (Q/K).

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Q = K gives ln 1 = 0, so ΔG = 0.

33. Match List I with List II. List I: a. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) b. H₂(g) + I₂(g) ⇌ 2HI(g) c. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) d. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) List II: i. Kp = Kc ii. Kp = Kc(RT) iii. Kp = Kc(RT)⁻² iv. Kp = Kc(RT)⁻¹

    1. a-iv, b-i, c-ii, d-iii
    2. a-iii, b-i, c-ii, d-iv
    3. a-iii, b-ii, c-i, d-iv
    4. a-i, b-iii, c-ii, d-iv
    Answer

    B. a-iii, b-i, c-ii, d-iv

    Δn is −2, 0, +1 and −1 respectively.

34. Match List I with List II. List I: a. Q < K b. Q > K c. Q = K d. K > 10³ List II: i. Net backward reaction ii. System at equilibrium iii. Net forward reaction iv. Reaction goes nearly to completion

    1. a-ii, b-i, c-iii, d-iv
    2. a-iii, b-i, c-ii, d-iv
    3. a-iii, b-i, c-iv, d-ii
    4. a-i, b-iii, c-ii, d-iv
    Answer

    B. a-iii, b-i, c-ii, d-iv

    Q < K gives forward change, Q > K backward change, Q = K equilibrium.

35. Match List I with List II. List I: a. Increase of pressure b. Increase of temperature c. Addition of a catalyst d. Addition of NH₃ List II: i. Shifts forward ii. Shifts backward and K decreases iii. No change in position iv. Shifts backward and K is unchanged

    1. a-ii, b-i, c-iii, d-iv
    2. a-iii, b-ii, c-i, d-iv
    3. a-i, b-iv, c-iii, d-ii
    4. a-i, b-ii, c-iii, d-iv
    Answer

    D. a-i, b-ii, c-iii, d-iv

    For N₂ + 3H₂ ⇌ 2NH₃ (exothermic): higher pressure favours NH₃; higher temperature shifts backward and lowers K; a catalyst does not shift it; extra NH₃ shifts it backward without changing K.

36. Match List I with List II. List I: a. CaCO₃(s) ⇌ CaO(s) + CO₂(g) b. Ni(s) + 4CO(g) ⇌ Ni(CO)₄(g) c. 2NO₂(g) ⇌ N₂O₄(g) d. H₂O(l) ⇌ H₂O(g) List II: i. [CO₂] ii. [Ni(CO)₄] / [CO]⁴ iii. [N₂O₄] / [NO₂]² iv. [H₂O(g)]

    1. a-iv, b-ii, c-iii, d-i
    2. a-i, b-ii, c-iii, d-iv
    3. a-i, b-ii, c-iv, d-iii
    4. a-ii, b-i, c-iii, d-iv
    Answer

    B. a-i, b-ii, c-iii, d-iv

    Pure solids and pure liquids are omitted from Kc.

37. Match List I with List II. List I: a. Ice ⇌ water b. Water ⇌ water vapour in a closed vessel c. Sugar(s) ⇌ sugar in a saturated solution d. CO₂(g) ⇌ CO₂(aq) List II: i. Equilibrium vapour pressure ii. Solubility at a given temperature iii. Both phases coexist at 273 K and 1 atm iv. Dissolved amount proportional to gas pressure

    1. a-iii, b-i, c-iv, d-ii
    2. a-iii, b-i, c-ii, d-iv
    3. a-iii, b-ii, c-i, d-iv
    4. a-i, b-iii, c-ii, d-iv
    Answer

    B. a-iii, b-i, c-ii, d-iv

    Each physical equilibrium is linked to its constant property.

38. One mole each of A(g) and B(g) are placed in a 1 L vessel and react: A + B ⇌ C + D, Kc = 4. At equilibrium [C] is nearest to

    1. 0.67 mol L⁻¹
    2. 0.33 mol L⁻¹
    3. 0.50 mol L⁻¹
    4. 0.80 mol L⁻¹
    Answer

    A. 0.67 mol L⁻¹

    x²/(1 − x)² = 4, so x/(1 − x) = 2 and x = 2/3 = 0.67.

39. For PCl₅ ⇌ PCl₃ + Cl₂, Kc = 0.04 mol L⁻¹. At equilibrium [PCl₃] = [Cl₂] = 0.2 mol L⁻¹. The concentration of PCl₅ is

    1. 0.04 mol L⁻¹
    2. 0.2 mol L⁻¹
    3. 1.0 mol L⁻¹
    4. 5.0 mol L⁻¹
    Answer

    C. 1.0 mol L⁻¹

    [PCl₅] = (0.2 × 0.2)/0.04 = 1.0 mol L⁻¹.

40. For N₂O₄(g) ⇌ 2NO₂(g), at equilibrium the partial pressures are 0.5 atm for N₂O₄ and 2 atm for NO₂. The value of Kp is

    1. 8 atm
    2. 0.125 atm
    3. 4 atm
    4. 2 atm
    Answer

    A. 8 atm

    Kp = (pNO₂)²/pN₂O₄ = 4/0.5 = 8 atm.

41. For H₂ + I₂ ⇌ 2HI, Kc = 36. A mixture has [H₂] = [I₂] = 1 mol L⁻¹ and [HI] = 4 mol L⁻¹. The reaction quotient and the direction are

    1. Q = 4; reaction goes forward
    2. Q = 16; reaction goes forward
    3. Q = 16; reaction goes backward
    4. Q = 4; reaction goes backward
    Answer

    B. Q = 16; reaction goes forward

    Q = 4²/(1 × 1) = 16 < 36, so more HI forms.

42. In PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), ΔH is positive. Dissociation of PCl₅ is increased by

    1. adding Cl₂
    2. raising the pressure
    3. raising the temperature
    4. adding a catalyst
    Answer

    C. raising the temperature

    Heat favours the endothermic (forward) direction; more pressure, extra Cl₂ and a catalyst do not increase dissociation.

43. Argon is added to an equilibrium mixture of N₂, H₂ and NH₃ held in a cylinder at constant pressure with a movable piston. The equilibrium

    1. shifts and Kc becomes larger
    2. shifts towards NH₃
    3. does not shift
    4. shifts towards N₂ and H₂
    Answer

    D. shifts towards N₂ and H₂

    The volume increases, so partial pressures fall and the side with more moles of gas (N₂ + H₂) is favoured; K is unchanged.

44. For N₂ + O₂ ⇌ 2NO, K₁ = 4 × 10⁻⁴ and for 2NO + O₂ ⇌ 2NO₂, K₂ = 5 × 10². The equilibrium constant for N₂ + 2O₂ ⇌ 2NO₂ is

    1. 0.2
    2. 2 × 10⁻³
    3. 8 × 10⁻⁷
    4. 1.25 × 10⁶
    Answer

    A. 0.2

    Adding the two reactions: K = K₁ × K₂ = 4 × 10⁻⁴ × 5 × 10² = 0.2.