Equilibrium - MCQs with Explanations

1. In the lime kiln, the reversible reaction \(CaCO_3 \rightleftharpoons CaO + CO_2\) proceeds to completion because of:

• (a) high temperature
• (b) \(CO_2\) escapes
• (c) \(CaO\) is removed
• (d) low temperature.

Correct Answer: (b)

Explanation: According to the source, if \(CO_2\) escapes, the backward reaction cannot take place, pushing the reaction to completion.

2. In which one of the following reactions, \(K_p\) is less than \(K_c\)?

• (a) \(2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}\)
• (b) \(N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\)
• (c) \(PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}\)
• (d) \(H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}\)

Correct Answer: (b)

Explanation: The relationship is \(K_p = K_c(RT)^{\Delta n_g}\). \(K_p < K_c\) when \(\Delta n_g\) is negative. For reaction (b), \(\Delta n_g = 2 - (1+3) = -2\).

3. The equilibrium constants for the reactions \(Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}\) and \(Cu_{(s)} + 2Ag^{+}_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2Ag_{(s)}\) are \(K_1\) and \(K_2\) respectively. The equilibrium constant for \(Zn_{(s)} + 2Ag^{+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + 2Ag_{(s)}\) will be:

• (a) \(K_1 + K_2\)
• (b) \(K_1 \times K_2\)
• (c) \(K_1 / K_2\)
• (d) \(K_1 - K_2\)

Correct Answer: (b)

Explanation: The source states that the overall equilibrium constant is the product of equilibrium constants of step reactions.

4. For the system \(A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)}\) the equilibrium concentrations are \([A]=0.06\), \([B]=0.12\), \([C]=0.216\). The \(K_{eq}\) is:

• (a) 250
• (b) 416
• (c) \(4 \times 10^{-3}\)
• (d) 125

Correct Answer: (a)

Explanation: \(K = \frac{[C]}{[A][B]^2} = \frac{0.216}{0.06 \times (0.12)^2} = 250\).

5. If 0.2 mol of \(H_{2(g)}\) and 2.0 mol of \(S_{(s)}\) are mixed in a \(1\text{ dm}^3\) vessel at \(90^\circ C\), the partial pressure of \(H_2S_{(g)}\) formed (\(K_p = 6.8 \times 10^{-2}\)) would be:

• (a) 0.19 atm
• (b) 0.38 atm
• (c) 0.6 atm
• (d) 0.072 atm

Correct Answer: (b)

Explanation: Solving for the moles of \(H_2S\) formed yields \(x \approx 0.0127\text{ mol}\). Then, \(P = \frac{nRT}{V} = \frac{0.0127 \times 0.0821 \times 363}{1} = 0.38\text{ atm}\).

6. Which one of the following mixtures will give a solution with pH greater than 7?

• (a) 50 mL of 0.1 M HCl + 50 mL of 0.2 M NaCl
• (b) 50 mL of 0.1 M HCl + 50 mL of 0.1 M \(CH_3COONa\)
• (c) 50 mL of 0.1 M \(CH_3COOH\) + 50 mL of 0.1 M NaOH
• (d) 50 mL of 0.2 M \(H_2SO_4\) + 50 mL of 0.3 M NaOH

Correct Answer: (c)

Explanation: Mixture (c) results in a salt of a weak acid and a strong base (\(CH_3COONa\)), which undergoes anionic hydrolysis to produce a basic solution.

7. 40 mL of 0.1 M ammonium hydroxide is mixed with 20 mL of 0.1 M HCl. What is the pH of the mixture? (\(pK_b\) of ammonia solution is 4.74)

• (a) 4.74
• (b) 2.26
• (c) 9.26
• (d) 5

Correct Answer: (c)

Explanation: This forms a basic buffer where ammonium hydroxide is 50% neutralised, so \([Salt] = [Base]\). Thus, \(pOH = pK_b = 4.74\). \(pH = 14 - 4.74 = 9.26\).

8. What is the decreasing order of strength of the bases \(OH^{-}\), \(NH_2^{-}\), \(H-C \equiv C^{-}\) and \(CH_3CH_2^{-}\)?

• (a) \(CH_3CH_2^{-} > NH_2^{-} > H-C \equiv C^{-} > OH^{-}\)
• (b) \(H-C \equiv C^{-} > CH_3CH_2^{-} > NH_2^{-} > OH^{-}\)
• (c) \(OH^{-} > NH_2^{-} > H-C \equiv C^{-} > CH_3CH_2^{-}\)
• (d) \(NH_2^{-} > H-C \equiv C^{-} > OH^{-} > CH_3CH_2^{-}\)

Correct Answer: (a)

Explanation: Strength is determined by the weakness of their conjugate acids (\(H_2O\), \(NH_3\), \(C_2H_2\), \(C_2H_6\)). A weak acid has a strong conjugate base.

9. The pH of a solution is 7.00. Base is added to increase the pH to 12.0. The increase in OH⁻ ion concentration is:

• (a) 5 times
• (b) 1000 times
• (c) \(10^5\) times
• (d) 4 times

Correct Answer: (c)

Explanation: At pH 7, \([OH^-] = 10^{-7}\). At pH 12, \(pOH = 2\), so \([OH^-] = 10^{-2}\). The increase is \(10^5\) times.

10. Order of solubility of \(AgCl\) in (1) water (\(S_1\)), (2) 0.01 M \(CaCl_2\) (\(S_2\)), (3) 0.01 M \(NaCl\) (\(S_3\)), (4) 0.05 M \(AgNO_3\) (\(S_4\)):

• (a) \(S_1 > S_2 > S_3 > S_4\)
• (b) \(S_1 > S_3 > S_2 > S_4\)
• (c) \(S_1 > S_2 = S_3 > S_4\)
• (d) \(S_1 > S_3 > S_4 > S_2\)

Correct Answer: (b)

Explanation: Solubility is inversely proportional to the concentration of common ions. \(S_1\) (no common ion) is highest. \(S_3\) (\(0.01\text{ M }Cl^-\)) > \(S_2\) (\(0.02\text{ M }Cl^-\)) > \(S_4\) (\(0.05\text{ M }Ag^+\)).

11. The pH of an aqueous solution of \(CH_3COONa\) of concentrated C (M) is given by:

• (a) \(7 - \frac{1}{2}pK_a + \frac{1}{2}\log C\)
• (b) \(\frac{1}{2}pK_w + \frac{1}{2}pK_b + \frac{1}{2}\log C\)
• (c) \(\frac{1}{2}pK_w - \frac{1}{2}pK_b - \frac{1}{2}\log C\)
• (d) \(\frac{1}{2}pK_w + \frac{1}{2}pK_a + \frac{1}{2}\log C\)

Correct Answer: (d)

Explanation: The source identifies (d) as the formula for the pH of a hydrolysed salt of a strong base and a weak acid.

12. Relations for equilibria \(SO_2 + \frac{1}{2}O_2 \rightleftharpoons SO_3\) (\(K_1\)) and \(2SO_3 \rightleftharpoons 2SO_2 + O_2\) (\(K_2\)):

• (a) \(K_2 = (1/K_1)^2\)
• (b) \(K_2 = (1/K_1)\)
• (c) \(K_1 = (1/K_2)^3\)
• (d) \(K_2 = (K_1)^2\)

Correct Answer: (a)

Explanation: \(K_2\) corresponds to the inverse and doubling of reaction 1, so \(K_2 = 1/K_1^2\).

13. pH value of which one of the following is not equal to one?

• (a) 0.1 M \(CH_3COOH\)
• (b) 0.1 M \(HNO_3\)
• (c) 0.05 M \(H_2SO_4\)
• (d) \(50\text{ cm}^3\text{ 0.4 M HCl} + 50\text{ cm}^3\text{ 0.2 M NaOH}\)

Correct Answer: (a)

Explanation: \(CH_3COOH\) is a weak acid and does not dissociate completely, so its pH will not be 1 even at 0.1 M.

14. Buffer: 0.1 mol sodium acetate in \(1000\text{ cm}^3\) of 0.1 M acetic acid. If 0.1 mol sodium acetate is added, the new pH is:

• (a) \(pK_a\)
• (b) \(pK_a + 2\)
• (c) \(pK_a - \log 2\)
• (d) \(pK_a + \log 2\)

Correct Answer: (d)

Explanation: New \([Salt] = 0.2\) and \([Acid] = 0.1\). Using Henderson's equation: \(pH = pK_a + \log(0.2/0.1) = pK_a + \log 2\).

15. \(H_2S\) passed into solution of 0.1 mol \(Zn^{2+}\) and 0.01 mol \(Cu^{2+}\) till \([S^{2-}] = 8.1 \times 10^{-19}\). \(K_{sp}\) ZnS (\(3 \times 10^{-22}\)) and CuS (\(8 \times 10^{-36}\)). What is true?

• (a) Only ZnS precipitates
• (b) Both CuS and ZnS precipitate
• (c) Only CuS precipitates
• (d) No precipitation occurs

Correct Answer: (b)

Explanation: Precipitation occurs if the ionic product exceeds \(K_{sp}\). For both ions, the products (\(8.1 \times 10^{-20}\) for Zn and \(8.1 \times 10^{-21}\) for Cu) are greater than their respective \(K_{sp}\) values.

16. 0.023 g of sodium metal is reacted with \(100\text{ cm}^3\) of water. The pH is:

• (a) 10
• (b) 8
• (c) 9
• (d) 12

Correct Answer: (d)

Explanation: 0.001 mol Na produces 0.001 mol NaOH. In 100 mL, \([OH^-] = 0.01\text{ M}\). \(pOH = 2\), so \(pH = 12\).

17. Exothermic reversible \(SO_2 + Cl_2 \rightarrow SO_2Cl_2\). If extra \(SO_2\) is introduced at constant volume:

• (a) Pressure will not change
• (b) Temperature will not change
• (c) Temperature will increase
• (d) Temperature will decrease

Correct Answer: (c)

Explanation: Extra reactant shifts equilibrium forward. Since the reaction is exothermic, temperature increases.

18. If pH of a saturated solution of \(Ba(OH)_2\) is 12, the value of its \(K_{sp}\) is:

• (a) \(4.0 \times 10^{-4}\text{ M}^3\)
• (b) \(5.0 \times 10^{-6}\text{ M}^3\)
• (c) \(5.0 \times 10^{-7}\text{ M}^3\)
• (d) \(4.0 \times 10^{-6}\text{ M}^3\)

Correct Answer: (c)

Explanation: \(pOH = 2\), so \([OH^-] = 10^{-2}\text{ M}\). \([Ba^{2+}] = 0.5 \times 10^{-2}\text{ M}\). \(K_{sp} = [Ba^{2+}][OH^-]^2 = 5 \times 10^{-7}\).

19. If reactant concentrations are doubled, the equilibrium constant will:

• (a) also be doubled
• (b) be halved
• (c) become one-fourth
• (d) remain the same.

Correct Answer: (d)

Explanation: The equilibrium constant is independent of initial concentrations.

20. Relationship between \(K\), \(\beta_1\), \(\beta_2\) and \(\beta_3\) for sequential steps leading to \(MCl_3\):

• (a) \(pK = p\beta_1 + p\beta_2 + p\beta_3\)
• (b) \(K = \beta_1 \beta_2 \beta_3\)
• (c) \(\log K = \log \beta_1 + \log \beta_2 + \log \beta_3\)
• (d) all of the above.

Correct Answer: (d)

Explanation: Multiply the steps: \(K = \beta_1 \beta_2 \beta_3\). Taking log and negative log yield the other forms.

21. \(H_2S\) in HCl precipitates Group II cations but not Group IV because:

• (a) presence of HCl decreases the sulphide ion concentration
• (b) sulphides of group IV cations are unstable in HCl
• (c) \(K_{sp}\) of group II sulphides is more than group IV
• (d) presence of HCl increases the sulphide ion concentration.

Correct Answer: (a)

Explanation: Adding HCl suppresses \(H_2S\) dissociation. Only Group II cations, which have very low \(K_{sp}\) values, can precipitate at low \([S^{2-}]\).

22. If \(Q > K_c\), the reaction proceeds:

• (a) from left to right
• (b) from right to left
• (c) in both directions
• (d) no net reaction.

Correct Answer: (b)

Explanation: When \(Q > K_c\), the system proceeds in the backward direction.

23. Which can act as both Bronsted acid and Bronsted base?

• (a) \(Na_2CO_3\)
• (b) \(OH^-\)
• (c) \(HCO_3^-\)
• (d) \(NH_3\)

Correct Answer: (c)

Explanation: The bicarbonate ion can either lose a proton (\(\rightarrow CO_3^{2-}\)) or gain one (\(\rightarrow H_2CO_3\)).

24. pH of mixing 5 mL 0.1 M \(NH_4OH\) with 250 mL 0.1 M \(NH_4Cl\). (\(K_b = 1.8 \times 10^{-5}\))

• (a) 5.0223
• (b) 6.3562
• (c) 7.5563
• (d) 8.6210

Correct Answer: (c)

Explanation: Calculated using the basic buffer formula \(pOH = pK_b + \log([Salt]/[Base])\). \(pOH \approx 6.44\), so \(pH \approx 7.56\).

25. For \(2NO \rightleftharpoons N_2 + O_2\) (\(\Delta H = -43.5\)), for the inverse \(N_2 + O_2 \rightleftharpoons 2NO\):

• (a) K is independent of T
• (b) K increases as T decreases
• (c) K decreases as T decreases
• (d) K varies with addition of NO.

Correct Answer: (c)

Explanation: For the endothermic inverse reaction, K decreases as temperature decreases.

26. Equilibrium constant is 1.732 at 298 K. In the vessel mixture point P is on the line representing \(Q = \tan(75^\circ) = 3.73\). Predict:

• (a) Reaction goes forward
• (b) Immediate equilibrium
• (c) Reaction goes backward
• (d) Nothing can be said.

Correct Answer: (c)

Explanation: Since \(Q (3.73) > K_c (1.732)\), the reaction proceeds in the backward direction.

27. pH of strong acid is 5.0. Diluted 100 times, the pH is:

• (a) 6.6
• (b) 2.7
• (c) 7
• (d) 3.01

Correct Answer: (a)

Explanation: At high dilution (\(10^{-7}\)), the contribution of \(H^+\) from water must be added to the acid concentration, yielding a pH slightly below 7.

28. 0.00001 M HCl diluted 1000 folds. Resulting pH:

• (a) 5
• (b) 7
• (c) 8
• (d) 6.98

Correct Answer: (d)

Explanation: Similar to Q27, at this dilution, \([H^+]\) from water dominates, keeping the pH slightly below 7.

29. Solution: 0.05 M each NaCl and \(Na_2CrO_4\). \(AgNO_3\) is added. What is true?

• (a) \(CrO_4^{2-}\) precipitated first
• (b) \(Cl^-\) precipitated first
• (c) Both precipitated together
• (d) Second ion precipitates when first is half gone.

Correct Answer: (b)

Explanation: Based on \(K_{sp}\) values, the amount of \(Ag^+\) required to start \(AgCl\) precipitation is much lower than for \(Ag_2CrO_4\).

30. \(P + Q \rightleftharpoons R + S\). \(k_f = 2.0 \times 10^{-3}\). At equilibrium, P is twice that of R. The \(k_b\) is:

• (a) \(1.5 \times 10^2\)
• (b) \(5.0 \times 10^{-4}\)
• (c) \(8.0 \times 10^{-3}\)
• (d) none.

Correct Answer: (c)

Explanation: Equilibrium moles lead to \(K_c = 0.25\). Thus, \(k_b = k_f / K_c = (2.0 \times 10^{-3}) / 0.25 = 8 \times 10^{-3}\).

31. Hypo reaction \(A + B \rightleftharpoons C + D\). log K vs \(T^{-1}\) is linear with \(\theta = \tan^{-1}(0.5)\) and intercept 10. K at 298 and 798 K:

• (a) \(K_1 = 9.96 \times 10^9\), \(K_2 = 2.76 \times 10^{10}\)
• (b) \(K_1 = 3.79 \times 10^8\), \(K_2 = 9.98 \times 10^9\)
• (c) \(K_1 = 7.96 \times 10^9\), \(K_2 = 7.96 \times 10^9\)
• (d) \(K_1 = 9.98 \times 10^9\), \(K_2 = 9.96 \times 10^9\)

Correct Answer: (d)

Explanation: Calculated K at 298 is \(9.98 \times 10^9\). Since \(\Delta H\) is stated to be independent of T, K remains almost constant at \(9.96 \times 10^9\) even at 798 K.

32. \(N_2O_4 \rightleftharpoons 2NO_2\). Variation of \(x\) with \(D/d\) where \(x=0\) at point Y. What is \(D/d\) at Y?

• (a) 0
• (b) 1.5
• (c) 1
• (d) 0.5

Correct Answer: (c)

Explanation: Using \(x = (D/d) - 1\), when \(x=0\), \(D/d = 1\).

33. Aniline hydrochloride is X due to hydrolysis of Y. X and Y are:

• (a) basic, \(C_6H_5NH_3^+\)
• (b) acidic, \(C_6H_5NH_3^+\)
• (c) basic, \(Cl^-\)
• (d) acidic, \(Cl^-\)

Correct Answer: (b)

Explanation: Hydrolysis of the phenylammonium cation (\(C_6H_5NH_3^+\)) produces an acidic solution.

34. Highest pH of 0.1 M \(Mg^{2+}\) without precipitating \(Mg(OH)_2\) (\(K_{sp} = 1.2 \times 10^{-11}\)):

• (a) 4.96
• (b) 6.96
• (c) 7.54
• (d) 9.04

Correct Answer: (d)

Explanation: Calculated \([OH^-] \approx 1.1 \times 10^{-5}\), which yields a pH of 9.04.

35. Relations between reactions I, II, III:

• (a) \(K_2 K_3 = K_1\)
• (b) \(K_3 = K_1 K_2\)
• (c) \(K_3 K_2^3 = K_1^2\)
• (d) \(K_1 \sqrt{K_2} = K_3\)

Correct Answer: (b)

Explanation: Multiplying the constants for I and II results in the constant for reaction III.

36. Acidic buffer can be prepared by mixing:

• (a) ammonium acetate and acetic acid
• (b) ammonium chloride and ammonium hydroxide
• (c) sulphuric acid and sodium sulphate
• (d) sodium chloride and sodium hydroxide.

Correct Answer: (a)

Explanation: An acidic buffer contains a weak acid and its salt with a strong base.

37. Solution A has pH 3 and B has pH 6. Which is true?

• (a) A is twice as acidic as B.
• (b) B is twice as acidic as A.
• (c) A is 1000 times more acidic than B.
• (d) B is 1000 times more acidic than A.

Correct Answer: (c)

Explanation: A unit change in pH is a 10-fold change in \([H^+]\). Three units means \(10^3 = 1000\) times.

38. Equilibrium: 1 mol \(PCl_3\), 2 mol \(Cl_2\) in 3.0 L. 0.70 mol \(PCl_3\) remains. \(K_c\) for \(PCl_3 + Cl_2 \rightarrow PCl_5\) is:

• (a) 0.76
• (b) 0.076
• (c) 0.38
• (d) 1.52

Correct Answer: (a)

Explanation: Moles reacted = 0.3. Equilibrium moles are 0.7, 1.7, and 0.3. \(K_c = \frac{0.3/3}{(0.7/3) \times (1.7/3)} = 0.76\).

39. pH of resultant: 1.0 M acetic acid mixed with 0.01 mol/L HCl. (\(K_a = 1.8 \times 10^{-5}\))

• (a) 1.846
• (b) 1.04
• (c) 2.36
• (d) 2.79

Correct Answer: (a)

Explanation: Added \([H^+]\) from acetic acid dissociation (\(4.2 \times 10^{-3}\)) to initial \([H^+]\) from HCl yields total \([H^+] \approx 0.0142\text{ M}\), pH 1.846.

40. \(K_p/K_c\) for \(CO + \frac{1}{2} O_2 \rightleftharpoons CO_2\):

• (a) \(\sqrt{RT}\)
• (b) \(\sqrt{RT}/2\)
• (c) 1
• (d) none.

Correct Answer: (d)

Explanation: \(\Delta n_g = -1/2\). Thus \(K_p/K_c = (RT)^{-1/2} = 1/\sqrt{RT}\).

41. Saturated \(Ca_3(PO_4)_2\). \([PO_4^{3-}] = 3.3 \times 10^{-7}\). \(K_{sp}\) is:

• (a) \(1.32 \times 10^{-31}\)
• (b) \(1.32 \times 10^{-32}\)
• (c) \(1.32 \times 10^{-33}\)
• (d) none.

Correct Answer: (b)

Explanation: Based on stoichiometry, \([Ca^{2+}] = \frac{3}{2} [PO_4^{3-}]\). \(K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2 = 1.32 \times 10^{-32}\).

42. Units of \(K_c\) for \(2SO_2 + O_2 \rightleftharpoons 2SO_3\):

• (a) \(L \text{ mol}^{-1}\)
• (b) \(mol \text{ L}^{-1}\)
• (c) \((mol \text{ L}^{-1})^2\)
• (d) \((L \text{ mol}^{-1})^2\)

Correct Answer: (a)

Explanation: \(K_c = \frac{M^2}{M^2 \times M} = M^{-1} = L \text{ mol}^{-1}\).

43. pH of which is not affected by dilution?

• (a) 0.01 M \(NaHCO_3\)
• (b) 0.01 M \(NaH_2PO_4\)
• (c) 0.01 M \(CH_3COONH_4\)
• (d) All of these.

Correct Answer: (d)

Explanation: Buffers and solutions containing amphiprotic anions do not change pH significantly on dilution.

44. 4.5 moles each \(H_2\), \(I_2\) heated in 10 L. 3 moles HI found at equilibrium. \(K_{eq}\) is:

• (a) 1
• (b) 10
• (c) 5
• (d) 0.33

Correct Answer: (a)

Explanation: Moles reacted = 1.5. Equilibrium moles: 3, 3, 3. \(K = \frac{3^2}{3 \times 3} = 1\).

45. Acid solution pH 6 diluted 100 times. Resultant pH:

• (a) 5.95
• (b) 6.95
• (c) 7
• (d) 8

Correct Answer: (b)

Explanation: At \(10^{-8}\text{ M}\), water's ionization contributes significant \(H^+\), keeping pH near 7 (6.95).