CBSE Class 11 Chemistry HOTs Equilibrium Set 02

Higher Order Thinking Questions

Question 1. Assertion (A): Addition of a catalyst does not change the value of the equilibrium constant of a reaction.
Reason (R): A catalyst increases the rates of the forward and backward reactions equally.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) A catalyst lowers the activation energy of the forward and backward reactions by the same amount, so both rates increase equally. The equilibrium composition and K stay the same, and equilibrium is reached sooner. R explains A.
In simple words: A catalyst speeds up both directions equally, so the final balance does not change.

Teacher's Note:
1. K depends only on temperature.
2. A catalyst shortens the time to reach equilibrium and nothing else.

Question 2. Assertion (A): A reaction with a very large equilibrium constant (K_c > 10³) is always a very fast reaction.
Reason (R): The equilibrium constant is the ratio of the rate constants of the forward and the backward reactions.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (d) A is false: K shows how far a reaction goes, not how fast it goes. A reaction with a large K may still be extremely slow (for example, H₂ + O₂ at room temperature without a spark). R is true, since K_c = k_f ÷ k_b, but it does not make the reaction fast.
In simple words: A large K means a high yield, not a high speed.

Teacher's Note:
1. Rate belongs to kinetics; extent belongs to equilibrium.
2. A reaction can be thermodynamically favourable and still very slow.

Question 3. Assertion (A): Adding an inert gas to the equilibrium N₂ + 3H₂ ⇌ 2NH₃ at constant volume does not shift the equilibrium.
Reason (R): The concentrations (partial pressures) of the reacting gases remain unchanged.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) At constant volume the moles per litre of N₂, H₂ and NH₃ stay the same, so Q_c = K_c still. At constant pressure the volume would increase and the equilibrium would shift toward the side with more moles (backward here).
In simple words: The inert gas takes up no part in the reaction and does not change the concentrations of the others.

Teacher's Note:
1. Always say whether the inert gas is added at constant volume or at constant pressure.
2. Only a change in partial pressures or concentrations can shift an equilibrium.

Question 4. Assertion (A): The pH of a 10⁻⁸ M solution of HCl is about 6.98 and not 8.
Reason (R): The H⁺ ions from the ionisation of water must also be considered in very dilute solutions of an acid.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) HCl gives 10⁻⁸ M H⁺, which is less than the 10⁻⁷ M from water, so water cannot be ignored. Solving x(x + 10⁻⁸) = 10⁻¹⁴ gives x = 9.5 × 10⁻⁸ M, so [H⁺] = 1.05 × 10⁻⁷ M and pH = 6.98. An acid cannot make a solution basic.
In simple words: Very dilute acid can only push the pH a little below 7.

Teacher's Note:
1. Check whether the acid concentration is near 10⁻⁷ M.
2. Add the H⁺ from the acid and from water.

Question 5. Assertion (A): An aqueous solution of NaCl is neutral.
Reason (R): Neither Na⁺ nor Cl⁻ undergoes hydrolysis.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) NaCl is the salt of a strong acid (HCl) and a strong base (NaOH). Na⁺ and Cl⁻ are the conjugate acid of a very strong base and the conjugate base of a very strong acid, so they do not react with water and [H⁺] = [OH⁻].
In simple words: The ions are too weak as an acid and as a base to affect water.

Teacher's Note:
1. Compare with NH₄Cl (acidic) and CH₃COONa (basic).
2. Neutral here means pH 7 at 298 K.

Question 6. Assertion (A): The pH of a buffer solution changes only slightly when it is diluted with a little water.
Reason (R): The pH of a buffer depends on the ratio of the concentrations of salt and acid, and this ratio does not change on dilution.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) By the Henderson equation, pH = pK_a + log([salt] ÷ [acid]). Dilution reduces [salt] and [acid] by the same factor, so the ratio and the pH stay almost the same.
In simple words: Both components are diluted equally, so their ratio, and the pH, hardly changes.

Teacher's Note:
1. Very large dilution weakens the buffer capacity.
2. Capacity depends on how much acid and salt are present.

Question 7. Assertion (A): AgCl is less soluble in 0.1 M NaCl than in pure water.
Reason (R): The solubility product of AgCl decreases in the presence of NaCl.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (c) A is true: Cl⁻ is a common ion, so AgCl dissolves less (1.8 × 10⁻⁹ M against 1.3 × 10⁻⁵ M). R is false: K_sp depends only on temperature and stays constant. The solubility changes, not the K_sp.
In simple words: The common ion changes the solubility, but not the solubility product.

Teacher's Note:
1. Do not confuse solubility with K_sp.
2. K_sp changes only with temperature.

Question 8. Ammonia is made by an exothermic reaction, yet the Haber process is run at about 700 K and not at a much lower temperature. Explain this and say why a very high pressure is used.
Answer:
N₂ + 3H₂ ⇌ 2NH₃ is exothermic (ΔH = −92.38 kJ mol⁻¹), so a low temperature gives a higher equilibrium yield (K is larger).
At a low temperature, however, the reaction is far too slow. About 700 K with an iron catalyst is a compromise between a reasonable rate and an acceptable yield.
The forward reaction decreases the moles of gas (4 to 2), so a high pressure (about 200 atm) increases the yield. Continuous removal of NH₃ also shifts the equilibrium forward.
In simple words: Yield improves when it is cold, speed improves when it is hot, so the plant uses a middle temperature with a catalyst.

Teacher's Note:
1. Thermodynamics gives the yield; kinetics gives the speed.
2. Mention both temperature and pressure.

Question 9. For N₂O₄(g) ⇌ 2NO₂(g), K_p = 0.14 atm at 298 K. Find the degree of dissociation of N₂O₄ at a total pressure of 1 atm.
Answer:
Starting with 1 mol, the equilibrium amounts are (1 − α) mol of N₂O₄ and 2α mol of NO₂; the total is (1 + α) mol.
p(N₂O₄) = (1 − α)P ÷ (1 + α) and p(NO₂) = 2αP ÷ (1 + α), so K_p = 4α²P ÷ (1 − α²).
α² = K_p ÷ (4P + K_p) = 0.14 ÷ 4.14 = 0.0338, so α = 0.184 (about 18%).
In simple words: About 18 out of every 100 N₂O₄ molecules dissociate at 1 atm.

Teacher's Note:
1. At a higher pressure α becomes smaller.
2. Write the expressions in terms of α for dissociation problems.

Question 10. Which of the following changes will increase the degree of dissociation of PCl₅ in PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), ΔH = +124 kJ mol⁻¹? (more than one may be correct)
(a) Decreasing the total pressure
(b) Adding Cl₂ at constant volume
(c) Increasing the temperature
(d) Adding an inert gas at constant pressure
Answer:
Ans. (a), (c) and (d)
Lower pressure favours the side with more moles of gas (the products). The reaction is endothermic, so a higher temperature shifts it forward. An inert gas at constant pressure increases the volume, which has the same effect as lowering the pressure.
Adding Cl₂ (a product) shifts the equilibrium backward and decreases the dissociation.
In simple words: Anything that favours the side with more gas moles, or the endothermic direction, raises the dissociation.

Teacher's Note:
1. Check each change separately.
2. Increasing the number of gas moles is favoured at low pressure.

Question 11. Which of the following species can act both as a Brønsted acid and as a Brønsted base? (more than one may be correct)
(a) H₂O
(b) HCO₃⁻
(c) NH₃
(d) CO₃²⁻
Answer:
Ans. (a), (b) and (c)
H₂O: H₂O + H₂O ⇌ H₃O⁺ + OH⁻ (it donates and accepts). HCO₃⁻ can give H⁺ to form CO₃²⁻ or accept H⁺ to form H₂CO₃. NH₃ can accept H⁺ to give NH₄⁺ or donate H⁺ to give NH₂⁻.
CO₃²⁻ has no proton to give, so it can only be a base.
In simple words: A species with a transferable H and a lone pair can be both.

Teacher's Note:
1. Such species are called amphoteric (amphiprotic).
2. Check whether the species has a proton to donate.

Question 12. Ag₂CrO₄ (K_sp = 1.1 × 10⁻¹²) has a much smaller K_sp than AgCl (K_sp = 1.8 × 10⁻¹⁰). Yet Ag₂CrO₄ is more soluble than AgCl in water. Show this by calculation and explain.
Answer:
AgCl: S² = 1.8 × 10⁻¹⁰, so S = 1.34 × 10⁻⁵ mol L⁻¹.
Ag₂CrO₄: 4S³ = 1.1 × 10⁻¹², so S³ = 2.75 × 10⁻¹³ and S = 6.5 × 10⁻⁵ mol L⁻¹.
S(Ag₂CrO₄) is about five times larger. K_sp values can be compared directly only for salts of the same type (the same number of ions), because K_sp is a different function of S for each type.
In simple words: The formula that links K_sp and S depends on the number of ions in the salt.

Teacher's Note:
1. Always compute S before comparing salts of different types.
2. For the same type, a larger K_sp means a larger solubility.

Question 13. In blood, the pH is 7.4 and the pK_a1 of carbonic acid (H₂CO₃) is 6.1. Calculate the ratio [HCO₃⁻] ÷ [H₂CO₃] in blood and say what happens to this ratio when acid enters the blood. (log 20 = 1.30)
Answer:
By the Henderson equation, 7.4 = 6.1 + log([HCO₃⁻] ÷ [H₂CO₃]), so log([HCO₃⁻] ÷ [H₂CO₃]) = 1.3.
[HCO₃⁻] ÷ [H₂CO₃] = antilog 1.3 = 20.
Added H⁺ reacts with HCO₃⁻ to form H₂CO₃ (which breaks down to CO₂ and is breathed out), so the ratio falls and the pH drops slightly. Blood holds much more of the base form, which helps it neutralise acids.
In simple words: Blood keeps about 20 parts of bicarbonate for every 1 part of carbonic acid.

Teacher's Note:
1. A fall in blood pH below 7.35 is called acidosis.
2. Breathing out CO₂ helps to restore the ratio.