NCERT Exemplar Equilibrium - Chemistry Class 11 - NEET PDF Download
| Table of contents | |
| Assertion and Reason Type Questions | |
| Long Answer Type Questions |
Multiple Choice Questions - I
Q.1. We know that the relationship between Kc and Kp is Kp = Kc (RT)∆n What would be the value of ∆n for the reaction NH4Cl (s) ⇌ NH3 (g) + HCl (g) (1) 1 (2) 0.5 (3) 1.5 (4) 2 Ans. (4)
Solution.
The relationship between Kp and Kc is Kp = Kc (RT)∆n Where ∆n = (number of moles of gaseous products) - (number of moles of gaseous reactants) For the reaction, NH4Cl (s) ⇌ NH3 (g) + HCl (g) ∆n = 2 - 0 = 2
Q.2. For the reaction H2(g) + I2(g) ⇌ 2HI (g), the standard free energy is ΔGθ > 0. The equilibrium constant (K) would be __________. (1) K = 0 (2) K > 1 (3) K = 1 (4) K < 1 Ans. (4)
Solution.
ΔG° = -RT ln K. ΔG° > 0 means ΔG° is positive. This occurs only if ln K is negative, i.e., K < 1.
Q.3. Which of the following is not a general characteristic of equilibria involving physical processes? (1) Equilibrium is possible only in a closed system at a given temperature. (2) All measurable properties of the system remain constant. (3) All the physical processes stop at equilibrium. (4) The opposing processes occur at the same rate and there is dynamic but stable condition. Ans. (3)
Solution.
All the physical processes like melting of ice and freezing of water do not stop at equilibrium; they continue at equal rates giving a dynamic equilibrium. Hence statement (3) is incorrect.
Q.4. PCl5, PCl3 and Cl2 are at equilibrium at 500 K in a closed container and their concentrations are 0.8 × 10-3 mol L-1, 1.2 × 10-3 mol L-1 and 1.2 × 10-3 mol L-1 respectively. The value of Kc for the reaction PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) will be (i) 1.8 × 103 mol L-1 (ii) 1.8 × 10-3 (iii) 1.8 × 10-3 L mol-1 (iv) 0.55 × 104 Ans. (ii)
Solution.
PCl5(g) ⇌ PCl3(g) + Cl2(g) Kc = [PCl3][Cl2]/[PCl5] = (1.2 × 10-3 × 1.2 × 10-3)/(0.8 × 10-3) = 1.44 × 10-6 / 0.8 × 10-3 = 1.8 × 10-3
Q.5. Which of the following statements is incorrect? (1) In equilibrium mixture of ice and water kept in perfectly insulated flask mass of ice and water does not change with time. (2) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate. (3) On addition of catalyst the equilibrium constant value is not affected. (4) Equilibrium constant for a reaction with negative ∆H value decreases as the temperature increases. Ans. (2)
Solution.
Fe3+ + SCN- ⇌ FeSCN2+ (red) When oxalic acid is added, it complexes Fe3+ to form [Fe(C2O4)3]3-, decreasing free Fe3+ concentration, so intensity of red colour decreases (not increases).
Q.6. When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture it becomes pink. On the basis of this information mark the correct answer.
Solution.
The equilibrium shifts backward on cooling; the backward reaction must be exothermic. Therefore the forward reaction is endothermic, so ΔH > 0.
Q.7. The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H+ ions and OH- ions are equal. What will be the pH of pure water at 60°C? (1) Equal to 7.0 (2) Greater than 7.0 (3) Less than 7.0 (4) Equal to zero Ans. (3)
Solution.
At higher temperature Kw increases, so [H+] = [OH-] becomes greater than 10-7 M. Therefore pH < 7.
Q.8. The ionisation constant of an acid, Ka, is the measure of strength of an acid. The Ka values of acetic acid, hypochlorous acid and formic acid are 1.74 × 10-5, 3.0 × 10-8 and 1.8 × 10-4 respectively. Which of the following orders of pH of 0.1 mol dm-3 solutions of these acids is correct? (1) Acetic acid > hypochlorous acid > formic acid (2) Hypochlorous acid > acetic acid > formic acid (3) Formic acid > hypochlorous acid > acetic acid (4) Formic acid > acetic acid > hypochlorous acid Ans. (4)
Solution.
Larger Ka ⇒ larger [H3O+] ⇒ lower pH. The acids in decreasing Ka are formic (1.8×10-4) > acetic (1.74×10-5) > hypochlorous (3.0×10-8). Hence pH order: formic acid < acetic acid < hypochlorous acid, so pH decreasing order reversed gives option (4).
For the reaction H2S ⇌ H+ + HS-
Q.10. Acidity of BF3 can be explained on the basis of which of the following concepts? (1) Arrhenius concept (2) Bronsted-Lowry concept (3) Lewis concept (4) Bronsted-Lowry as well as Lewis concept. Ans. (3)
Solution.
According to Lewis concept, an electron-deficient species acts as a Lewis acid. BF3 is electron deficient (B has only six electrons) and thus acts as a Lewis acid.
Q.11. Which of the following will produce a buffer solution when mixed in equal volumes? (1) 0.1 mol dm-3 NH4OH and 0.1 mol dm-3 HCl (2) 0.05 mol dm-3 NH4OH and 0.1 mol dm-3 HCl (3) 0.1 mol dm-3 NH4OH and 0.05 mol dm-3 HCl (4) 0.1 mol dm-3 CH3COONa and 0.1 mol dm-3 NaOH Ans. (3)
Solution.
In option (3), mixing equal volumes of 0.1 M NH4OH and 0.05 M HCl: HCl will be completely neutralised by some NH4OH to form NH4Cl, leaving excess NH4OH. The final mixture contains both NH4OH (base) and NH4Cl (its conjugate acid); this constitutes a buffer.
Q.12. In which of the following solvents is silver chloride most soluble? (i) 0.1 mol dm-3 AgNO3 solution (ii) 0.1 mol dm-3 HCl solution (iii) H2O (iv) Aqueous ammonia Ans. (iv)
Solution.
AgCl forms a soluble complex with aqueous ammonia: AgCl + 2NH3 → [Ag(NH3)2]+ + Cl-
Q.13. What will be the value of pH of 0.01 mol dm-3 CH3COOH (Ka = 1.74 × 10-5)? (1) 3.4 (2) 3.6 (3) 3.9 (4) 3.0 Ans. (1)
Solution.
Q.14. Ka for CH3COOH is 1.8 × 10-5 and Kb for NH4OH is 1.8 × 10-5. The pH of ammonium acetate will be (1) 7.005 (2) 4.75 (3) 7.0 (4) Between 6 and 7 Ans. (3)
Solution.
Ammonium acetate is a salt formed from a weak acid and a weak base with equal strengths (Ka = Kb). Therefore, the solution is essentially neutral, pH ≈ 7.0.
Q.15. Which of the following options will be correct for the stage of half completion of the reaction A ⇌ B. (1) ΔGθ = 0 (2) ΔGθ > 0 (3) ΔGθ < 0 (4) ΔGθ = -RT ln 2 Ans. (1)
Solution.
At half completion, [A] = [B], so equilibrium constant K = 1. Since ΔG° = -RT ln K, ΔG° = 0.
Q.16. On increasing the pressure, in which direction will the gas phase reaction proceed to re-establish equilibrium, is predicted by applying the Le Chatelier's principle. Consider the reaction.
Solution.
N2(g) + 3H2(g) ⇌ 2NH3(g) At constant temperature, a change in pressure alters the equilibrium composition but does not change the equilibrium constant K (which depends only on temperature).
Q.17. What will be the correct order of vapour pressure of water, acetone and ether at 30°C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point? (1) Water < ether < acetone (2) Water < acetone < ether (3) Ether < acetone < water (4) Acetone < ether < water Ans. (2)
Solution.
Greater the boiling point, lower the vapour pressure. Given water (highest b.p.) < acetone < ether (lowest b.p.), so vapour pressure order: water < acetone < ether.
Q.18. At 500 K, equilibrium constant, Kc, for the following reaction is 5.
Solution.
For the given relationships between the reactions (see image placeholders), apply the rules: reversing inverts K, multiplying equation raises K to power, etc. (Details shown in diagram placeholders.)
Q.19. In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume? (1) H2 (g) + I2 (g) ⇌ 2HI (g) (2) PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) (3) N2(g) + 3H2 (g) ⇌ 2NH3 (g) (4) The equilibrium will remain unaffected in all the three cases. Ans. (4)
Solution.
Addition of a small amount of inert gas at constant volume does not change partial pressures of reacting species; therefore equilibrium remains unaffected in all three cases.
Multiple Choice Questions - II
In the following questions two or more options may be correct.
Q.20. For the reaction N2O4 (g) ⇌ 2NO2 (g), the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct? (1) The reaction is endothermic (2) The reaction is exothermic (3) If NO2 (g) and N2O4(g) are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more N2O4 (g) will be formed. (4) The entropy of the system increases. Ans. (1, 3, 4)
Solution.
(1) As K increases with temperature, forward reaction is favoured at higher T, indicating the forward reaction is endothermic. (3) At 400 K, reaction quotient Q = pNO22/pN2O4 = (20)2/2 = 400/2 = 200 > K (50). So equilibrium shifts left forming more N2O4. (4) The reaction increases the number of gas moles (1 → 2), so entropy increases.
Q.21. At a particular temperature and atmospheric pressure, the solid and liquid phases of a pure substance can exist in equilibrium. Which of the following term defines this temperature? (1) Normal melting point (2) Equilibrium temperature (3) Boiling point (4) Freezing point Ans. (1, 4)
Solution.
Measured at atmospheric pressure, the temperatures are called normal melting point (or freezing point), so both (1) and (4) apply.
Short Answer Type Questions
Q.22. The ionisation of hydrochloric in water is given below: HCl(aq) + H2O (l) ⇌ H3O+ (aq) + Cl- (aq) Label two conjugate acid-base pairs in this ionisation.
Ans.
| HCl (acid) | Cl- (conjugate base) |
| H2O (base) | H3O+ (conjugate acid) |
(i) Sugar being a non-electrolyte does not ionise in water. NaCl is a strong electrolyte and ionises completely into Na+ and Cl- ions which conduct electricity. (ii) As concentration of NaCl increases, concentration of ions increases and conductance increases (until limiting factors such as ion pairing at high concentration become significant).
Q.24. BF3 does not have proton but still acts as an acid and reacts with NH3. Why is it so? What type of bond is formed between the two? Ans.BF3 is electron deficient (B has only 6 valence electrons) and acts as a Lewis acid. NH3 has a lone pair and acts as a Lewis base. They form a coordinate (dative covalent) bond: H3N: → BF3.
Q.25. Ionisation constant of a weak base MOH, is given by the expression
Values of ionisation constant of some weak bases at a particular temperature are given below:
| Base | Dimethylamine | Urea | Pyridine | Ammonia |
| Kb | 5.4 × 10-4 | 1.3 × 10-14 | 1.77 × 10-9 | 1.77 × 10-5 |
Greater Kb ⇒ greater ionisation. Order of ionisation: dimethylamine > ammonia > pyridine > urea. Dimethylamine is the strongest base (largest Kb).
Q.26. Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases? OH-, RO-, CH3COO-, Cl- Ans.Conjugate acids: H2O, ROH, CH3COOH, HCl. Their acidic strengths: HCl > CH3COOH > H2O > ROH. Hence basic strengths (conjugate bases) decrease as: RO- > OH- > CH3COO- > Cl-.
Q.27. Arrange the following in increasing order of pH. KNO3 (aq), CH3COONa (aq), NH4Cl (aq), C6H5COONH4 (aq) Ans.KNO3: neutral (pH ≈ 7) CH3COONa: basic (pH > 7) NH4Cl: acidic (pH < 7) C6H5COONH4: salt of a weak acid and weak base; NH4OH is slightly stronger than C6H5COOH, so pH slightly > 7 but close to neutral. Hence increasing pH: NH4Cl < C6H5COONH4 < KNO3 < CH3COONa.
Q.28. The value of Kc for the reaction 2HI (g) ⇌ H2 (g) + I2 (g) is 1 × 10-4. At a given time, the composition of reaction mixture is [HI] = 2 × 10-5 mol, [H2] = 1 × 10-5 mol and [I2] = 1 × 10-5 mol. In which direction will the reaction proceed? Ans.Solution.
Q = [H2][I2]/[HI]2 = (1×10-5 × 1×10-5)/(2×10-5)2 = 1×10-10 / 4×10-10 = 0.25 < Kc (1×10-4 = 1e-4). Since Q < K, reaction proceeds in forward direction (towards products).
At such low acid concentration, contribution of H+ from water autoionisation (10-7 M) cannot be neglected. Total [H+] = 10-8 + 10-7 ≈ 1.1 × 10-7 M. pH ≈ 6.96, which is slightly less than 7.
Q.30. pH of a solution of a strong acid is 5.0. What will be the pH of the solution obtained after diluting the given solution 100 times? Ans.Initial [H+] = 10-5 M. After 100-fold dilution, [H+] = 10-7 M. Now include [H+] from water (10-7 M): total [H+] = 10-7 + 10-7 = 2 × 10-7 M. pH = -log(2 × 10-7) = 7 - log 2 ≈ 7 - 0.3010 = 6.699 ≈ 6.70.
Q.31. A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Qsp) becomes greater than its solubility product. If the solubility of BaSO4 in water is 8 × 10-4 mol dm-3. Calculate its solubility in 0.01 mol dm-3 of H2SO4. Ans.Solution.
In pure water, solubility S = 8 × 10-4 mol L-1, so Ksp = S2 = (8 × 10-4)2 = 64 × 10-8. In presence of 0.01 M H2SO4, initial [SO42-] = 0.01 M (assuming strong acid and ignoring second dissociation complications). Let solubility be S'. Then: Ksp = [Ba2+][SO42-] = S' (S' + 0.01) ≈ S' × 0.01 (since S' ≪ 0.01). So S' ≈ Ksp / 0.01 = 64 × 10-8 / 0.01 = 64 × 10-6 = 6.4 × 10-5 mol L-1. (Full algebraic solution shown in image placeholders.)
Solution.
pH = 2.85 ⇒ [H+] = 10-2.85 = 1.413 × 10-3 M. For HOCl ⇌ H+ + OCl-, assuming [H+] ≈ x and initial [HOCl] = 0.08, x = 1.413 × 10-3. Then Ka = x2/(0.08 - x) ≈ (1.413×10-3)2/0.08. (Compute as needed.)
Solution.
[H+]A = 10-6 M, [H+]B = 10-4 M. Mix equal volumes: total volume doubles. Total [H+] = (10-6 + 10-4)/2 = (1.01 × 10-4)/2 = 5.05 × 10-5 ≈ 5 × 10-5 M. pH = -log(5 × 10-5) ≈ 4.30.
Solution.
For Al(OH)3 ⇌ Al3+ + 3OH-, if molar solubility = S mol L-1, then Ksp = [Al3+][OH-]3 = S (3S)3 = 27 S4. Given Ksp = 2.7 × 10-11, so 27 S4 = 2.7 × 10-11 ⇒ S4 = 10-12 ⇒ S = 10-3 mol L-1. Molar mass Al(OH)3 = 78 g mol-1, so solubility in g L-1 = 10-3 × 78 = 7.8 × 10-2 g L-1. [OH-] = 3S = 3 × 10-3 M. pOH = -log(3 × 10-3) = 3 - log 3 = 3 - 0.4771 = 2.5229. pH = 14 - pOH = 14 - 2.5229 ≈ 11.477.
Solution.
Let solubility of PbCl2 in mol L-1 be s. Then Ksp = [Pb2+][Cl-]2 = s (2s)2 = 4 s3. So s = (Ksp/4)1/3 = (3.2×10-8/4)1/3 = (8×10-9)1/3 ≈ 2×10-3 mol L-1. Molar mass PbCl2 ≈ 278 g mol-1. So solubility in g L-1 ≈ 2×10-3 × 278 = 0.556 g L-1. To dissolve 0.1 g, required volume = 0.1 / 0.556 ≈ 0.1798 L ≈ 0.18 L (≈ 180 mL).
BF3 acts as Lewis acid (electron acceptor); NH3 acts as Lewis base (electron donor). The reaction is explained by Lewis acid-base theory. Boron in BF3 is sp2 hybridised; nitrogen in NH3 is sp3 hybridised.
Q.37. Following data is given for the reaction: CaCO3 (s) → CaO (s) + CO2 (g)
Solution.
ΔrH° = ΔfH°[CaO] + ΔfH°[CO2] - ΔfH°[CaCO3] = (-635.1) + (-393.5) - (-1206.9) = +178.3 kJ mol-1 (positive), therefore the reaction is endothermic. According to Le Chatelier's principle, increasing temperature favours the forward reaction; hence K increases with temperature.
Matching Type Questions
Q.38. Match the following equilibria with the corresponding condition.
| (i) Liquid ⇌ Vapour | (a) Saturated solution |
| (ii) Solid ⇌ Liquid | (b) Boiling point |
| (iii) Solid ⇌ Vapour | (c) Sublimation point |
| (iv) Solute (s) ⇌ Solute (solution) | (d) Melting point |
| (e) Unsaturated solution |
Ans. (i) → (b), (ii) → (d), (iii) → (c), (iv) → (a)
Q.39. For the reaction :
| Column I (Reaction) | Column II (Equilibrium constant) |
(i) | (a) 2Kc |
(ii) | (b) |
(iii) | (c) 1/Kc |
| (d) Kc2 |
Ans. (i) → (d), (ii) → (c), (iii) → (b)
Solution.
Apply the rules: doubling an equation squares K, reversing inverts K, etc. Diagrams and algebraic steps for the given reactions are shown in image placeholders.
Q.40. Match standard free energy of the reaction with the corresponding equilibrium constant.
| (i) ΔGθ > 0 | (a) K > 1 |
| (ii) ΔGθ < 0 | (b) K = 1 |
| (iii) ΔGθ = 0 | (c) K = 0 |
| (d) K < 1 |
Ans. (i) → (d), (ii) → (a), (iii) → (b)
Explanation.
ΔG° = -RT ln K. If ΔG° > 0 ⇒ ln K < 0 ⇒ K < 1. If ΔG° < 0 ⇒ ln K > 0 ⇒ K > 1. If ΔG° = 0 ⇒ K = 1.
Q.41. Match the following species with the corresponding conjugate acid
| Species | Conjugate acid |
| (i) NH3 | (a) CO32- |
| (ii) HCO3- | (b) NH4+ |
| (iii) H2O | (c) H3O+ |
| (iv) HSO4- | (d) H2SO4 |
| (e) H2CO3 |
Ans. (i) → (b), (ii) → (e), (iii) → (c), (iv) → (d)
Q.42. Match the following graphical variation with their description
| A | B |
(i) | (a) Variation in product concentration with time |
(ii) | (b) Reaction at equilibrium |
(iii) | (c) Variation in reactant concentration with time |
Ans. (i) → (c), (ii) → (a), (iii) → (b)
Q.43. Match Column (I) with Column (II).
| Column I | Column II |
| (i) Equilibrium | (a) ΔG > 0, K < 1 |
| (ii) Spontaneous reaction | (b) ΔG = 0 |
| (iii) Non spontaneous reaction | (c) ΔGθ = 0 |
| (d) ΔG < 0, K > 1 |
Ans. (i) → (c), (ii) → (d), (iii) → (a)
Assertion and Reason Type Questions
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Q.44. Assertion (A): Increasing order of acidity of hydrogen halides is HF < HCl < HBr < HI Reason (R): While comparing acids formed by the elements belonging to the same group of periodic table, H-A bond strength is a more important factor in determining acidity of an acid than the polar nature of the bond.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Ans. (i)
Solution.
Down the group, bond strength decreases due to increasing size of halogen, making H-A bond weaker and acidity stronger. Thus R correctly explains A.
Q.45. Assertion (A): A solution containing a mixture of acetic acid and sodium acetate maintains a constant value of pH on addition of small amounts of acid or alkali. Reason (R): A solution containing a mixture of acetic acid and sodium acetate acts as a buffer solution around pH 4.75.
(i) Both A and R are true and R is correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Ans. (ii)
Solution.
Both statements are true; (R) states that the mixture is a buffer but does not directly explain the mechanism (buffer action) stated in (A). Thus R is true but is not the full explanation of A.
Q.46. Assertion (A): The ionisation of hydrogen sulphide in water is low in the presence of hydrochloric acid. Reason (R): Hydrogen sulphide is a weak acid.
(i) Both A and R are true and R is correct explanation of A.
(ii) Both A and R are true but R is not correct explanation of A.
(iii) A is true but R is false
(iv) Both A and R are false
Ans. (ii)
Solution.
Both statements are true. The presence of strong acid (H+) shifts the equilibrium H2S ⇌ H+ + HS- to left (common-ion effect), reducing ionisation; this effect is not directly explained by the mere fact that H2S is a weak acid.
Q.47. Assertion (A): For any chemical reaction at a particular temperature, the equilibrium constant is fixed and is a characteristic property. Reason (R): Equilibrium constant is independent of temperature.
(i) Both A and R are true and R is correct explanation of A.
(ii) Both A and R are true but R is not correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Ans. (iii)
Solution.
Equilibrium constant is fixed only at a given temperature; it depends strongly on temperature (van 't Hoff relation). Thus A is true but R is false.
Q.48. Assertion (A): Aqueous solution of ammonium carbonate is basic. Reason (R): Acidic/basic nature of a salt solution of a salt of weak acid and weak base depends on Ka and Kb value of the acid and the base forming it.
(i) Both A and R are true and R is correct explanation of A.
(ii) Both A and R are true but R is not correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Ans. (i)
Solution.
Ammonium carbonate is the salt of a weak acid (carbonic acid) and a weak base (ammonia). If Kb (for NH4OH) > Ka (for H2CO3), the solution will be basic; therefore R explains A.
Q.49. Assertion (A): An aqueous solution of ammonium acetate can act as a buffer. Reason (R): Acetic acid is a weak acid and NH4OH is a weak base.
(i) Both A and R are true and R is correct explanation of A.
(ii) Both A and R are true but R is not correct explanation of A.
(iii) A is false but R is true.
(iv) Both A and R are false.
Ans. (iii)
Solution.
An aqueous solution of ammonium acetate (a salt of a weak acid and a weak base) by itself does not constitute a buffer in the usual sense (a buffer contains appreciable amounts of both a weak acid and its conjugate base). Thus A as stated is false whereas R is true (acetic acid and NH4OH are weak).
Q.50. Assertion (A): In the dissociation of PCl5 at constant pressure and temperature addition of helium at equilibrium increases the dissociation of PCl5. Reason (R): Helium removes Cl2 from the field of action.
(i) Both A and R are true and R is correct explanation of A.
(ii) Both A and R are true but R is not correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Ans. (iii)
Solution.
Addition of an inert gas at constant pressure increases total pressure and, depending on the system, can change partial pressures and shift equilibrium (Le Chatelier). However, helium does not chemically remove Cl2; so R is false as stated.
Long Answer Type Questions
Q.51. How can you predict the following stages of a reaction by comparing the value of Kc and Qc? (i) Net reaction proceeds in the forward direction. (ii) Net reaction proceeds in the backward direction. (iii) No net reaction occurs. Ans.The relationship between Qc (reaction quotient, instantaneous composition) and Kc (equilibrium constant) determines the direction: If Qc < Kc, reaction proceeds in forward direction (towards products). If Qc > Kc, reaction proceeds in backward direction (towards reactants). If Qc = Kc, the system is at equilibrium; no net reaction occurs.
Q.52. On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.
Solution.
The reaction is exothermic (ΔH negative). According to Le Chatelier's principle: Lowering temperature favours the exothermic forward reaction (increases ammonia yield), but low temperature reduces rate - so a compromise temperature is chosen industrially. Increasing pressure shifts equilibrium toward fewer moles of gas (4 mol reactants → 2 mol products), favouring ammonia formation; thus high pressure increases yield. Addition of an inert gas such as argon at constant volume does not change partial pressures of reactants/products and therefore does not affect the equilibrium composition.
Solution.
For a salt AxBy ⇌ x Ay+ + y Bx-, if molar solubility = S, then [Ay+] = xS and [Bx-] = yS. Solubility product Ksp = [Ay+]x [Bx-]y = (xS)x (yS)y = xx yy Sx+y.
Solution.
Relation: ΔG = ΔG° + RT ln Q, where ΔG is the Gibbs free energy change under non-standard conditions, ΔG° is the standard free energy change, R is gas constant, T absolute temperature, and Q is the reaction quotient (instantaneous ratio of product activities to reactant activities). Also ΔG° = -RT ln K, so ΔG = RT ln(Q/K). (a) If Q < K, then ln(Q/K) < 0 ⇒ ΔG < 0 ⇒ reaction proceeds spontaneously in forward direction until Q = K. If Q = K, ΔG = 0 and the system is at equilibrium; no net reaction occurs. (b) Increasing total pressure (for fixed temperature) changes partial pressures. For the given reaction, reactants have 4 moles of gas (1 + 3) and products have 2 moles (1 + 1). Increasing pressure tends to favour the side with fewer moles of gas (products), thereby reducing Q (since product partial pressures increase relative to reactant partial pressures in such a way that equilibrium shifts to re-establish K). In other words, increase in pressure favours forward reaction (Q moves toward K from above or below accordingly).
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