Equilibrium
Welcome to HSLC Guru! This page provides complete English-medium notes, question-answers, MCQs, fill-in-the-blanks, true/false, glossary and a formula table for Class 11 Chemistry Chapter 7 — Equilibrium as per the ASSEB (Assam State School Education Board) syllabus. The chapter introduces physical and chemical equilibrium, the law of mass action, equilibrium constants Kc and Kp, Le Chatelier’s principle, and ionic equilibrium with detailed pH, Ka, Kb, buffer and Ksp calculations.
Summary
Equilibrium is a state in which two opposing processes occur at equal rates so that the observable properties of the system do not change with time. Physical equilibrium involves only a change of phase — for example, the equilibrium between liquid water and its vapour in a closed container, ice and water at 0 °C, or a saturated solution of sugar in water. Chemical equilibrium, in contrast, involves chemical reactions and is dynamic in nature: the forward and backward reactions continue at the molecular level, but their rates are equal so that the concentrations of reactants and products remain constant. A reaction can attain equilibrium only in a closed vessel.
The law of mass action, proposed by Guldberg and Waage, states that the rate of a chemical reaction at a given temperature is directly proportional to the product of the active masses (molar concentrations) of the reactants raised to powers equal to their stoichiometric coefficients. Applying this law to a reversible reaction aA + bB ⇌ cC + dD gives the equilibrium constant in terms of concentrations.
$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$
For gaseous reactions, the equilibrium constant can also be expressed in terms of partial pressures, Kp. The two are related by the change in number of moles of gas (Δng = moles of gaseous products − moles of gaseous reactants).
$K_p = K_c (RT)^{\Delta n_g}$
The reaction quotient Q has the same form as Kc but is calculated using the concentrations at any moment. If Q < Kc the forward reaction proceeds; if Q > Kc the reverse reaction proceeds; if Q = Kc the system is at equilibrium. ΔG° = −RT ln K relates the standard Gibbs energy change to the equilibrium constant. Le Chatelier’s principle states that if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to undo the change. Increase of pressure favours the side with fewer moles of gas; increase of temperature favours the endothermic direction; addition of an inert gas at constant volume does not affect the equilibrium; a catalyst does not shift the equilibrium but helps the system to attain equilibrium faster.
Ionic equilibrium deals with the equilibrium between ions and unionised molecules of weak electrolytes in aqueous solution. Three classical concepts of acids and bases are covered: Arrhenius (acid gives H+, base gives OH− in water), Brønsted–Lowry (acid is a proton donor, base is a proton acceptor — they form conjugate acid–base pairs), and Lewis (acid is an electron-pair acceptor, base is an electron-pair donor). pH is defined as pH = −log10[H+]; similarly pOH = −log[OH−], and at 25 °C pH + pOH = 14. The ionic product of water Kw = [H+][OH−] = 10−14 at 25 °C.
$pH = -\log_{10}[H^+]$
$K_a \cdot K_b = K_w = 10^{-14}\ \text{at }25^\circ\text{C}$
For a weak acid HA of concentration c with degree of dissociation α, Ostwald’s dilution law gives Ka = cα²/(1 − α); for very weak electrolytes (α << 1) this reduces to α ≈ √(Ka/c). The common-ion effect is the suppression of dissociation of a weak electrolyte by adding a strong electrolyte that has an ion in common with it. A buffer solution resists changes in pH on adding small amounts of acid or base; an acidic buffer is a mixture of a weak acid and its salt with a strong base (e.g. CH3COOH/CH3COONa), while a basic buffer is a mixture of a weak base and its salt with a strong acid (e.g. NH4OH/NH4Cl). The pH of a buffer is given by the Henderson–Hasselbalch equation.
$K_a = \frac{c\alpha^2}{1-\alpha}$
$pH = pK_a + \log\frac{[A^-]}{[HA]}$
For a sparingly soluble salt A+B−, the solubility product is the product of the ionic concentrations in a saturated solution. A precipitate forms when the ionic product exceeds Ksp.
$K_{sp} = [A^+][B^-]$
1-Mark Question Answers
Q1. Define chemical equilibrium.
Answer: It is the state of a reversible reaction in which the rates of the forward and backward reactions become equal so that the concentrations of reactants and products remain constant with time.
Q2. What is meant by the term “dynamic equilibrium”?
Answer: It means that although the macroscopic properties of the system do not change, the forward and backward reactions are still going on at the molecular level at equal rates.
Q3. Write the unit of Kc for the reaction N2 + 3H2 ⇌ 2NH3.
Answer: Δn = 2 − 4 = −2, so the unit of Kc is mol−2 L2 (or L2 mol−2).
Q4. Define pH.
Answer: pH is the negative logarithm to the base 10 of the molar concentration of hydrogen ions: pH = −log10[H+].
Q5. What is the value of Kw at 25 °C?
Answer: Kw = 1.0 × 10−14 mol2 L−2 at 25 °C.
Q6. State whether NaCl is acidic, basic or neutral in aqueous solution.
Answer: NaCl is the salt of a strong acid (HCl) and a strong base (NaOH); its aqueous solution is neutral with pH ≈ 7.
Q7. Write the conjugate base of HSO4−.
Answer: SO42−.
Q8. Define the solubility product (Ksp).
Answer: It is the product of the molar concentrations of the ions of a sparingly soluble salt in its saturated solution, each raised to the power of its stoichiometric coefficient.
Q9. Give one example of a Lewis acid that is not a Brønsted acid.
Answer: BF3 (it can accept a lone pair but cannot donate a proton).
Q10. What is a buffer solution?
Answer: A solution that resists a change in its pH on the addition of a small amount of a strong acid or a strong base, or on dilution.
Q10a. Write one example each of physical and chemical equilibrium.
Answer: Physical: H2O(l) ⇌ H2O(g) in a closed bottle. Chemical: N2(g) + 3H2(g) ⇌ 2NH3(g).
Q10b. Write the value of [H+] in pure water at 25 °C.
Answer: [H+] = 1.0 × 10−7 mol L−1.
2–3 Marks Question Answers
Q11. State and explain the law of mass action.
Answer: The law of mass action (Guldberg and Waage, 1864) states that the rate of a chemical reaction at a constant temperature is directly proportional to the product of the molar concentrations (active masses) of the reactants, each raised to the power equal to its stoichiometric coefficient. For aA + bB → products, rate = k[A]a[B]b. When applied to both forward and reverse reactions of an equilibrium, it leads to the equilibrium-constant expression Kc = [C]c[D]d/[A]a[B]b.
Q12. Distinguish between Kc and Kp. How are they related?
Answer: Kc is the equilibrium constant in terms of molar concentrations, while Kp is in terms of partial pressures of gaseous species. They are related by Kp = Kc(RT)Δng, where Δng = (moles of gaseous products) − (moles of gaseous reactants). When Δng = 0, Kp = Kc.
Q13. What is the reaction quotient Q? How does its comparison with Kc predict the direction of a reaction?
Answer: Q has the same algebraic form as Kc but is calculated with the concentrations existing at any moment, not necessarily at equilibrium. (i) If Q < Kc, the forward reaction is favoured. (ii) If Q > Kc, the reverse reaction is favoured. (iii) If Q = Kc, the system is at equilibrium and there is no net change.
Q14. State Le Chatelier’s principle and apply it to the synthesis of ammonia: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ.
Answer: Le Chatelier’s principle: when a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to nullify the change. (i) High pressure favours the forward reaction (4 mol → 2 mol). (ii) Low temperature favours the forward (exothermic) reaction, but is too slow industrially, so a moderate temperature (~700 K) plus a catalyst (Fe with Mo) is used. (iii) Removal of NH3 drives the equilibrium to the right.
Q15. Compare the Arrhenius, Brønsted–Lowry and Lewis concepts of acids and bases.
Answer: Arrhenius: acid gives H+ in water (HCl); base gives OH− in water (NaOH). Brønsted–Lowry: acid is a proton donor, base is a proton acceptor — they form conjugate pairs (HCl/Cl−, H2O/OH−). Lewis: acid accepts a lone pair (BF3, AlCl3), base donates a lone pair (NH3, H2O). The Lewis concept is the most general.
Q16. Explain the common-ion effect with an example.
Answer: The common-ion effect is the suppression of the dissociation of a weak electrolyte by adding a strong electrolyte having an ion in common with it. Example: in CH3COOH ⇌ CH3COO− + H+, addition of CH3COONa (which is fully ionised to give CH3COO−) shifts the equilibrium to the left, decreasing [H+] and increasing pH. Used in the qualitative analysis of cations and in buffer preparation.
Q16a. Distinguish between an acidic buffer and a basic buffer with one example each.
Answer: An acidic buffer is a mixture of a weak acid and the salt of that acid with a strong base; it maintains pH below 7. Example: CH3COOH + CH3COONa (pH ≈ 4.74). A basic buffer is a mixture of a weak base and the salt of that base with a strong acid; it maintains pH above 7. Example: NH4OH + NH4Cl (pH ≈ 9.25).
Q16b. Derive Ostwald’s dilution law for a weak monobasic acid.
Answer: Consider HA ⇌ H+ + A− with initial concentration c and degree of dissociation α. At equilibrium: [HA] = c(1 − α), [H+] = [A−] = cα. Therefore Ka = [H+][A−]/[HA] = (cα)(cα)/[c(1 − α)] = cα²/(1 − α). For weak electrolytes α << 1, so Ka ≈ cα², giving α = √(Ka/c) — i.e. degree of dissociation increases on dilution.
Q16c. Why does pure water show a small electrical conductivity?
Answer: Pure water self-ionises to a very small extent: 2H2O(l) ⇌ H3O+(aq) + OH−(aq). Even though [H+] = [OH−] = 10−7 M only at 25 °C, these ions cause the small but measurable conductivity of pure water.
5–7 Marks Question Answers (Numericals)
Q17. For the reaction H2(g) + I2(g) ⇌ 2HI(g), 1.0 mol H2 and 1.0 mol I2 are placed in a 1 L flask at 731 K. At equilibrium, 1.56 mol of HI is found. Calculate Kc.
Answer: Let x mol of H2 react. Then 2x = 1.56 → x = 0.78 mol. At equilibrium [H2] = [I2] = (1 − 0.78) = 0.22 M; [HI] = 1.56 M. Kc = [HI]2/([H2][I2]) = (1.56)2/(0.22 × 0.22) = 2.4336/0.0484 ≈ 50.3 (dimensionless, since Δn = 0).
Q18. For the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g) at 500 K, Kc = 1.8 × 10−3 mol L−1. Calculate Kp. (R = 0.0821 L·atm·K−1·mol−1)
Answer: Δng = (1 + 1) − 1 = 1. Kp = Kc(RT)1 = 1.8 × 10−3 × (0.0821 × 500) = 1.8 × 10−3 × 41.05 = 7.39 × 10−2 atm.
Q19. Calculate the pH of (a) 0.001 M HCl, (b) 0.01 M NaOH, (c) 0.1 M CH3COOH (Ka = 1.8 × 10−5).
Answer: (a) HCl is fully dissociated, [H+] = 10−3, pH = 3. (b) [OH−] = 10−2, pOH = 2, pH = 14 − 2 = 12. (c) [H+] = √(Ka × c) = √(1.8 × 10−5 × 0.1) = √(1.8 × 10−6) ≈ 1.34 × 10−3; pH = −log(1.34 × 10−3) ≈ 2.87.
Q20. A buffer is prepared from 0.20 M CH3COOH and 0.30 M CH3COONa. Calculate its pH (pKa of CH3COOH = 4.74).
Answer: Using the Henderson–Hasselbalch equation: pH = pKa + log([salt]/[acid]) = 4.74 + log(0.30/0.20) = 4.74 + log 1.5 = 4.74 + 0.176 = 4.92.
Q21. The Ksp of AgCl at 25 °C is 1.6 × 10−10. Find (a) its molar solubility in pure water, (b) its solubility in 0.10 M NaCl.
Answer: (a) Let s = solubility. Ksp = s × s = s2 → s = √(1.6 × 10−10) = 1.26 × 10−5 mol L−1. (b) Now [Cl−] ≈ 0.10 M (from NaCl). Ksp = s × 0.10 → s = 1.6 × 10−10/0.10 = 1.6 × 10−9 mol L−1. The solubility is reduced by the common-ion effect.
Q22. The degree of dissociation of 0.05 M NH4OH is 1.88 × 10−2. Calculate Kb and pOH of the solution.
Answer: Kb = cα²/(1 − α) ≈ cα² (since α is small) = 0.05 × (1.88 × 10−2)2 = 0.05 × 3.534 × 10−4 = 1.77 × 10−5. [OH−] = cα = 0.05 × 1.88 × 10−2 = 9.4 × 10−4; pOH = −log(9.4 × 10−4) ≈ 3.03; pH ≈ 10.97.
Q23. For 2SO2(g) + O2(g) ⇌ 2SO3(g), at 1000 K the equilibrium concentrations are [SO2] = 0.20 M, [O2] = 0.10 M and [SO3] = 0.40 M. Calculate Kc and predict the direction of reaction if a fresh mixture has [SO2] = [O2] = [SO3] = 0.10 M.
Answer: Kc = [SO3]2/([SO2]2[O2]) = (0.40)2/((0.20)2(0.10)) = 0.16/(0.04 × 0.10) = 0.16/0.004 = 40 M−1. For the new mixture, Q = (0.10)2/((0.10)2(0.10)) = 1/0.10 = 10 M−1. Since Q (=10) < Kc (=40), the reaction proceeds in the forward direction until equilibrium is restored.
Q24. Calculate the pH of a solution obtained by mixing 50 mL of 0.10 M HCl with 50 mL of 0.20 M NaOH.
Answer: Moles HCl = 0.050 × 0.10 = 0.005; moles NaOH = 0.050 × 0.20 = 0.010. After neutralisation, excess NaOH = 0.010 − 0.005 = 0.005 mol in total volume 100 mL = 0.10 L. [OH−] = 0.005/0.10 = 0.05 M. pOH = −log 0.05 = 1.30; pH = 14 − 1.30 = 12.70.
Multiple Choice Questions (MCQ)
1. The unit of Kc for 2NO2(g) ⇌ N2O4(g) is —
(a) mol L−1 (b) mol−1 L (c) dimensionless (d) mol2 L−2
Answer: (b) mol−1 L.
2. If Q < Kc, the reaction —
(a) is at equilibrium (b) goes backward (c) goes forward (d) stops
Answer: (c) goes forward.
3. For N2 + 3H2 ⇌ 2NH3, Kp/Kc equals —
(a) RT (b) (RT)−2 (c) (RT)2 (d) 1
Answer: (b) (RT)−2.
4. The pH of 10−8 M HCl is —
(a) 8 (b) 7 (c) slightly less than 7 (d) 6
Answer: (c) slightly less than 7 (water contributes H+).
5. Conjugate base of NH4+ is —
(a) NH3 (b) NH2− (c) N2H4 (d) NH4OH
Answer: (a) NH3.
6. Which is a Lewis acid? —
(a) NH3 (b) BF3 (c) H2O (d) OH−
Answer: (b) BF3.
7. An aqueous solution of NH4Cl is —
(a) acidic (b) basic (c) neutral (d) amphoteric
Answer: (a) acidic.
8. A catalyst at equilibrium —
(a) shifts to forward side (b) shifts to reverse side (c) does not affect K (d) decreases ΔH
Answer: (c) does not affect K.
9. The pH of pure water at 25 °C is —
(a) 0 (b) 7 (c) 14 (d) 1
Answer: (b) 7.
10. If Ksp of BaSO4 is 1.0 × 10−10, its solubility in mol L−1 is —
(a) 10−10 (b) 10−5 (c) 10−7 (d) 10−4
Answer: (b) 10−5.
Fill in the Blanks
1. Chemical equilibrium is __________ in nature.
Answer: dynamic.
2. Kp = Kc(RT)__________.
Answer: Δng.
3. The product Ka × Kb for a conjugate acid–base pair equals __________ at 25 °C.
Answer: 10−14.
4. A buffer of CH3COOH and CH3COONa is an __________ buffer.
Answer: acidic.
5. A precipitate forms when the ionic product __________ Ksp.
Answer: exceeds.
True / False
1. A catalyst changes the value of the equilibrium constant. — False.
2. For the reaction H2 + I2 ⇌ 2HI, Kp = Kc. — True.
3. Stronger the acid, smaller the value of Ka. — False.
4. Pure liquids and solids do not appear in the Kc expression. — True.
5. Addition of an inert gas at constant volume shifts the equilibrium. — False.
Glossary
| Term | Meaning |
|---|---|
| Equilibrium | State in which forward and backward rates are equal. |
| Reversible reaction | Reaction proceeding in both directions under the same conditions. |
| Active mass | Molar concentration (or partial pressure for gases). |
| Equilibrium constant | Ratio of product to reactant concentration terms at equilibrium. |
| Reaction quotient (Q) | Same form as Kc but at any moment. |
| Le Chatelier’s principle | System opposes any disturbance to its equilibrium. |
| Brønsted acid/base | Proton donor / acceptor. |
| Lewis acid/base | Electron-pair acceptor / donor. |
| Conjugate pair | Acid–base pair differing by one proton. |
| pH | −log10[H+]. |
| Kw | Ionic product of water = 10−14 at 25 °C. |
| Ostwald’s law | Relates Ka, c and α for a weak electrolyte. |
| Common-ion effect | Suppression of dissociation by a common ion. |
| Buffer | Solution that resists pH change. |
| Henderson–Hasselbalch | pH = pKa + log([A−]/[HA]). |
| Solubility product | Product of ionic concentrations of a sparingly soluble salt. |
| Hydrolysis | Reaction of salt ions with water giving acidic / basic solution. |
Formula Table
| Quantity | Formula |
|---|---|
| Equilibrium constant Kc | [C]c[D]d / [A]a[B]b |
| Relation between Kp and Kc | Kp = Kc(RT)Δng |
| Standard Gibbs energy | ΔG° = −RT ln K |
| pH | pH = −log10[H+] |
| pOH and pH | pH + pOH = 14 (at 25 °C) |
| Kw | Kw = [H+][OH−] = 10−14 |
| Ka · Kb | Ka × Kb = Kw |
| Ostwald’s dilution law | Ka = cα²/(1 − α) ≈ cα² |
| [H+] of weak acid | [H+] = √(Ka · c) |
| Henderson–Hasselbalch (acidic buffer) | pH = pKa + log([salt]/[acid]) |
| Henderson (basic buffer) | pOH = pKb + log([salt]/[base]) |
| Ksp of A+B− | Ksp = [A+][B−] = s² |
| Ksp of AB2 | Ksp = 4s³ |
| Hydrolysis constant | Kh = Kw/Ka (or Kw/Kb) |
This concludes the complete English-medium notes for Class 11 Chemistry Chapter 7 — Equilibrium as per the ASSEB syllabus. Practice the numerical problems on Kc, Kp, pH, buffer pH and Ksp, and revise Le Chatelier’s principle and the three acid–base concepts thoroughly for the examination. Visit HSLC Guru for the next chapter.