Chapter 6: Equilibrium Long Questions | chemistry Class 11 CBSE

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Chapter 6: Equilibrium Long Questions | chemistry Class 11 CBSE | ByteNotes.in

Long Answer

Medium difficulty • Structured explanation

Medium

Question 1

Long Form

Analyse the general characteristics of physical equilibria and explain how each type of physical equilibrium (solid-liquid, liquid-vapour, solid-vapour, and dissolution equilibria) demonstrates these characteristics.

All physical equilibria share five common characteristics: they are possible only in a closed system at a given temperature; opposing processes occur simultaneously at the same rate; all measurable properties remain constant; each equilibrium is characterised by a constant value of a specific parameter at a given temperature; and the magnitude of that parameter indicates the extent to which the process has proceeded.
In solid-liquid equilibrium (e.g., ice-water at 273 K), the rates of melting and freezing are equal, mass of both phases remains constant, and the temperature remains fixed at the normal melting point for a given pressure. This illustrates that only one temperature exists for solid-liquid coexistence at a given pressure.
In liquid-vapour equilibrium, the equilibrium vapour pressure is constant at a given temperature and increases with temperature. More volatile liquids with weaker intermolecular forces have higher vapour pressure; equilibrium cannot be reached in an open system because vapour molecules are dispersed and the condensation rate is too low.
In solid-vapour equilibrium (e.g., solid iodine in a closed vessel), the rate of sublimation equals the rate of condensation; the intensity of colour becomes constant when equilibrium is attained, confirming that both processes continue at equal rates.
For dissolution equilibria, solubility of solids in liquids is constant at a given temperature (confirmed by radioactive sugar experiments), and for gases in liquids, Henry's law governs equilibrium: the dissolved gas concentration is proportional to the pressure of the gas above the liquid.
In all cases, the equilibrium is dynamic at the molecular level; this is why radioactive tracers can detect exchange between phases even when macroscopic properties appear unchanged. These characteristics collectively define physical equilibrium as a dynamic, temperature-specific, steady state.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 2

Long Form

Derive the relationship between Kp and Kc for a general gaseous reaction. Illustrate with the example of ammonia synthesis and explain when Kp equals Kc.

For the general reaction aA + bB ⇌ cC + dD involving gases, Kp is expressed in terms of partial pressures and Kc in terms of molar concentrations. Using the ideal gas equation p = [gas]RT, each partial pressure can be replaced: pA = [A]RT, pB = [B]RT, etc.
Substituting into Kp = (pCc × pDd)/(pAa × pBb), we get Kp = ([C]c[D]d/[A]a[B]b) × (RT)(c+d)–(a+b) = Kc × (RT)Δn, where Δn = (c+d) – (a+b), the difference in moles of gaseous products and reactants.
For N2(g) + 3H2(g) ⇌ 2NH3(g), Δn = 2 – (1+3) = –2; hence Kp = Kc(RT)–2, meaning Kp < Kc for this reaction at all temperatures above 0 K.
Kp = Kc only when Δn = 0, i.e., when the number of moles of gaseous reactants equals the number of moles of gaseous products, as in H2(g) + I2(g) ⇌ 2HI(g) where Δn = 0.
Pressure must be expressed in bar for this relationship because the standard state for pressure is 1 bar; 1 Pa = 1 N m–2 and 1 bar = 105 Pa. Using other units would give numerically different Kp values.
This relationship is essential for interconverting Kp and Kc, which represent the same equilibrium from different experimental perspectives; both are equally valid expressions of the same chemical equilibrium.

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Long Answer

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Hard

Question 3

Long Form

Analyse the three types of salt hydrolysis (strong acid-strong base, weak acid-strong base, and strong acid-weak base) and explain how the pH of their solutions is determined.

Salts of strong acids and strong bases (e.g., NaCl, KBr) do not undergo hydrolysis because cations like Na+, K+ and anions like Cl–, Br– simply get hydrated in water without reacting; their solutions are neutral with pH = 7 at 298 K.
Salts of weak acids and strong bases (e.g., CH3COONa) completely ionize in water. The acetate ion undergoes hydrolysis: CH3COO–(aq) + H2O(l) ⇌ CH3COOH(aq) + OH–(aq), because acetic acid is a weak acid (Ka = 1.8 × 10–5) and remains largely unionized.
The result is an increase in [OH–] concentration, making the solution alkaline with pH > 7. The degree of hydrolysis depends on the Ka of the weak acid; a weaker acid produces a more alkaline solution.
Salts of strong acids and weak bases (e.g., NH4Cl) give acidic solutions. NH4+ undergoes hydrolysis: NH4+(aq) + H2O(l) ⇌ NH4OH(aq) + H+(aq); since NH4OH is a weak base (Kb = 1.77 × 10–5) and remains mainly unionized, [H+] increases and pH < 7.
For salts of weak acids and weak bases (e.g., CH3COONH4), both cation and anion hydrolyze. The pH is given by pH = 7 + ½(pKa – pKb); if pKa > pKb, pH > 7 (basic); if pKa < pKb, pH < 7 (acidic); if pKa = pKb, pH = 7.
The degree of hydrolysis for salts of weak acid and weak base is independent of solution concentration, which distinguishes it from the other types; pH changes for all types depend only on the relative strengths of the acid and base involved.

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Long Answer

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Hard

Question 4

Long Form

Compare the Arrhenius, Brønsted-Lowry, and Lewis theories of acids and bases, highlighting the scope and limitations of each with appropriate examples.

Arrhenius theory defines acids as substances that give H+ ions in aqueous solution (e.g., HCl → H+ + Cl–) and bases as substances that give OH– ions (e.g., NaOH → Na+ + OH–). Its limitation is that it applies only to aqueous solutions and fails to explain the basicity of NH3, which has no OH group.
Brønsted-Lowry theory defines an acid as a proton donor and a base as a proton acceptor, introducing conjugate acid-base pairs. For example, in HCl + H2O → H3O+ + Cl–, HCl is the acid, H2O is the base, H3O+ is the conjugate acid, and Cl– is the conjugate base. This theory is applicable in non-aqueous solvents and explains the amphoteric nature of water.
The Brønsted-Lowry concept also explains that strong acids (HClO4, HCl, H2SO4) have very weak conjugate bases (ClO4–, Cl–, HSO4–), while weak acids like HF and CH3COOH have relatively stronger conjugate bases (F–, CH3COO–).
Lewis theory is the broadest: a Lewis acid accepts an electron pair and a Lewis base donates an electron pair. For example, BF3 (electron-deficient) acts as a Lewis acid by accepting a lone pair from NH3 (Lewis base) to form BF3:NH3. This definition includes as acids many species that lack protons, such as AlCl3, Co3+, and Mg2+.
Species like H2O, NH3, and OH– are Lewis bases because they can donate lone pairs; a bare proton H+ is a Lewis acid because it can accept a lone pair. Lewis theory thus accommodates a far wider range of acid-base reactions.
In terms of scope: Arrhenius < Brønsted-Lowry < Lewis. However, for ionization equilibria in aqueous solutions (Ka, Kb, pH calculations), the Arrhenius and Brønsted-Lowry frameworks remain the most practical and directly applicable.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 5

Long Form

Explain with suitable examples how the magnitude of the equilibrium constant predicts the extent of a reaction. Discuss the relationship between Kc, Gibbs energy, and reaction spontaneity.

The magnitude of Kc indicates the extent of the reaction: if Kc > 103, products predominate and the reaction proceeds nearly to completion (e.g., H2 + O2 at 500 K has Kc = 2.4 × 1047); if Kc < 10–3, reactants predominate (e.g., N2 + O2 ⇌ 2NO at 298 K has Kc = 4.8 × 10–31); if Kc is between 10–3 and 103, appreciable concentrations of both are present.
The equilibrium constant does not give information about the rate at which equilibrium is reached; a reaction can have a very large Kc but still proceed very slowly in the absence of a catalyst.
The Gibbs energy relationship is ΔG = ΔG° + RT lnQ; at equilibrium, ΔG = 0 and Q = Kc, giving ΔG° = –RT lnK, or equivalently K = e–ΔG°/RT.
If ΔG° < 0, then –ΔG°/RT is positive, e–ΔG°/RT > 1, so K > 1, indicating a spontaneous reaction where products predominate; if ΔG° > 0, K < 1, indicating the reverse reaction is spontaneous; if ΔG° = 0, K = 1.
This thermodynamic interpretation shows that chemical equilibrium is the state of minimum Gibbs energy for the system; the system moves spontaneously toward lower Gibbs energy until ΔG = 0 at equilibrium.
Practically, the relationship ΔG° = –RT lnK allows calculation of Kc from thermodynamic data, as illustrated by the phosphorylation of glucose in glycolysis where ΔG° = 13.8 kJ/mol gives Kc = 3.81 × 10–3, confirming that the reaction is non-spontaneous under standard conditions.

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Long Answer

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Hard

Question 6

Long Form

Analyse the factors affecting the ionization of weak acids, including the role of H-A bond strength, polarity, and position in the periodic table. Support with examples.

The extent of ionization of an acid HA depends on the strength and polarity of the H-A bond. When the H-A bond strength decreases (lower energy required to break the bond), HA becomes a stronger acid because proton release is easier.
For elements in the same group of the periodic table, bond strength is the dominant factor. As atomic size of A increases down the group, H-A bond length increases and bond strength decreases, so acid strength increases: HF << HCl << HBr << HI; similarly, H2S is a stronger acid than H2O.
For elements in the same period (row) of the periodic table, H-A bond polarity becomes the deciding factor. As the electronegativity of A increases across the period, the H-A bond becomes more polar and charge separation facilitates proton release: CH4 < NH3 < H2O < HF in terms of acid strength.
Additional factors include solvation of ions: greater solvation of the conjugate base A– by water stabilizes it and increases the tendency of HA to ionize. HCl, despite a stronger H-Cl bond than HI, is essentially completely ionized due to extensive solvation of Cl– in water.
For polyprotic acids, successive ionization constants decrease (Ka1 > Ka2 > Ka3) because removing further protons from an already negatively charged ion requires overcoming electrostatic repulsion, making the process progressively more difficult.
Indicators like phenolphthalein and bromothymol blue behave as weak acids (HIn) with characteristic colours in acid and base forms; their Ka values determine the pH range of colour change, and this property is exploited in acid-base titrations.

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Long Answer

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Hard

Question 7

Long Form

Explain how changes in concentration, pressure, and temperature affect the equilibrium of the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). How are these principles applied in the industrial synthesis of ammonia?

Adding N2 or H2 (reactants) to the equilibrium mixture decreases Qc below Kc, causing the forward reaction to proceed to restore equilibrium, increasing NH3 yield. Removing NH3 (product) also maintains Qc < Kc, driving the reaction forward; this is industrially exploited by liquifying and removing NH3 continuously from the reaction vessel.
This reaction has Δn = 2 – 4 = –2; increasing pressure shifts equilibrium toward the product side (fewer moles of gas), increasing NH3 yield. At high pressure (~200 atm), the yield of ammonia is significantly enhanced.
Temperature effect: ΔH = –92.38 kJ mol–1, so the reaction is exothermic. According to Le Chatelier's principle, raising temperature shifts equilibrium to the left, decreasing Kc and reducing NH3 yield. Low temperature is thermodynamically favourable.
However, at very low temperatures, the reaction rate becomes too slow to achieve equilibrium in a reasonable time. A compromise temperature of about 500°C is used, balancing reasonable reaction rate with acceptable equilibrium yield.
Fritz Haber discovered that an iron catalyst lowers the activation energy equally for forward and reverse reactions, enabling the reaction to reach equilibrium at a satisfactory rate at 500°C, where the equilibrium concentration of NH3 is reasonably favourable.
The optimum industrial conditions for ammonia synthesis are approximately 500°C and 200 atm with an iron catalyst; continuous removal of ammonia product keeps Qc < Kc and drives the reaction forward, achieving high overall conversion in continuous flow systems.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 8

Long Form

Describe the pH scale, derive the relationship pH + pOH = 14 at 298 K, and explain how pH is measured experimentally. Calculate the pH of a solution with [H+] = 3.8 × 10–3 M.

The pH scale is a logarithmic scale defined as pH = –log(aH+) = –log([H+]/mol L–1); in dilute solutions (< 0.01 M), activity approximates molarity, so pH ≈ –log[H+]. At 25°C, pure water has [H+] = 10–7 M, giving pH = 7. Acidic solutions have pH < 7, basic solutions pH > 7, and neutral solutions pH = 7.
The ionic product of water, Kw = [H3O+][OH–] = 10–14 at 298 K. Taking the negative logarithm: –log Kw = –log[H3O+] – log[OH–], giving pKw = pH + pOH = 14. This relationship is always satisfied in aqueous solutions at 298 K regardless of the nature of the solute.
A change of one pH unit corresponds to a tenfold change in [H+]; a change of two pH units corresponds to a hundredfold change. This logarithmic nature makes the pH scale practical for expressing the wide range of H+ concentrations encountered.
pH can be measured roughly using pH paper, which shows different colours at different pH values and can determine pH in the range 1–14 with an accuracy of about ±0.5. For greater accuracy, a pH meter measures the pH-dependent electrical potential of the test solution to a precision of ±0.001.
For the calculation: pH = –log[3.8 × 10–3] = –{log 3.8 + log 10–3} = –{0.58 + (–3.0)} = –{–2.42} = 2.42. Since pH < 7, the solution is acidic; this is consistent with the given high concentration of H+ ions.
pH measurements have critical applications in biological systems (human blood pH = 7.4), cosmetics, pharmaceuticals, and food science; deviations from normal pH in body fluids indicate malfunctioning, which is why maintaining physiological pH is essential.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 9

Long Form

Explain the solubility product constant (Ksp) and derive the expression for molar solubility in terms of Ksp for a salt of type MxXy. How does the common ion effect reduce the solubility of Ni(OH)2 in 0.10 M NaOH?

The solubility product constant Ksp is the equilibrium constant for the dissolution of a sparingly soluble salt; for MxXy(s) ⇌ xMp+(aq) + yXq–(aq), where x × p+ = y × q–, the expression is Ksp = [Mp+]x[Xq–]y, excluding the pure solid.
For a general salt MxXy with molar solubility S: [Mp+] = xS and [Xq–] = yS; substituting, Ksp = (xS)x(yS)y = xx·yy·S(x+y), giving S = (Ksp / xx·yy)1/(x+y). For BaSO4 (x = y = 1): S = √Ksp = √(1.1 × 10–10) = 1.05 × 10–5 mol L–1.
The common ion effect: when NaOH (0.10 M) is present, [OH–] = 0.10 M initially. For Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH–(aq), if solubility is S, [Ni2+] = S and [OH–] = 0.10 + 2S ≈ 0.10 (since Ksp is very small).
Substituting: Ksp = [Ni2+][OH–]2 = S × (0.10)2 = 2.0 × 10–15, giving S = 2.0 × 10–15 / 0.01 = 2.0 × 10–13 M. In pure water, S = (Ksp/4)1/3 ≈ 7.9 × 10–6 M, so solubility is dramatically reduced by the common ion.
This principle is used analytically: adding a common ion to precipitate a sparingly soluble salt almost completely for gravimetric estimation (e.g., precipitating Ag+ as AgCl, Fe3+ as Fe(OH)3, Ba2+ as BaSO4).
The solubility of salts of weak acids increases at lower pH because the anion is protonated; reducing [X–] shifts dissolution equilibrium forward. Solubility S = {Ksp([H+] + Ka)/Ka}1/2, showing S increases as [H+] increases (lower pH). This is important in biological mineral dissolution processes.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 10

Long Form

Explain the concept of homogeneous and heterogeneous equilibria with examples. Write the Kc expressions for the thermal decomposition of CaCO3, the Haber process, and the esterification of acetic acid with ethanol.

Homogeneous equilibrium exists when all reactants and products are in the same phase. Examples include gaseous reactions like N2(g) + 3H2(g) ⇌ 2NH3(g) and solution reactions like Fe3+(aq) + SCN–(aq) ⇌ [Fe(SCN)]2+(aq). In homogeneous equilibria, all species are included in the Kc expression.
Heterogeneous equilibrium involves species in more than one phase. The equilibrium between liquid water and its vapour (H2O(l) ⇌ H2O(g)) is a heterogeneous equilibrium involving liquid and gas phases. For such equilibria, pure solids and liquids are excluded from the Kc expression.
For thermal decomposition of calcium carbonate: CaCO3(s) ⇌ CaO(s) + CO2(g); since CaCO3 and CaO are pure solids with constant molar concentrations, K'c = [CO2(g)], and Kp = pCO2. At 1100 K, pCO2 = 2 × 105 Pa, so Kp = 2.00.
For the Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g); Kc = [NH3]2 / ([N2][H2]3). This is a homogeneous gaseous equilibrium; at 500 K, Kc = 3.55 × 102 (as computed in the textbook example).
For esterification: CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l); since water is not in excess and not a solvent here, it is included: Kc = [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH]. This is a homogeneous liquid-phase equilibrium.
For Ni(s) + 4CO(g) ⇌ Ni(CO)4(g), the solid Ni is excluded: Kc = [Ni(CO)4] / [CO]4. In all heterogeneous cases, the key rule is that the concentration of any pure solid or liquid is constant and absorbed into the equilibrium constant.

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Chapter 6: Equilibrium | Long Questions

Analyse the general characteristics of physical equilibria and explain how each type of physical equilibrium (solid-liquid, liquid-vapour, solid-vapour, and dissolution equilibria) demonstrates these characteristics.

Derive the relationship between Kp and Kc for a general gaseous reaction. Illustrate with the example of ammonia synthesis and explain when Kp equals Kc.

Analyse the three types of salt hydrolysis (strong acid-strong base, weak acid-strong base, and strong acid-weak base) and explain how the pH of their solutions is determined.

Compare the Arrhenius, Brønsted-Lowry, and Lewis theories of acids and bases, highlighting the scope and limitations of each with appropriate examples.

Explain with suitable examples how the magnitude of the equilibrium constant predicts the extent of a reaction. Discuss the relationship between Kc, Gibbs energy, and reaction spontaneity.

Analyse the factors affecting the ionization of weak acids, including the role of H-A bond strength, polarity, and position in the periodic table. Support with examples.

Explain how changes in concentration, pressure, and temperature affect the equilibrium of the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). How are these principles applied in the industrial synthesis of ammonia?

Describe the pH scale, derive the relationship pH + pOH = 14 at 298 K, and explain how pH is measured experimentally. Calculate the pH of a solution with [H+] = 3.8 × 10–3 M.

Explain the solubility product constant (Ksp) and derive the expression for molar solubility in terms of Ksp for a salt of type MxXy. How does the common ion effect reduce the solubility of Ni(OH)2 in 0.10 M NaOH?

Explain the concept of homogeneous and heterogeneous equilibria with examples. Write the Kc expressions for the thermal decomposition of CaCO3, the Haber process, and the esterification of acetic acid with ethanol.

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