Chapter 6: Equilibrium Case Study Questions | chemistry Class 11 CBSE

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A chemistry teacher sets up a demonstration using a transparent sealed box containing a manometer. After removing a drying agent, a watch glass containing water is placed inside the box. Initially, the mercury level in the right limb of the manometer slowly rises as water evaporates, increasing the pressure inside the box. After some time, the mercury level becomes constant, indicating that the pressure inside has stabilized. The teacher explains that at this stage, the rate of evaporation equals the rate of condensation. The experiment is repeated with methyl alcohol, acetone, and ether, and it is observed that different liquids reach different constant pressures at the same temperature. The liquid that reaches the highest constant pressure is noted to have the lowest boiling point among the four liquids tested.

Question 1: What does the constant mercury level in the manometer indicate about the system inside the box?

The constant mercury level indicates that the pressure inside the box has stopped increasing, meaning the system has reached equilibrium — the rate of evaporation of water equals the rate of condensation of water vapour.
At this stage, the number of water molecules leaving the liquid surface per unit time equals the number of vapour molecules returning to the liquid per unit time; the vapour pressure has reached its equilibrium value at the given temperature.
This pressure is called the equilibrium vapour pressure of water at that temperature; it remains constant as long as temperature does not change, which is a characteristic property of every pure liquid.

Question 2: Why is it impossible to reach liquid-vapour equilibrium when the watch glass is kept open to the atmosphere instead of inside the sealed box?

In an open system, water molecules that evaporate are dispersed into the large volume of the room; the concentration of water vapour above the liquid surface remains very low because molecules continuously move away.
As a result, the rate of condensation (which depends on the concentration of vapour molecules striking the liquid surface) remains much lower than the constant rate of evaporation; the two rates never become equal.
Equilibrium in physical processes is possible only in a closed system where the vapour phase is confined; in an open system, the vapour escapes and the liquid eventually disappears completely without ever establishing equilibrium.

Question 3: Among the four liquids (water, methyl alcohol, acetone, ether), if ether shows the highest equilibrium vapour pressure at room temperature, what can be concluded about its boiling point, intermolecular forces, and volatility? How does this relate to Le Chatelier's principle when temperature is increased?

A higher equilibrium vapour pressure at a given temperature indicates weaker intermolecular forces (van der Waals or hydrogen bonding) in ether compared to the other liquids; less energy is needed for molecules to escape the liquid phase, so more molecules are in the vapour phase at equilibrium.
A higher vapour pressure means that the liquid reaches the condition 'vapour pressure = atmospheric pressure' at a lower temperature; therefore ether has the lowest boiling point among the four liquids and is the most volatile.
When temperature is increased, the equilibrium H2O(l) ⇌ H2O(vap) is disturbed. By Le Chatelier's principle, the system shifts in the endothermic direction (evaporation is endothermic) to counteract the temperature rise; more liquid evaporates and the vapour pressure increases. This is why vapour pressure of all liquids increases with temperature, and why the time for complete evaporation decreases in a warmer room.

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In an industrial plant producing ammonia by the Haber process, engineers need to decide the optimal conditions for maximizing yield. The reaction is N2(g) + 3H2(g) ⇌ 2NH3(g); ΔH = −92.38 kJ mol−1. Kp values at different temperatures are: at 298 K, Kp = 6.8 × 105; at 400 K, Kp = 41; at 500 K, Kp = 3.6 × 10−2. The plant operates at approximately 500°C and 200 atm pressure with an iron catalyst. Ammonia produced is continuously liquified and removed from the reaction vessel. The engineers note that without the catalyst, the reaction barely proceeds even at 500°C, but with the catalyst, equilibrium is reached within minutes. Annual world production of ammonia is about hundred million tonnes, primarily used as fertiliser.

Question 1: What does the trend in Kp values with increasing temperature reveal about the nature of the ammonia synthesis reaction?

The Kp values decrease sharply as temperature increases: from 6.8 × 105 at 298 K to 41 at 400 K to 3.6 × 10−2 at 500 K. This means the equilibrium shifts away from products (NH3) as temperature rises.
This behaviour is characteristic of an exothermic reaction (ΔH = −92.38 kJ mol−1); by Le Chatelier's principle, increasing temperature favours the reverse (endothermic) reaction, decreasing the equilibrium constant and the yield of ammonia.
Thus, low temperature is thermodynamically favourable for high yield of ammonia; however, very low temperatures slow the reaction rate, which is why a compromise temperature (~500°C) with a catalyst is used industrially.

Question 2: Explain why ammonia is continuously liquified and removed from the reaction vessel in the Haber process. Use the concept of reaction quotient Qc to justify this practice.

When ammonia is removed from the equilibrium mixture by liquification, the concentration of NH3 in the gas phase decreases; this reduces the numerator of the reaction quotient Qc = [NH3]2/([N2][H2]3), making Qc < Kc.
By Le Chatelier's principle (and quantitatively by the Qc < Kc condition), the forward reaction is favoured to restore equilibrium, causing more N2 and H2 to react and produce more NH3.
Continuous removal maintains Qc perpetually below Kc, preventing equilibrium from being established and ensuring the reaction keeps proceeding in the forward direction; this dramatically increases the overall conversion of reactants to ammonia beyond what a single-pass equilibrium would achieve.

Question 3: Calculate Kc for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) at 500 K given Kp = 3.6 × 10−2. Also explain why a catalyst is essential despite the large theoretical yield at 298 K, and why very high pressure (200 atm) is used.

Δn = 2 − (1+3) = −2; using Kp = Kc(RT)Δn: 3.6 × 10−2 = Kc × (0.0831 × 500)−2 = Kc × (41.55)−2 = Kc/1726.4; therefore Kc = 3.6 × 10−2 × 1726.4 = 62.15 ≈ 62.
Although Kp is very large at 298 K (6.8 × 105), the activation energy for N≡N bond breaking is very high, making the reaction rate negligibly slow at room temperature even with high Kc. Without a catalyst, thousands of years would be needed to reach equilibrium; Fritz Haber discovered that iron catalyst lowers activation energy equally for forward and reverse reactions, enabling equilibrium to be reached in minutes at 500°C.
High pressure (200 atm) is used because Δn = −2; by Le Chatelier's principle, high pressure shifts equilibrium toward the side with fewer moles of gas (product side, 2 mol vs. 4 mol reactant), significantly increasing the equilibrium yield of NH3. The combination of catalyst (for rate), moderate temperature (~500°C, compromise between rate and yield), and high pressure (200 atm) optimizes industrial ammonia production.

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A student studying ionic equilibrium prepares four solutions: (A) 0.1 M HCl, (B) 0.1 M CH3COOH, (C) 0.1 M NaOH, and (D) 0.1 M NH3. The student measures the pH of each solution and also tests their electrical conductivity. Solution A shows high conductivity and pH = 1. Solution B shows lower conductivity and pH ≈ 2.87. Solution C shows high conductivity and pH = 13. Solution D shows lower conductivity and pH ≈ 11.12. The student then adds 0.1 M CH3COONa to solution B and observes that the pH rises to approximately 4.74. Ka of acetic acid = 1.74 × 10−5; Kb of NH3 = 1.77 × 10−5.

Question 1: Why does solution A (0.1 M HCl) show higher conductivity and lower pH than solution B (0.1 M CH3COOH) at the same concentration?

HCl is a strong acid that completely dissociates in water: HCl → H+ + Cl−; all 0.1 mol L−1 of HCl produces 0.1 mol L−1 of H+ and Cl− ions, giving pH = −log(0.1) = 1 and very high ionic concentration for conductivity.
CH3COOH is a weak acid that partially dissociates: CH3COOH ⇌ H+ + CH3COO−; only a small fraction (α ≈ 1.3%) ionizes, giving [H+] ≈ 1.35 × 10−3 M, pH ≈ 2.87. The low ionic concentration means lower conductivity.
The fundamental difference is the extent of ionization: strong electrolytes are completely dissociated (nearly 100%), while weak electrolytes establish an equilibrium with a dynamic balance between ionized and unionized forms; conductivity depends on total ion concentration, which is much lower for weak acids at the same nominal concentration.

Question 2: Explain why adding 0.1 M CH3COONa to solution B (0.1 M CH3COOH) raises the pH from 2.87 to 4.74.

Sodium acetate completely ionises to give Na+ and CH3COO− ions. The CH3COO− ion is the conjugate base of acetic acid; its addition to the acetic acid equilibrium CH3COOH ⇌ H+ + CH3COO− increases [CH3COO−], shifting equilibrium to the left by the common ion effect.
This suppresses the ionization of acetic acid, drastically reducing [H+] from ~1.35 × 10−3 M (pH 2.87) to about Ka = 1.74 × 10−5 M (pH 4.74) when [salt] = [acid]; the mixture now acts as an acetate buffer described by the Henderson-Hasselbalch equation: pH = pKa + log([CH3COO−]/[CH3COOH]) = 4.76 + log(0.1/0.1) = 4.76 ≈ 4.74.
The solution is now a buffer: it resists further pH changes on addition of small amounts of acid or base. Adding more H+ ions shifts the equilibrium to form more CH3COOH (consuming H+), while adding OH− ions reacts with CH3COOH to form CH3COO−, resisting the pH rise.

Question 3: Compare the pH values of solutions B (0.1 M CH3COOH) and D (0.1 M NH3). Calculate [OH−] in solution D and explain using Ka × Kb = Kw why the pKa + pKb = 14 relationship is important for these two solutions.

Solution B (weak acid, pH 2.87): [H+] = 10−2.87 ≈ 1.35 × 10−3 M; degree of ionization α = 1.35 × 10−3/0.1 = 1.35%. Solution D (weak base, pH 11.12): pOH = 14 − 11.12 = 2.88; [OH−] = 10−2.88 ≈ 1.32 × 10−3 M; degree of ionization of NH3 = 1.32 × 10−3/0.1 = 1.32%.
For CH3COOH and NH3 at 0.1 M: Ka(CH3COOH) = 1.74 × 10−5; Kb(NH3) = 1.77 × 10−5; [H+] ≈ √(Ka × c) and [OH−] ≈ √(Kb × c). Since Ka ≈ Kb, [H+] from CH3COOH ≈ [OH−] from NH3, explaining the nearly mirror-image pH values about 7.
For any conjugate acid-base pair, Ka × Kb = Kw = 10−14 (at 298 K), giving pKa + pKb = 14. For CH3COOH/CH3COO− pair: Ka = 1.74 × 10−5, so Kb(CH3COO−) = 10−14/1.74 × 10−5 = 5.75 × 10−10. For NH4+/NH3 pair: Kb(NH3) = 1.77 × 10−5, so Ka(NH4+) = 10−14/1.77 × 10−5 = 5.65 × 10−10. This relationship enables calculation of the strength of conjugate species and is essential for hydrolysis and buffer calculations.

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During a laboratory practical, students investigate the equilibrium H2(g) + I2(g) ⇌ 2HI(g) at 731 K in a sealed vessel. Six experiments are performed: experiments 1−4 start with only H2 and I2 at various initial concentrations, while experiments 5 and 6 start with only HI. After sufficient time, intensity of the purple colour of iodine vapour becomes constant. Students record equilibrium concentrations and calculate the expression [HI]2/([H2][I2]) for each experiment. All six experiments yield values close to 46.4, regardless of initial conditions or whether the equilibrium was approached from the reactant or product side. The teacher uses this data to introduce the Law of Chemical Equilibrium.

Question 1: What conclusion can be drawn from the fact that the expression [HI]2/([H2][I2]) gives a constant value of 46.4 in all six experiments?

The constant value of [HI]2/([H2][I2]) = 46.4 regardless of initial concentrations demonstrates that at a given temperature, the ratio of the product of equilibrium concentrations of products to reactants (each raised to stoichiometric coefficients) is always constant — this is the equilibrium constant Kc = 46.4 for this reaction at 731 K.
The fact that the same Kc is obtained whether starting from pure reactants (experiments 1−4) or pure products (experiments 5−6) proves that chemical equilibrium can be approached from either direction and is independent of initial concentrations.
This data validates the Law of Chemical Equilibrium (Law of Mass Action), proposed by Guldberg and Waage in 1864: Kc = [C]c[D]d/[A]a[B]b for the general reaction aA + bB ⇌ cC + dD at constant temperature.

Question 2: If the equilibrium constant Kc = 46.4 for H2(g) + I2(g) ⇌ 2HI(g), what is the Kc for (i) the reverse reaction 2HI(g) ⇌ H2(g) + I2(g), and (ii) the reaction ½H2(g) + ½I2(g) ⇌ HI(g)?

For the reverse reaction 2HI(g) ⇌ H2(g) + I2(g), the equilibrium constant is the inverse of the forward reaction: K'c = 1/Kc = 1/46.4 = 0.0216. This is because the expression inverts (products become reactants and vice versa).
For the reaction multiplied by ½: ½H2(g) + ½I2(g) ⇌ HI(g), the equilibrium constant is the square root of the original: K''c = (Kc)1/2 = (46.4)1/2 = 6.81. When the equation is multiplied by factor n, Kc is raised to the power n.
These relationships confirm that the value of the equilibrium constant is specific to the form of the balanced chemical equation; it is therefore essential to always specify the equation when reporting a Kc value.

Question 3: At some point during experiment 1, the concentrations are [H2] = 0.10 M, [I2] = 0.20 M, and [HI] = 0.40 M (Kc = 46.4 at 700 K). Calculate Qc and predict the direction of the reaction. After equilibrium is reached, if more I2 is added, predict and explain the new equilibrium position.

Qc = [HI]2/([H2][I2]) = (0.40)2/(0.10 × 0.20) = 0.16/0.02 = 8.0. Since Qc (8.0) < Kc (46.4), the reaction mixture is not at equilibrium and will proceed in the forward direction (left to right) — more HI will form as H2 and I2 are consumed, until Qc reaches 46.4.
After equilibrium is established, adding more I2 increases [I2], making Qc = [HI]2/([H2][I2]) decrease below Kc (denominator increases while numerator is momentarily unchanged). This signals the forward reaction to proceed.
By Le Chatelier's principle, the system responds to the increased [I2] by consuming it through the forward reaction; more HI is produced and [H2] decreases until a new equilibrium is established. Kc remains unchanged (temperature constant), but the equilibrium concentrations shift: [HI] increases, [H2] decreases, and [I2] is higher than original but lower than immediately after addition.

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Case Set 5

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A student prepares a saturated solution of sugar in water and then adds a small amount of radioactive sugar (same chemical compound, but containing a radioactive carbon isotope) to the saturated solution. Initially, no radioactivity is detected in the solid sugar at the bottom. After some time, the student measures radioactivity both in the solution and in the solid sugar at the bottom of the beaker. Radioactivity is now detected in both phases, and the ratio of radioactive to non-radioactive sugar molecules in solution rises until it reaches a constant value. The student records these observations as evidence for the dynamic nature of dissolution equilibrium.

Question 1: What does the appearance of radioactivity in the solid sugar prove about the nature of equilibrium in a saturated solution?

Radioactivity appearing in the solid sugar proves that sugar molecules from the solution are continuously crystallizing out (depositing onto the solid phase), even though the total mass of solid sugar remains visually unchanged — this demonstrates that the equilibrium is dynamic, not static.
If the equilibrium were static (all processes stopped), radioactive sugar molecules added to the solution could never enter the solid phase; their appearance in the solid confirms continuous exchange of sugar molecules between solid and solution phases.
The equilibrium Sugar(solid) ⇌ Sugar(solution) involves simultaneous and equal rates of dissolution and crystallization; the constancy of the radioactive-to-non-radioactive ratio at a final value confirms that a true dynamic equilibrium has been established.

Question 2: What would happen to the solubility of sugar in this saturated solution if the temperature were increased? Explain in terms of dissolution equilibrium.

Increasing temperature generally increases the solubility of most solid solutes in water. For the dissolution equilibrium Sugar(solid) ⇌ Sugar(solution), if dissolution is endothermic (which it is for most sugars), increasing temperature shifts the equilibrium to the right (forward direction), dissolving more solid sugar.
By Le Chatelier's principle, the system counteracts the temperature increase by proceeding in the endothermic direction (dissolution), absorbing heat and dissolving more sugar; the concentration of sugar in solution at the new equilibrium is higher.
Practically, this is why making a thick sugar syrup requires dissolving sugar at high temperature; when the syrup is cooled to room temperature, the equilibrium shifts left and sugar crystals separate out, because the lower temperature reduces solubility.

Question 3: Compare the dissolution equilibria of solids in liquids and gases in liquids in terms of the effect of temperature and pressure, using sugar and carbon dioxide as respective examples. How does the Henry's law equilibrium for CO2 explain the fizzing of soda water?

Solid-liquid dissolution (sugar in water): Solubility is constant at a given temperature and generally increases with temperature (endothermic dissolution). Pressure has negligible effect on the solubility of solids in liquids because solids and liquids are nearly incompressible.
Gas-liquid dissolution (CO2 in water): Governed by Henry's law — mass of dissolved gas is proportional to the pressure of gas above the solvent. CO2(g) ⇌ CO2(aq); [CO2(aq)]/[CO2(g)] is constant at a given temperature. Solubility of gases decreases with increasing temperature (exothermic dissolution) and increases with increasing pressure.
Soda water is sealed under high CO2 pressure, maintaining high dissolved CO2 by Henry's law. When the bottle is opened, CO2 partial pressure above the liquid drops to its atmospheric level (~0.0004 atm); the equilibrium shifts left (dissolved CO2 >> equilibrium value for low pressure), so excess dissolved CO2 escapes as bubbles (fizzing). The rate and volume of fizzing increase at higher temperatures because CO2 solubility decreases further with warming, releasing even more CO2.

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Chapter 6: Equilibrium | Case Study Questions

Passage

Question 1: What does the constant mercury level in the manometer indicate about the system inside the box?

Question 2: Why is it impossible to reach liquid-vapour equilibrium when the watch glass is kept open to the atmosphere instead of inside the sealed box?

Passage

Question 1: What does the trend in Kp values with increasing temperature reveal about the nature of the ammonia synthesis reaction?

Question 2: Explain why ammonia is continuously liquified and removed from the reaction vessel in the Haber process. Use the concept of reaction quotient Qc to justify this practice.

Question 3: Calculate Kc for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) at 500 K given Kp = 3.6 × 10−2. Also explain why a catalyst is essential despite the large theoretical yield at 298 K, and why very high pressure (200 atm) is used.

Passage

Question 1: Why does solution A (0.1 M HCl) show higher conductivity and lower pH than solution B (0.1 M CH3COOH) at the same concentration?

Question 2: Explain why adding 0.1 M CH3COONa to solution B (0.1 M CH3COOH) raises the pH from 2.87 to 4.74.

Question 3: Compare the pH values of solutions B (0.1 M CH3COOH) and D (0.1 M NH3). Calculate [OH−] in solution D and explain using Ka × Kb = Kw why the pKa + pKb = 14 relationship is important for these two solutions.

Passage

Question 1: What conclusion can be drawn from the fact that the expression [HI]2/([H2][I2]) gives a constant value of 46.4 in all six experiments?

Question 2: If the equilibrium constant Kc = 46.4 for H2(g) + I2(g) ⇌ 2HI(g), what is the Kc for (i) the reverse reaction 2HI(g) ⇌ H2(g) + I2(g), and (ii) the reaction ½H2(g) + ½I2(g) ⇌ HI(g)?

Question 3: At some point during experiment 1, the concentrations are [H2] = 0.10 M, [I2] = 0.20 M, and [HI] = 0.40 M (Kc = 46.4 at 700 K). Calculate Qc and predict the direction of the reaction. After equilibrium is reached, if more I2 is added, predict and explain the new equilibrium position.

Passage

Question 1: What does the appearance of radioactivity in the solid sugar prove about the nature of equilibrium in a saturated solution?

Question 2: What would happen to the solubility of sugar in this saturated solution if the temperature were increased? Explain in terms of dissolution equilibrium.

Question 3: Compare the dissolution equilibria of solids in liquids and gases in liquids in terms of the effect of temperature and pressure, using sugar and carbon dioxide as respective examples. How does the Henry's law equilibrium for CO2 explain the fizzing of soda water?

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