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Equilibrium

NCERT Class 11 · Chemistry Based on NCERT Class 11 Chemistry textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter explores the dynamic nature of equilibrium in both physical and chemical processes, introducing the law of chemical equilibrium and the equilibrium constant. It covers homogeneous and heterogeneous equilibria, the relationship between Kp and Kc, and applications of equilibrium constants to predict reaction extent and direction. The big idea is that equilibrium is a dynamic state where forward and reverse reaction rates are equal, and the equilibrium constant quantifies the relative concentrations of reactants and products at a given temperature.
Key Concepts (14)
Dynamic equilibrium
A state in a reversible reaction where the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant over time, though both reactions continue to occur.
e.g. In a closed container, water evaporates and condenses at the same rate, so the amount of liquid water and water vapour remains constant.
Equilibrium mixture
The mixture of reactants and products present in the equilibrium state of a reversible reaction.
e.g. In the reaction H2(g) + I2(g) ⇌ 2HI(g), the equilibrium mixture contains H2, I2, and HI at constant concentrations.
Law of chemical equilibrium (Law of mass action)
At a given temperature, the product of the concentrations of the reaction products raised to their respective stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value.
e.g. For the reaction aA + bB ⇌ cC + dD, Kc = [C]^c[D]^d / [A]^a[B]^b.
Equilibrium constant (Kc)
The constant value obtained from the equilibrium constant expression for a given reversible reaction at a specific temperature, expressed in terms of molar concentrations.
e.g. For H2(g) + I2(g) ⇌ 2HI(g), Kc = [HI]^2 / [H2][I2] = 57.0 at 700 K.
Equilibrium constant (Kp)
The equilibrium constant expressed in terms of partial pressures of gaseous reactants and products, used for reactions involving gases.
e.g. For N2(g) + 3H2(g) ⇌ 2NH3(g), Kp = (pNH3)^2 / (pN2)(pH2)^3.
Homogeneous equilibrium
An equilibrium in which all reactants and products are in the same phase (e.g., all gases or all in aqueous solution).
e.g. N2(g) + 3H2(g) ⇌ 2NH3(g) is a homogeneous gaseous equilibrium.
Heterogeneous equilibrium
An equilibrium in which the reactants and products are in more than one phase.
e.g. CaCO3(s) ⇌ CaO(s) + CO2(g) is a heterogeneous equilibrium involving solid and gas phases.
Reaction quotient (Qc)
A quantity defined in the same way as the equilibrium constant Kc, but using concentrations that are not necessarily at equilibrium, used to predict the direction of a reaction.
e.g. For H2(g) + I2(g) ⇌ 2HI(g), Qc = [HI]t^2 / [H2]t[I2]t at any time t.
Vapour pressure
The constant pressure exerted by the vapour of a liquid in equilibrium with its liquid phase at a given temperature.
e.g. Water in a closed container at 25°C exerts a vapour pressure of about 23.8 mm Hg.
Normal boiling point
The temperature at which the liquid and vapour phases of a pure substance are in equilibrium at one atmospheric pressure (1.013 bar).
e.g. The normal boiling point of water is 100°C at 1.013 bar.
Normal melting point
The temperature at which the solid and liquid phases of a pure substance are in equilibrium at one atmospheric pressure.
e.g. The normal melting point of ice is 0°C at 1 atm.
Henry's law
The mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the pressure of the gas above the solvent.
e.g. In a sealed soda water bottle, CO2 is dissolved under high pressure; when opened, pressure drops and CO2 escapes.
Saturated solution
A solution in which no more solute can be dissolved at a given temperature, and a dynamic equilibrium exists between the dissolved solute and the undissolved solid.
e.g. A sugar syrup at room temperature that has undissolved sugar crystals at the bottom is a saturated solution.
Standard state
A reference state for a substance: for a pure gas, it is 1 bar pressure; for a solute in solution, it is 1 molar concentration.
e.g. When calculating Kp, pressures are expressed relative to 1 bar, making Kp dimensionless.
Formulas & Equations (12)
Kc = [C]^c[D]^d / [A]^a[B]^b
Reads as: Kc equals concentration of C raised to c times concentration of D raised to d, divided by concentration of A raised to a times concentration of B raised to b.
Variables: [A], [B], [C], [D] are equilibrium molar concentrations (mol/L); a, b, c, d are stoichiometric coefficients.
Use when: Writing the equilibrium constant expression for any reversible reaction at equilibrium.
⚠ Forgetting to raise concentrations to the power of their stoichiometric coefficients, or using initial concentrations instead of equilibrium concentrations.
Kp = (pC)^c (pD)^d / (pA)^a (pB)^b
Reads as: Kp equals partial pressure of C raised to c times partial pressure of D raised to d, divided by partial pressure of A raised to a times partial pressure of B raised to b.
Variables: pA, pB, pC, pD are equilibrium partial pressures (usually in bar); a, b, c, d are stoichiometric coefficients.
Use when: Writing the equilibrium constant for gaseous reactions in terms of partial pressures.
⚠ Using concentrations instead of partial pressures, or forgetting that solids and pure liquids do not appear in the expression.
Kp = Kc (RT)^(Δn)
Reads as: Kp equals Kc times (RT) raised to delta n.
Variables: R = 0.0831 bar L mol⁻¹ K⁻¹, T = temperature in Kelvin, Δn = (moles of gaseous products) – (moles of gaseous reactants).
Use when: Converting between Kc and Kp for a gaseous reaction.
⚠ Using the wrong value of R (e.g., 0.0821 L atm mol⁻¹ K⁻¹ without converting units), or miscalculating Δn (e.g., forgetting to count only gaseous species).
Qc = [C]^c[D]^d / [A]^a[B]^b
Reads as: Qc equals concentration of C raised to c times concentration of D raised to d, divided by concentration of A raised to a times concentration of B raised to b.
Variables: Same as Kc, but concentrations are at any given time, not necessarily at equilibrium.
Use when: Predicting the direction of a reaction by comparing Qc with Kc.
⚠ Assuming Qc = Kc always, or using Qc to calculate equilibrium concentrations directly without setting up an ICE table.
p = cRT
Reads as: Pressure equals concentration times R times T.
Variables: p = partial pressure (bar), c = concentration (mol/L), R = 0.0831 bar L mol⁻¹ K⁻¹, T = temperature (K).
Use when: Converting between concentration and partial pressure of a gas.
⚠ Forgetting to use consistent units (e.g., using R = 0.0821 L atm mol⁻¹ K⁻¹ when pressure is in bar).
H2O(l) ⇌ H2O(g)
Reads as: Liquid water is in equilibrium with water vapour.
Variables: Represents a physical equilibrium between liquid and vapour phases.
Use when: Describing liquid-vapour equilibrium or vapour pressure.
⚠ Thinking this is a static equilibrium; it is dynamic with equal rates of evaporation and condensation.
N2(g) + 3H2(g) ⇌ 2NH3(g)
Reads as: Nitrogen gas plus three moles of hydrogen gas are in equilibrium with two moles of ammonia gas.
Variables: Gaseous reactants and products; Δn = 2 – (1+3) = –2.
Use when: Example of homogeneous gaseous equilibrium; used in Haber's process.
⚠ Writing Kc expression incorrectly: Kc = [NH3]^2 / [N2][H2]^3, not [NH3]^2 / [N2][H2].
H2(g) + I2(g) ⇌ 2HI(g)
Reads as: Hydrogen gas plus iodine gas are in equilibrium with two moles of hydrogen iodide gas.
Variables: Δn = 2 – (1+1) = 0, so Kp = Kc.
Use when: Classic example where Kp = Kc; used to illustrate law of mass action.
⚠ Forgetting that the equilibrium constant expression uses the square of [HI] because the coefficient is 2.
CaCO3(s) ⇌ CaO(s) + CO2(g)
Reads as: Solid calcium carbonate is in equilibrium with solid calcium oxide and carbon dioxide gas.
Variables: Heterogeneous equilibrium; solids do not appear in Kc or Kp expression.
Use when: Example of heterogeneous equilibrium; Kp = pCO2.
⚠ Including concentrations of CaCO3(s) and CaO(s) in the equilibrium expression.
CO2(g) + C(s) ⇌ 2CO(g)
Reads as: Carbon dioxide gas plus solid carbon are in equilibrium with two moles of carbon monoxide gas.
Variables: Heterogeneous equilibrium; solid carbon does not appear in Kp expression.
Use when: Example of heterogeneous equilibrium involving a solid and gases.
⚠ Including [C(s)] in the equilibrium constant expression.
K'c = 1/Kc (for reverse reaction)
Reads as: K prime c equals one over Kc.
Variables: Kc is the equilibrium constant for the forward reaction; K'c is for the reverse reaction.
Use when: Finding the equilibrium constant for the reverse reaction.
⚠ Forgetting to take the reciprocal; instead, students may write K'c = Kc.
K''c = (Kc)^n (when equation is multiplied by n)
Reads as: K double prime c equals Kc raised to the power n.
Variables: n is the factor by which the stoichiometric coefficients are multiplied.
Use when: When the balanced equation is multiplied by a factor, the new equilibrium constant is the original raised to that factor.
⚠ Multiplying Kc by n instead of raising it to the power n.
How It Works
Attainment of chemical equilibrium
  1. Step 1: Reactants are placed in a closed vessel at a constant temperature; the forward reaction begins, converting reactants to products.
  2. Step 2: As products accumulate, the reverse reaction starts, converting products back to reactants.
  3. Step 3: Over time, the rate of the forward reaction decreases (as reactant concentrations drop) and the rate of the reverse reaction increases (as product concentrations rise).
  4. Step 4: Eventually, the rates of the forward and reverse reactions become equal, and the concentrations of all species become constant — this is dynamic equilibrium.
💡 Equilibrium can be reached from either direction (starting with only reactants or only products), and the same equilibrium mixture is obtained if the total number of atoms is the same.
Using the reaction quotient (Qc) to predict direction
  1. Step 1: Calculate Qc using the current concentrations (not necessarily equilibrium) in the same form as the Kc expression.
  2. Step 2: Compare Qc with Kc.
  3. Step 3: If Qc < Kc, the forward reaction is favoured (net reaction goes left to right).
  4. Step 4: If Qc > Kc, the reverse reaction is favoured (net reaction goes right to left).
  5. Step 5: If Qc = Kc, the system is at equilibrium.
💡 Qc uses the same formula as Kc, but with any concentrations at a given time, not just equilibrium concentrations.
Calculating equilibrium concentrations using ICE table
  1. Step 1: Write the balanced chemical equation.
  2. Step 2: Create a table with rows for Initial concentration, Change in concentration, and Equilibrium concentration (ICE).
  3. Step 3: Define x as the change in concentration of one reactant or product, and use stoichiometry to express all changes in terms of x.
  4. Step 4: Write the equilibrium concentrations in terms of x and substitute into the Kc expression.
  5. Step 5: Solve for x (may require solving a quadratic equation).
  6. Step 6: Calculate the equilibrium concentrations using the value of x that makes chemical sense (positive, less than initial concentrations).
💡 Always check that the calculated x is physically possible (e.g., not leading to negative concentrations).
Establishing liquid-vapour equilibrium
  1. Step 1: Place a liquid (e.g., water) in a closed container with a manometer to measure pressure.
  2. Step 2: Initially, the space above the liquid contains no vapour; as evaporation occurs, the pressure increases.
  3. Step 3: The rate of evaporation is constant, but as vapour accumulates, the rate of condensation increases.
  4. Step 4: Eventually, the rate of evaporation equals the rate of condensation, and the pressure becomes constant — this is the equilibrium vapour pressure.
💡 Equilibrium vapour pressure depends only on temperature and the nature of the liquid, not on the amount of liquid or the size of the container.
Demonstrating dynamic nature of equilibrium using isotopes
  1. Step 1: Perform the Haber process (N2 + 3H2 ⇌ 2NH3) using deuterium (D2) instead of H2, reaching equilibrium with ND3.
  2. Step 2: Mix this equilibrium mixture with another equilibrium mixture obtained using ordinary H2 (containing NH3).
  3. Step 3: After some time, analyse the mixture using a mass spectrometer.
  4. Step 4: The presence of mixed isotopic species (e.g., NH2D, NHD2, HD) confirms that forward and reverse reactions continue even at equilibrium, proving dynamic nature.
💡 If equilibrium were static, no mixing of isotopes would occur; the scrambling of H and D atoms proves ongoing reactions.
Exam Traps — Don't Make These Mistakes
⚠ Including solids or pure liquids in equilibrium expression
✗ Kc = [CaO][CO2] / [CaCO3]
→ The concentrations of pure solids and pure liquids are constant and do not appear in the equilibrium constant expression. Only gases and aqueous species are included.
✓ Kc = [CO2] or Kp = pCO2 (for CaCO3(s) ⇌ CaO(s) + CO2(g)).
⚠ Confusing Kc and Kp when Δn = 0
✗ Kp = Kc(RT)^Δn, and for H2(g) + I2(g) ⇌ 2HI(g), Δn = 0, so Kp = Kc × 0 = 0.
→ Any number raised to the power 0 is 1, not 0. So (RT)^0 = 1, and Kp = Kc.
✓ Kp = Kc(RT)^0 = Kc × 1 = Kc.
⚠ Using initial concentrations instead of equilibrium concentrations in Kc
✗ Kc = [HI]^2 / [H2][I2] using the initial concentrations of H2, I2, and HI.
→ Kc is defined only using equilibrium concentrations. Initial concentrations are not at equilibrium unless Qc = Kc.
✓ Use the equilibrium concentrations (after the reaction has reached constant composition) in the Kc expression.
⚠ Forgetting to square or raise concentrations to stoichiometric coefficients
✗ Kc = [HI] / [H2][I2] for H2(g) + I2(g) ⇌ 2HI(g).
→ The stoichiometric coefficient of HI is 2, so its concentration must be squared in the equilibrium expression.
✓ Kc = [HI]^2 / [H2][I2].
⚠ Misinterpreting the magnitude of Kc for extent of reaction
✗ If Kc is very large (e.g., 10^47), the reaction does not occur at all.
→ A very large Kc means the products are strongly favoured; the reaction proceeds nearly to completion, not that it doesn't occur.
✓ If Kc > 10^3, products predominate; if Kc < 10^-3, reactants predominate; if between 10^-3 and 10^3, both are present appreciably.
⚠ Confusing Qc and Kc in direction prediction
✗ If Qc > Kc, the reaction proceeds in the forward direction.
→ If Qc > Kc, there are too many products relative to equilibrium, so the reverse reaction is favoured to reduce product concentrations.
✓ If Qc > Kc, net reaction proceeds from right to left (reverse direction). If Qc < Kc, net reaction proceeds from left to right (forward direction).
Compare & Contrast
Definition and units
Kc
Kp
Kc is expressed in terms of molar concentrations (mol/L), while Kp is expressed in terms of partial pressures (usually bar). They are related by Kp = Kc(RT)^(Δn).
Phases of reactants and products
Homogeneous equilibrium
Heterogeneous equilibrium
In homogeneous equilibrium, all reactants and products are in the same phase (e.g., all gases). In heterogeneous equilibrium, reactants and products are in different phases (e.g., solid and gas), and pure solids or liquids do not appear in the equilibrium constant expression.
When concentrations are measured
Reaction quotient (Qc)
Equilibrium constant (Kc)
Qc is calculated using concentrations at any given time (not necessarily at equilibrium), while Kc is calculated only using equilibrium concentrations. Comparing Qc with Kc predicts the direction of the reaction.
Direction of reaction
Forward reaction
Reverse reaction
The forward reaction converts reactants to products; the reverse reaction converts products back to reactants. At equilibrium, their rates are equal. The equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction.
Exam Focus
Marks pattern: Typically 2-3 MCQs (1 mark each), 1 short answer (2-3 marks) on writing Kc/Kp expressions, 1 long answer (5 marks) involving ICE table calculations or Kp-Kc conversion, and sometimes a 3-mark question on predicting direction using Qc.
Most asked: Writing equilibrium constant expressions (Kc and Kp) for given reactions · Relationship between Kp and Kc (Kp = Kc(RT)^Δn) · Predicting direction of reaction using Qc vs Kc · Calculating equilibrium concentrations using ICE table · Heterogeneous equilibrium and omission of solids/liquids · Effect of temperature on equilibrium (Le Chatelier's principle — though not explicitly in this text, it is a standard follow-up)
Don't skip: Numerical problems involving ICE table to find equilibrium concentrations and Kc/Kp; Conversion between Kp and Kc using Kp = Kc(RT)^Δn; Writing correct equilibrium expressions for heterogeneous reactions
One-Line Revision (read aloud before the exam)
  1. Equilibrium is dynamic: forward and reverse reactions occur at equal rates, not stopped.
  2. The law of mass action: Kc = [products]^coefficients / [reactants]^coefficients at equilibrium.
  3. Pure solids and pure liquids are omitted from equilibrium constant expressions.
  4. Kp = Kc(RT)^Δn, where Δn = moles of gaseous products – moles of gaseous reactants.
  5. For H2(g) + I2(g) ⇌ 2HI(g), Δn = 0, so Kp = Kc.
  6. If Qc < Kc, the forward reaction is favoured; if Qc > Kc, the reverse reaction is favoured.
  7. A very large Kc (>10^3) means the reaction proceeds nearly to completion.
  8. A very small Kc (<10^-3) means the reaction hardly proceeds; reactants dominate.
  9. Equilibrium can be reached from either direction (starting with reactants or products).
  10. Vapour pressure is the constant pressure of vapour in equilibrium with its liquid at a given temperature.
  11. Henry's law: mass of gas dissolved in a solvent is proportional to the pressure of the gas above the solvent.
  12. The equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction.
  13. When a balanced equation is multiplied by n, the new equilibrium constant is (Kc)^n.
  14. In heterogeneous equilibrium, only gases and aqueous species appear in the Kc/Kp expression.
  15. The normal boiling point is the temperature at which liquid and vapour are in equilibrium at 1 atm pressure.

MCQs — 10 Questions with Answers

Q1. At 273 K and 1 atm pressure, ice and water are in equilibrium. Which statement correctly describes this system?

  • A. Molecules from ice and water exchange at the boundary, but the mass of each phase remains constant. ✓
  • B. All molecular motion ceases and the system is completely static.
  • C. The rate of melting is greater than the rate of freezing.
  • D. No exchange of molecules occurs between the two phases.

Answer: A — At the melting point, dynamic equilibrium means molecules continuously exchange between phases at equal rates, so masses remain constant despite ongoing molecular activity.

Q2. In a closed container, water is placed and left undisturbed. Initially, the rate of evaporation is greater than the rate of condensation. Which of the following correctly explains what happens next?

  • A. Evaporation continues to exceed condensation indefinitely.
  • B. As water vapour accumulates, condensation rate increases until it equals evaporation rate. ✓
  • C. Both rates decrease but evaporation always remains faster.
  • D. Condensation rate exceeds evaporation rate once equilibrium is reached.

Answer: B — In a closed container, accumulating vapour molecules increase collision with liquid surface, raising condensation rate until equilibrium is achieved when both rates are equal.

Q3. Acetone, ethyl alcohol, and water are each placed in watch glasses and exposed to the atmosphere at the same temperature. The order of evaporation rates at equilibrium is: acetone > ethyl alcohol > water. This implies which sequence is correct?

  • A. Vapour pressure: water > ethyl alcohol > acetone
  • B. Volatility: acetone > ethyl alcohol > water ✓
  • C. Boiling point: acetone > ethyl alcohol > water
  • D. Kinetic energy of molecules: water > ethyl alcohol > acetone

Answer: B — Higher evaporation rate indicates higher vapour pressure and greater volatility; volatile substances have weaker intermolecular forces and escape more readily.

Q4. At a given temperature, the vapour pressure of water is 23.7 mmHg. A student claims this value will decrease if the temperature is increased to 30°C while keeping the container closed. Is this statement correct?

  • A. Yes, because condensation increases with temperature.
  • B. No, vapour pressure increases with temperature because more molecules gain kinetic energy to escape. ✓
  • C. Yes, because evaporation decreases at higher temperatures.
  • D. No, because the container is closed and volume is fixed.

Answer: B — Vapour pressure is a temperature-dependent property; higher temperature provides molecules with greater kinetic energy, increasing evaporation rate and equilibrium vapour pressure.

Q5. Which of the following is NOT a characteristic of a system in dynamic equilibrium?

  • A. Forward and reverse processes occur simultaneously.
  • B. Macroscopic properties (concentration, pressure) remain constant.
  • C. All molecular motion in the system stops completely. ✓
  • D. Rate of forward reaction equals rate of reverse reaction.

Answer: C — Dynamic equilibrium is defined by continuous molecular activity at equal rates; complete cessation of molecular motion would indicate a static state, not dynamic equilibrium.

Q6. An experimenter uses a drying agent to remove water vapour from a closed box, then quickly places a watch glass containing water inside. Initially, the mercury level in the manometer rises rapidly, but the rate of rise decreases over time until it becomes zero. Explain this observation using equilibrium concepts.

  • A. Evaporation rate decreases because water molecules are depleted.
  • B. Evaporation rate remains constant, but condensation rate increases until both rates are equal, reaching equilibrium. ✓
  • C. Water molecules lose kinetic energy as the box cools down.
  • D. The manometer reading becomes constant because evaporation stops completely.

Answer: B — Constant evaporation rate produces vapour; increasing vapour density raises condensation rate until equilibrium is achieved, stopping further pressure increase despite ongoing molecular exchange.

Q7. If a closed container with liquid water is heated from 20°C to 60°C, what happens to the equilibrium vapour pressure of water in the container?

  • A. It decreases because temperature increases molecular spacing.
  • B. It increases because more water molecules have sufficient kinetic energy to evaporate. ✓
  • C. It remains constant because the container is closed and sealed.
  • D. It decreases because evaporation becomes less favourable at higher temperatures.

Answer: B — Vapour pressure is an intensive property dependent on temperature; higher temperature increases molecular kinetic energy, enhancing evaporation and raising equilibrium vapour pressure.

Q8. Two statements are given: (I) At the melting point, the rate of melting of ice equals the rate of freezing of water. (II) At the melting point under atmospheric pressure, no change in temperature occurs even when heat is supplied to the ice-water mixture. Which of the following is correct?

  • A. Both statements are true; statement (II) explains why (I) is observed. ✓
  • B. Only statement (I) is true; statement (II) is incorrect.
  • C. Both statements are true but independent of each other.
  • D. Only statement (II) is true; statement (I) is incorrect.

Answer: A — Statement (I) defines dynamic equilibrium at the melting point; statement (II) reflects that at equilibrium, supplied heat converts ice to water at constant temperature without changing the equilibrium temperature.

Q9. In a sealed container at 25°C, liquid acetone is in equilibrium with its vapour. The vapour pressure is 246 mmHg. If the same mass of acetone is placed in an identical sealed container at 25°C but the container volume is doubled, what will happen to the final equilibrium vapour pressure?

  • A. It will decrease to 123 mmHg because there are fewer molecules per unit volume.
  • B. It will remain approximately 246 mmHg because vapour pressure is independent of volume and depends only on temperature. ✓
  • C. It will increase because more space allows more evaporation.
  • D. It will decrease gradually until it reaches zero as the container expands.

Answer: B — Vapour pressure is an intensive property determined solely by temperature and the nature of the liquid; container volume does not affect the equilibrium pressure at a given temperature.

Q10. HOTS: A student observes that in a closed room, a wet cloth dries faster in summer (35°C) than in winter (15°C). Using concepts of dynamic equilibrium and vapour pressure, explain why this occurs and predict what would happen if the room were not closed (open to atmosphere). (i) In summer, the higher temperature increases the vapour pressure of water and the rate of evaporation. (ii) In winter, the lower temperature decreases both vapour pressure and evaporation rate. (iii) In an open room, the dried air would be continuously replaced by fresh air, favouring further evaporation regardless of season.

  • A. Statements (i) and (ii) are correct; (iii) explains the open-room scenario.
  • B. Only statement (i) is correct; the cloth dries identically in closed and open rooms.
  • C. Statements (ii) and (iii) are correct; statement (i) is contradictory.
  • D. All three statements are correct and together explain the phenomenon comprehensively. ✓

Answer: D — Temperature directly affects vapour pressure and evaporation rate (i, ii); in open systems, continuous removal of saturated air prevents equilibrium and maintains faster drying (iii)—all three address the complete picture.

Flashcards

What is dynamic equilibrium?

A state where forward and reverse reactions occur simultaneously at equal rates, resulting in constant concentrations of reactants and products, though molecular exchange continues.

Define vapour pressure.

The pressure exerted by water or liquid molecules in the gaseous state when evaporation and condensation rates are equal at a given temperature.

What is the normal melting point?

The temperature at atmospheric pressure at which solid and liquid phases of a pure substance are in equilibrium; for ice and water, it is 273 K or 0°C.

Why is equilibrium called 'dynamic' and not 'static'?

Because molecules continue to move and exchange between phases or products and reactants at the molecular level, even though macroscopic properties appear unchanging.

What happens to vapour pressure as temperature increases?

Vapour pressure increases with temperature because more liquid molecules have sufficient kinetic energy to escape into the gaseous phase.

Which condition is essential for a physical or chemical equilibrium to establish?

A closed system (no exchange of matter or energy with surroundings) is essential so that forward and reverse processes can balance each other.

Explain why rate of evaporation equals rate of condensation at equilibrium using molecular kinetics.

Initially evaporation rate exceeds condensation rate, but as vapour accumulates, more molecules return to liquid phase until both rates match, preventing further net change.

What is the relationship between volatility and boiling point?

A liquid with higher vapour pressure is more volatile and has a lower boiling point because molecules escape more easily at lower temperatures.

Why does the mass of ice and water remain constant at 273 K and 1 atm pressure?

The rate at which water molecules freeze equals the rate at which ice molecules melt, so there is no net change in the amount of either phase.

What does the double half-arrow symbol (⇌) represent in equilibrium equations?

It indicates that the reaction proceeds in both forward and reverse directions simultaneously, emphasizing the dynamic nature of the equilibrium.

Important Board Questions

Define dynamic equilibrium and distinguish it from static equilibrium with one example each. [2 marks]

State that dynamic equilibrium involves simultaneous forward-reverse processes at equal rates with molecular exchange; static equilibrium has no molecular-level activity. Example: ice⇌water at 273K for dynamic; completely frozen ice below 273K for static.

A watch glass containing water is placed inside a closed, dry box that was previously evacuated using a drying agent. The mercury level in an attached manometer rises initially but eventually becomes constant. Explain this observation in terms of evaporation and condensation rates, and describe what happens at the molecular level when equilibrium is reached. [5 marks]

Explain that initially, evaporation dominates (no vapour present); as vapour accumulates, condensation rate increases until both rates equal, stopping further pressure rise. At equilibrium, molecules continuously transfer between liquid and vapour phases at equal rates despite constant macroscopic pressure—this is dynamic equilibrium.

A student is asked why acetone evaporates faster than water at the same temperature in open containers, but when each liquid is placed in identical sealed containers at the same temperature, the final equilibrium vapour pressures are different yet both remain constant over time. Explain: (i) the difference in evaporation rates in open systems, (ii) why vapour pressures are different in sealed containers, and (iii) why these pressures remain constant once established despite continued molecular exchange. [6 marks]

For (i): acetone has higher vapour pressure (weaker intermolecular forces) so more molecules escape per unit time; in open systems, there is no equilibrium limiting factor. For (ii): vapour pressure depends on temperature and molecular properties (intermolecular forces), not on container size or amount of liquid—acetone's higher value reflects weaker interactions. For (iii): in sealed containers, evaporation and condensation rates become equal at the characteristic vapour pressure for that temperature, creating dynamic equilibrium where macro-level pressure stabilizes despite ongoing molecular transfers; if you attempt to increase pressure by adding more liquid, evaporation decreases and condensation increases until the same vapour pressure is re-established.

Assertion-Reason Questions for Equilibrium

Q1. Assertion (A):

At the normal melting point of ice (273 K, 1 atm), the ice and water phases are in dynamic equilibrium.

Reason (R):

At dynamic equilibrium, the rate of fusion equals the rate of freezing, and no net change in the amounts of ice and water occurs.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The reason correctly explains why ice and water can coexist without changing amounts—the forward and reverse processes occur at equal rates. This is the defining characteristic of dynamic equilibrium.

Q2. Assertion (A):

The vapor pressure of a liquid increases with increasing temperature.

Reason (R):

Higher temperature increases the average kinetic energy of molecules, allowing more of them to escape from the liquid surface.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true and causally related. Increased kinetic energy at higher temperatures directly causes more molecules to have sufficient energy to overcome intermolecular forces and enter the vapor phase, increasing vapor pressure.

Q3. Assertion (A):

In a closed container, the equilibrium vapor pressure of water remains constant at a given temperature regardless of the amount of liquid present.

Reason (R):

Vapor pressure depends on temperature and the nature of the substance, not on the quantity of liquid, making it an intensive property.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason is the correct explanation. Vapor pressure is indeed an intensive property—1 mL and 100 mL of water have the same equilibrium vapor pressure at the same temperature.

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Case-Based Questions (CBQ) for Equilibrium

Case Study:

A manufacturing facility in India processes water to remove moisture for industrial applications. In their laboratory, three identical sealed containers are filled with 100 mL of pure water at 25°C. Container X is kept at 25°C, Container Y is heated to 50°C, and Container Z is cooled to 10°C. A manometer (pressure measuring device) is attached to each container to monitor vapor pressure. After 2 hours, the technician records the pressure readings and observes the volume of liquid water remaining in each container.

Q1. Which container will show the highest pressure reading on the manometer after 2 hours?

  • A. Container X at 25°C
  • B. Container Y at 50°C ✓
  • C. Container Z at 10°C
  • D. All containers will show equal pressure

Answer: B

Q2. In Container Y heated to 50°C, compared to Container X at 25°C, the volume of liquid water will be:

  • A. Greater because more water evaporates
  • B. Lesser because more water evaporates ✓
  • C. The same because it is a sealed container
  • D. Unpredictable without knowing atmospheric pressure

Answer: B

Q3. If the manometer in Container X shows a reading of 23.76 mmHg at 25°C, what can the technician conclude?

  • A. The water has completely evaporated
  • B. More water will condense in the next hour
  • C. The container is leaking vapor
  • D. The system has reached liquid-vapor equilibrium at 25°C ✓

Answer: D

Q4. The technician wants to minimize water loss in Container Y. Which action would be most effective?

  • A. Increase the volume of the container
  • B. Increase the external pressure on the container
  • C. Add more water to dilute the vapor
  • D. Reduce the temperature back to 25°C, then seal it ✓

Answer: D

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