CBSE Class 11 Chemistry Equilibrium Notes Set 07

These Class 11 CBSE Chemistry notes for Chapter 6 Equilibrium (chemical equilibrium) start from physical equilibria such as ice and water, then build up the law of mass action, the equilibrium constants K_c and K_p, the reaction quotient, the link between K and Gibbs energy, and Le Chatelier's principle. Nine solved examples work through direction-of-reaction and equilibrium-concentration problems step by step.

Important Notes and Formulas

Equilibrium in Physical Processes

A system is at equilibrium when its measurable properties (pressure, concentration, colour, density and so on) no longer change with time. In physical processes the two opposing changes are physical, for example melting and freezing. Equilibrium is dynamic: both changes go on at the same rate, so nothing seems to change.

General characteristics of physical equilibria:

1. Equilibrium is possible only in a closed system at a given temperature.
2. Both opposing processes occur at the same rate, and the system is dynamic but stable.
3. All measurable properties of the system stay constant.
4. At equilibrium one parameter has a constant value at a given temperature, for example the vapour pressure of a liquid, the concentration of a saturated solution or the solubility of a gas at fixed pressure.
5. The magnitude of that quantity shows how far the process has gone before reaching equilibrium.

Equilibrium in Chemical Processes

A reaction that goes both ways under the same conditions is a reversible reaction, written with the sign ⇌. A reaction in which the products cannot form the reactants again is irreversible, for example AgNO₃ + NaCl → AgCl↓ + NaNO₃ (the solid AgCl leaves the solution). In a closed vessel the forward reaction slows down as reactants are used up, the backward reaction speeds up as products build up, and at one moment the two rates become equal. This is chemical equilibrium.

1. The concentrations of all reactants and products stay constant at equilibrium. The rates of the forward and backward reactions are equal and not zero.
2. Equilibrium is dynamic. When N₂ + 3H₂ ⇌ 2NH₃ is allowed to reach equilibrium with D₂ in place of H₂ and the two equilibrium mixtures are mixed, deuterated ammonia such as NH₂D, NHD₂ and ND₃ appears. The reactions are therefore still going on.
3. Equilibrium can be reached from either side, starting with only reactants or only products, and the final composition at a given temperature is the same.
4. It can be reached only in a closed system. If a gaseous product (for example CO₂ from CaCO₃) escapes, or a product leaves as a solid, the reaction does not reach equilibrium.
5. A catalyst speeds up the forward and backward reactions equally. It helps equilibrium to be reached sooner but does not change the equilibrium composition or the value of K.

Law of Mass Action and the Equilibrium Constant

The rate of a reaction is proportional to the product of the molar concentrations (active masses) of the reactants, each raised to the power of its stoichiometric coefficient. For a reversible reaction aA + bB ⇌ cC + dD, the forward rate is k_f[A]^a[B]^b and the backward rate is k_b[C]^c[D]^d. At equilibrium they are equal, so

\( K_c = \frac{k_f}{k_b} = \frac{[C]^c\,[D]^d}{[A]^a\,[B]^b} \)

This is the law of chemical equilibrium. K_c depends only on temperature. It does not depend on the starting concentrations, the direction from which equilibrium is approached, or the presence of a catalyst. For gases the equilibrium constant can also be written with partial pressures:

\( K_p = \frac{p_C^{\,c}\;p_D^{\,d}}{p_A^{\,a}\;p_B^{\,b}} \)

Relation between K_p and K_c: K_p = K_c(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants). Use R = 0.0831 L bar K⁻¹ mol⁻¹ (or 0.0821 L atm K⁻¹ mol⁻¹) with pressure in bar (or atm) and concentration in mol L⁻¹.

Units: the unit of K_c is (mol L⁻¹)^Δn and that of K_p is (bar)^Δn. K is dimensionless when Δn = 0.

Homogeneous and Heterogeneous Equilibria

In a homogeneous equilibrium all reactants and products are in the same phase, for example N₂(g) + 3H₂(g) ⇌ 2NH₃(g) or CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l). In a heterogeneous equilibrium more than one phase is present. The concentration of a pure solid or a pure liquid is constant, so it is left out of the expression of K.

Rules for Manipulating Equilibrium Constants

1. Reversed reaction: if K is the constant for the forward reaction, the constant for the reverse reaction is 1/K.
2. Multiplying the equation by n raises K to the power n: K′ = Kⁿ. For example, halving an equation gives √K.
3. Adding reactions: if a reaction can be written as the sum of two steps with constants K₁ and K₂, then K = K₁ × K₂.
4. K changes only with temperature. A catalyst, a change of concentration, pressure or volume does not change K.

Applications of Equilibrium Constants

1. Extent of a reaction.

1. If K_c > 10³, the products predominate at equilibrium and the reaction goes almost to completion.
2. If K_c < 10⁻³, the reactants predominate and the reaction hardly proceeds.
3. If K_c lies between 10⁻³ and 10³, appreciable amounts of both reactants and products are present at equilibrium.

A large K says nothing about how fast equilibrium is reached. K tells the extent of a reaction and not its rate.

2. Direction of reaction. The reaction quotient Q_c has the same form as K_c but uses the concentrations present at any moment (not only at equilibrium).

3. Equilibrium concentrations. Steps: (i) write the balanced equation; (ii) under each species write the initial concentration; (iii) let x mol L⁻¹ of reactant react and write each equilibrium concentration in terms of x; (iv) substitute in the expression of K_c; (v) solve for x, rejecting a root that gives a negative concentration; (vi) put x back to get each concentration and check by recalculating K_c.

Degree of dissociation (α) is the fraction of the molecules that have dissociated. If a gas dissociates to the extent α, then 100α per cent has dissociated. For PCl₅ ⇌ PCl₃ + Cl₂ starting with 1 mol, the equilibrium amounts are (1 − α), α and α, and the total is (1 + α) mol.

Relationship between K, Q and Gibbs Energy

\( \Delta G = \Delta G^\circ + RT\ln Q \)  and at equilibrium ΔG = 0, so  \( \Delta G^\circ = -RT\ln K = -2.303\,RT\log K \)

1. If ΔG° < 0, then K > 1 and the reaction is spontaneous in the forward direction (products favoured at equilibrium).
2. If ΔG° > 0, then K < 1 and the reaction is non-spontaneous in the forward direction (reactants favoured).
3. If ΔG° = 0, then K = 1.
4. ΔG° is the standard Gibbs energy change (all species at standard state), while ΔG at any moment depends on Q. A reaction can go forward when ΔG < 0 even if ΔG° > 0, provided Q is small enough.

Le Chatelier's Principle

If a change of concentration, pressure or temperature is imposed on a system at equilibrium, the system shifts in the direction that reduces the effect of the change, and a new equilibrium is set up.

Effect of temperature, an experiment: the equilibrium 2NO₂(g) ⇌ N₂O₄(g), ΔH = −57.2 kJ mol⁻¹, is exothermic in the forward direction. The brown colour of NO₂ deepens on heating and fades on cooling. In [Co(H₂O)₆]²⁺ (pink) + 4Cl⁻ ⇌ [CoCl₄]²⁻ (blue) + 6H₂O, the forward reaction is endothermic, so heating turns the solution blue and cooling turns it pink again.

Industrial applications. For the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.38 kJ mol⁻¹, high pressure (about 200 atm) and low temperature favour ammonia. A low temperature makes the reaction too slow, so a moderate temperature of about 700 K is used with an iron catalyst. For the contact process 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196.6 kJ mol⁻¹, a low temperature and a high pressure favour SO₃, and V₂O₅ is used as the catalyst at about 720 K. In both processes the product is removed to push the equilibrium forward.

Solved Examples

Example 1. The reaction CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −92.0 kJ mol⁻¹, has reached equilibrium. What will happen if (i) the volume of the vessel is suddenly halved, (ii) the partial pressure of hydrogen is suddenly doubled, (iii) an inert gas is added at constant volume?
Answer:
(i) Halving the volume doubles every concentration. Q_c = 2[CH₃OH] ÷ {2[CO] × (2[H₂])²} = ¼ K_c. Since Q_c < K_c, the reaction moves forward and more CH₃OH forms.
(ii) Doubling p(H₂) gives Q_p = p(CH₃OH) ÷ {p(CO) × (2p(H₂))²} = ¼ K_p. Again Q_p < K_p, so the reaction moves forward.
(iii) At constant volume an inert gas does not change the concentrations, so there is no effect on the equilibrium.

Example 2. HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI is 0.04 atm. Find K_p for 2HI(g) ⇌ H₂(g) + I₂(g).
Answer:
Pressure of HI that dissociated = 0.20 − 0.04 = 0.16 atm.
Two moles of HI give one mole each of H₂ and I₂, so p(H₂) = p(I₂) = 0.16 ÷ 2 = 0.08 atm.
K_p = (0.08 × 0.08) ÷ (0.04)² = 4.0

Example 3. One mole of H₂O and one mole of CO are heated in a 10 L vessel at 725 K. At equilibrium 40% of the water has reacted: H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g). Calculate K_c.
Answer:
Water reacted = 0.40 mol, so at equilibrium: H₂O = CO = 0.60 mol; H₂ = CO₂ = 0.40 mol.
Concentrations (V = 10 L): [H₂O] = [CO] = 0.06 mol L⁻¹ and [H₂] = [CO₂] = 0.04 mol L⁻¹.
K_c = (0.04 × 0.04) ÷ (0.06 × 0.06) = 0.444

Example 4. For H₂(g) + I₂(g) ⇌ 2HI(g), K_c = 54.8 at 700 K. Starting with pure HI, equilibrium is reached with [HI] = 0.5 mol L⁻¹. Find [H₂] and [I₂] at equilibrium.
Answer:
For the dissociation 2HI ⇌ H₂ + I₂ the constant is the reciprocal, K_c′ = 1 ÷ 54.8.
Let [H₂] = [I₂] = x. Then x² ÷ (0.5)² = 1 ÷ 54.8, so x = 0.5 ÷ √54.8 = 0.068 mol L⁻¹.

Example 5. Initially 0.78 M ICl is present. Find the equilibrium concentrations for 2ICl(g) ⇌ I₂(g) + Cl₂(g), K_c = 0.14.
Answer:
Let [I₂] = [Cl₂] = x. Then [ICl] = 0.78 − 2x.
K_c = x² ÷ (0.78 − 2x)² = 0.14, so x ÷ (0.78 − 2x) = √0.14 = 0.374.
x = 0.292 − 0.748x, so 1.748x = 0.292 and x = 0.167.
[I₂] = [Cl₂] = 0.167 M and [ICl] = 0.78 − 2(0.167) = 0.446 M.

Example 6. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), K_c = 0.061 at 500 K. A mixture contains 3.0 mol L⁻¹ N₂, 2.0 mol L⁻¹ H₂ and 0.5 mol L⁻¹ NH₃. Is the system at equilibrium? If not, in which direction does the reaction go?
Answer:
Q_c = [NH₃]² ÷ ([N₂][H₂]³) = (0.5)² ÷ (3.0 × 2.0³) = 0.25 ÷ 24 = 0.0104.
Q_c ≠ K_c, so the system is not at equilibrium.
Q_c < K_c (0.0104 < 0.061), so the reaction proceeds in the forward direction to form more NH₃.

Example 7. For CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), K_p = 0.1 at 400 °C. A vessel contains CO and steam at 4.0 bar each. What is the partial pressure of H₂ at equilibrium?
Answer:
Let p(H₂) = p bar. Then p(CO₂) = p and p(CO) = p(H₂O) = 4.0 − p.
K_p = p² ÷ (4.0 − p)² = 0.1, so p ÷ (4.0 − p) = √0.1 = 0.316.
p = 1.264 − 0.316p, so 1.316p = 1.264 and p = 0.96 bar.

Example 8. Which of these reactions has appreciable amounts of both reactants and products at equilibrium? (a) Cl₂(g) ⇌ 2Cl(g), K_c = 5 × 10⁻³⁹ (b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g), K_c = 3.7 × 10⁸ (c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g), K_c = 1.8
Answer:
Appreciable amounts of both sides are present only when K_c lies roughly between 10⁻³ and 10³.
(a) K_c is extremely small, so almost no Cl atoms form. (b) K_c is very large, so the reaction is almost complete.
In (c), K_c = 1.8, so reactants and products are both present in comparable amounts. The answer is (c).

Example 9. 0.10 mol of PCl₅ is heated in an 8.0 L flask at 540 K. The pressure of the equilibrium mixture is 1.0 atm. Calculate K_c and K_p for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). (R = 0.0821 L atm K⁻¹ mol⁻¹)
Answer:
Total moles at equilibrium, n = PV ÷ RT = (1.0 × 8.0) ÷ (0.0821 × 540) = 0.180 mol.
If x mol of PCl₅ dissociates, n = 0.10 − x + x + x = 0.10 + x, so x = 0.080 mol. Then PCl₅ = 0.020 mol and PCl₃ = Cl₂ = 0.080 mol.
K_c = [PCl₃][Cl₂] ÷ [PCl₅] = (0.080 ÷ 8)² ÷ (0.020 ÷ 8) = 0.04 mol L⁻¹ (about 0.041 with unrounded values).
K_p = K_c(RT)^Δn with Δn = 1: K_p = 0.041 × 0.0821 × 540 ≈ 1.8 atm.