SF Prep
04 Chemistry Notes 🕒 Updated 2026-10-07
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Chemical Equilibrium — Introduction

Chemical Equilibrium

1. Chemical Equilibrium — Introduction

  • For a reversible reaction R ⇌ P :
  • At equilibrium, Rateforward = Ratebackward.
  • Concentrations of reactants and products become constant (not necessarily equal).

Rate vs Time graph :

  • Rf decreases and Rb increases until they become equal — this is the only correct rate-vs-time graph at equilibrium.

Concentration vs Time graph :

  • [R] decreases and [P] increases, then both become constant. At equilibrium [R] may be > [P], [R] = [P], or [R] < [P] — all three are possible.

2. Law of Mass Action & Equilibrium Constant

  • Law of mass action : Rate ∝ [active mass (concentration)]stoichiometric coefficient.

Derivation of Keq :

Consider: A + 2B ⇌ C

  • Rateforward = kf [A]1 [B]2
  • Ratebackward = kb [C]
  • At equilibrium, kf[A][B]² = kb[C]
Keq = kfkb = [C]eq[A]eq [B]²eq

Two ways to calculate Keq :

  • (i) Keq = kf (forward rate constant)kb (backward rate constant)
  • (ii) Keq = [Products]s.c. at equilibrium[Reactants]s.c. at equilibrium

s.c. = stoichiometric coefficient.

Example :

2A(g) + B(g) ⇌ 3C(g)

Keq = kfkb = [C]³eq[A]²eq [B]eq

3. Kc — Equilibrium Constant in Terms of Concentration

  • When Keq is expressed using molar concentrations of species whose concentrations change during the reaction, it is called Kc.

For A(g) ⇌ B(g) + 2C(g) :

Kc = [B][C]²[A]

Pure solids & liquids :

  • Concentration of a pure solid or liquid = dM = constant, so it is excluded from Kc.

CaCO3(s) ⇌ CaO(s) + CO2(g)

Kc = [CO2]

Relation between concentration and pressure :

PV = nRT  ⇒  conc = nV = PRT

4. Relation between Kp and Kc

  • For a general gaseous equilibrium, e.g. A(g) + 2B(g) ⇌ C(g), with Δng = 1 − 3 = −2
  • Since concentration = nV = PRT  (from PV = nRT)
Kc = [C][A][B]² = PCPA · PB² · (RT)² = Kp · (RT)²
Kp = Kc (RT)Δng

where Δng = (moles of gaseous products) − (moles of gaseous reactants); R = 0.0821 L·atm·mol−1·K−1; T in Kelvin.

  • Kp is the equilibrium constant expressed in terms of partial pressures.

When is Kp = Kc ?

  • When Δng = 0, or
  • When RT = 1 (numerically)

Example – PCl5 decomposition :

  • PCl5(g) ⇌ PCl3(g) + Cl2(g)   ;   Δng = 2 − 1 = +1
Kc = [PCl3][Cl2][PCl5]  ;  Kp = PPCl₃ · PCl₂PPCl₅  ;  Kp = Kc(RT)1

5. Writing Kc Expressions

  • Only gaseous and aqueous species appear in the expression; pure solids and pure liquids are omitted (activity = 1).
EquilibriumKc
N2O4(g) ⇌ 2NO2(g)[NO2]²[N2O4]
PCl5(g) ⇌ PCl3(g) + Cl2(g)[PCl3][Cl2][PCl5]
CO2(g) + H2(g) ⇌ CO(g) + H2O(g)[CO][H2O][CO2][H2]

6. Writing Kp Expressions

EquilibriumKcKp
CaCO3(s) ⇌ CaO(s) + CO2(g)[CO2]PCO₂
NH2COONH4(s) ⇌ 2NH3(g) + CO2(g)[NH3]² [CO2]PNH₃² · PCO₂

Solids are excluded from both Kc and Kp expressions.

7. Types of Chemical Equilibrium

  • Homogeneous Equilibrium — all reactants and products in the same phase, e.g. PCl5(g) ⇌ PCl3(g) + Cl2(g)
  • Heterogeneous Equilibrium — species in more than one phase
  • Multiple Equilibrium
  • Simultaneous Equilibrium

8. Factors Affecting K — Stoichiometry

  • If a balanced equation is multiplied by a factor n, the new equilibrium constant is raised to the power n.
A ⇌ B   ;   K1 = [B][A]
2A ⇌ 2B   ;   K2 = [B]²[A]² = K1²
Knew = (Kold)multiplying factor
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Factors Affecting K — Reversing the Reaction

9. Factors Affecting K — Reversing the Reaction

  • If the reaction is written in the reverse direction, the new K is the reciprocal of the original.
A ⇌ B   ;   K1 = [B][A]
B ⇌ A   ;   K2 = [A][B] = 1K1

10. Equilibrium Constant Manipulation

  • When reactions are added, their equilibrium constants are multiplied.
A ⇌ B ; K1
B ⇌ C ; K2
⇒ A ⇌ C ; K = K1 × K2

11. Formation Constant

  • The overall stability (formation) constant β of a complex is the product of its stepwise stability constants.

12. Reaction Quotient (Q) & Direction of Reaction

  • Q has the same expression as K but uses instantaneous (non-equilibrium) concentrations.
  • At t = 0 for PCl5(g) ⇌ PCl3(g) + Cl2(g), if only PCl5 is present: Q = 0.
Q < K → reaction proceeds in the forward direction
Q > K → reaction proceeds in the backward direction
Q = K → system is at equilibrium
  • Q rises from its initial value and levels off at K as the system reaches equilibrium.

13. Properties of Equilibrium Constant

  • Kp (equilibrium constant in terms of partial pressure) depends only on Temperature.
  • It is independent of catalyst, total pressure, or initial amounts of reactants.
  • When K = 1, the equilibrium concentration of reactants is equal to the concentration of products.

14. Degree of Dissociation (α)

  • For reaction PCl5(g) ⇌ PCl3(g) + Cl2(g):
TimePCl5PCl3Cl2
t=0a00
t=teqa-xxx
  • Kc = [PCl3][Cl2][PCl5] = (x/V)(x/V)(a-x)/V
  • Degree of Dissociation (α) = moles dissociated at equilibriuminitial moles = xa

15. Equilibrium Expressions in terms of α and PT

Reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g)
Moles at eq: a(1-α)     aα     aα
Total moles = a(1+α)
Kp = α21-α2 × PT
  • General expression for A(g) ⇌ B(g) + C(g): Kp = α21-α2 × PT
  • For N2O4(g) ⇌ 2NO2(g):
Kp = 4α21-α2 × PT

16. Degree of Dissociation & Pressure Dependence

  • For reaction A(g) ⇌ B(g) + C(g) with total pressure p and degree of dissociation α:
  • Kp = [(αp)/(1+α)]2 / [(1-α)p/(1+α)] = (α2p) / (1-α2).
  • When total pressure p increases, α must decrease to keep Kp constant.

17. Dissociation of AB2

  • For the dissociation 2AB2(g) ⇌ 2AB(g) + B2(g), if the degree of dissociation 'x' is small compared to 1 (x << 1):
x ≈ (2KpP)1/3

Where Kp is the equilibrium constant and P is the total pressure.

18. Vapour Density and Degree of Dissociation

  • Consider the dissociation equilibrium: An ⇌ nA
SpeciesAnnA
Initial molesa0
Moles at eq.a(1 − α)anα
  • Total moles at equilibrium = a(1 − α) + anα = a[1 + (n − 1)α]
  • Since Vapour Density (V.D.) is inversely proportional to total moles for a given mass, the ratio of initial V.D. (D) to equilibrium V.D. (d) is:
Dd = a[1 + (n − 1)α]a
  • Solving for degree of dissociation (α):
α = D − dd(n − 1)

where D = initial V.D. (theoretical), d = equilibrium V.D. (observed), and n = total gaseous moles of product per mole of reactant.

  • Relation using Molar Masses (Mth = theoretical molar mass, Mobs = observed molar mass):
α = Mth − MobsMobs(n − 1)
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Graphs of Dissociation

19. Graphs of Dissociation

  • For the dissociation N2O4(g) ⇌ 2NO2(g), the relationship between α and the ratio (d/D) is represented by a straight line.
  • The graph shows α = 1 at (d/D) = 0.5 and α = 0 at (d/D) = 1. The degree of dissociation α decreases as the ratio (d/D) increases.

20. Heterogeneous Equilibrium

  • Equilibria involving solids are generally heterogeneous.
  • Example: CaCO3(s) ⇌ CaO(s) + CO2(g)
Kp = PCO2 = nCO2 · R · TV
Kc = [CO2]
  • Note: The equilibrium pressure of CO2 depends only on temperature.

21. Heterogeneous Equilibrium (Solid-Gas)

  • Whenever a solid alone is dissociated in a closed container producing only gaseous products, the equilibrium constant Kp is determined by the total pressure.

22. Simultaneous Equilibria: Dissociation of Two Solids

  • When two solids A and D dissociate simultaneously:
A(s) ⇌ B(g) + C(g) ; Kp1 = pBpC = x
D(s) ⇌ C(g) + E(g) ; Kp2 = pCpE = y
  • Let pC = p. Since C is produced by both reactions, the partial pressures are related by pB = x/p and pE = y/p.
  • Total pressure (PT) = pB + pC + pE = xp + p + yp = x + y + p2p.
  • C comes from both solids, so pC = pB + pE ⇒ p = x/p + y/p ⇒ pC = √(x + y) and PT = 2pC.

23. Factors Affecting K — Temperature (van 't Hoff Equation)

  • Rate constants: kF = AF · e−(Ea)F/RT  ;  kb = Ab · e−(Ea)b/RT
  • kFkb = AFAb · e−ΔH/RT
KT = AFAb · e−ΔH/RT
ln KT = ln A − ΔHRT  ;  log KT = log A − ΔH2.303 RT

Two-temperature form (van 't Hoff) :

ln KT₂KT₁ = ΔHR [ 1T₁ − 1T₂ ]
log KT₂KT₁ = ΔH2.303 R [ 1T₁ − 1T₂ ]

Compare with the Arrhenius two-temperature form: log kT₂kT₁ = Ea2.303 R [ 1T₁ − 1T₂ ]

24. Effect of Temperature on K — Endothermic vs Exothermic

Endothermic equilibrium (ΔH > 0) :

  • Temperature increases → KT₂ > KT₁ (K increases significantly — exponential dependence)
  • Temperature decreases → KT₂ < KT₁

Exothermic equilibrium (ΔH < 0) :

  • Temperature increases → KT₂ < KT₁
  • Temperature decreases → KT₂ > KT₁

25. Gibbs Energy & Equilibrium Constant

ΔG = ΔG° + RT ln Q
  • At equilibrium: ΔG = 0 and Q = K
ΔG° = −RT ln K  ⇒  K = e−ΔG°/RT
  • Substituting ΔG° = ΔH° − TΔS° :
ln K = −ΔH°RT + ΔS°R
log K = −ΔH°2.303 RT + ΔS°2.303 R

Graph of ln K vs 1/T :

  • Straight line: y = mx + c form
  • Slope = −ΔH°/R
  • y-intercept = ΔS°/R
1/T ln K ΔS°/R slope = −ΔH°/R

26. Intercepts and Slopes

EquationSlopeIntercept
ln K = -ΔH°RT + ΔS°R-ΔH°RΔS°R
log K = -ΔH°2.303R(1T) + ΔS°2.303R-ΔH°2.303RΔS°2.303R

27. Le Chatelier's Principle

  • If a system at equilibrium is subjected to a change of pressure, temperature, or number of moles of substance, there will be a tendency for a net reaction in the direction that tends to reduce the effect of this change.

28. Effect of Concentration Change

  • For reaction: PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
  • Increase in reactant concentration: Reaction moves in forward direction.
  • Increase in product concentration: Reaction moves in backward direction.
  • Key Note: If the reactant or product is a solid (s) or pure liquid (l), change in its concentration has no effect on the equilibrium position.
CaCO3 (s) ⇌ CaO (s) + CO2 (g)
Q = [CO2]
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Effect of Pressure Change

29. Effect of Pressure Change

  • For reaction: PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
  • Reaction quotient Q = [PCl3][Cl2][PCl5] = bca × 1V
  • Pressure increases (Volume decreases): Reaction moves backward.
  • Pressure decreases (Volume increases): Reaction moves forward.

30. Addition of Inert Gas

Constant Volume :

  • Addition of inert gas at constant volume has no effect on the equilibrium state.
  • Partial pressure of species: P(i) = X(i) × P(T).

Constant Pressure :

  • For reaction: PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
  • Addition of inert gas at constant pressure causes the volume to increase.
  • Since volume increases, the reaction moves in the forward direction.

31. Equilibrium under Inert Gas Addition & Pressure Change

  • Addition of inert gas at constant Volume: Total pressure increases but partial pressures of reactants/products remain constant; equilibrium does not shift.
  • Addition of inert gas at constant Pressure: Volume of the system increases. The equilibrium shifts towards the side with a greater number of gaseous moles.
  • Effect of Pressure change:
    • Increasing total pressure shifts the equilibrium towards the side with fewer gaseous moles.
    • Decreasing total pressure shifts the equilibrium towards the side with more gaseous moles.
  • Condensed phases (solids/liquids): For processes like ice ⇌ water, increasing pressure favors the formation of the denser phase (Le Chatelier's Principle).

32. Effect of Catalyst

  • Addition of a catalyst has no change on the equilibrium position.

33. Effect of Temperature on Equilibrium

  • Since Keq depends exponentially on T, the direction of equilibrium shift depends on the sign of ΔH :
ΔHTemperature ChangeKeq ChangeShift
(+)veT ↑K ↑Forward
(+)veT ↓K ↓Backward
(−)veT ↑K ↓Backward
(−)veT ↓K ↑Forward

34. Effect of Pressure on Physical Equilibrium

  • For any physical equilibrium, increasing the pressure shifts the equilibrium in the direction of the physical state having higher density.

Examples of Physical Equilibria :

H2O(s) ⇌ H2O(l) ; ΔH = (+)ve
  • T ↑ : Forward (more ice melts).
  • P ↑ : Forward (since density of H2O(l) > density of H2O(s), more ice melts).
H2O(l) ⇌ H2O(v) ; ΔH = (+)ve
  • T ↑ : Forward.
  • P ↑ : Backward.
C(graphite) ⇌ C(diamond) ; ΔH = (+)ve
  • T ↑ : Forward (endothermic).
  • P ↑ : Forward (diamond is denser than graphite).

35. Quick Formula Revision

QuantityFormula
Law of mass actionaA + bB ⇌ cC + dD : Kc = [C]c[D]d / [A]a[B]b
Kc from rate constantsKc = kf / kb
Kp – KcKp = Kc(RT)Δng ; Δng = gaseous moles (P) − gaseous moles (R)
KxKp = Kx(PT)Δng
Manipulating KReversed : 1/K ; multiplied by n : Kn ; reactions added : K1 × K2
DirectionQ < K : forward ; Q = K : equilibrium ; Q > K : backward
A ⇌ B + C (α, PT)Kp = α2PT / (1 − α2)
N2O4 ⇌ 2NO2Kp = 4α2PT / (1 − α2)
Vapour density methodα = (D − d) / [d(n − 1)] ; D = initial VD, d = equilibrium VD, n = moles of products from 1 mol
Van't Hoff equationlog(K2/K1) = ΔH° / 2.303R × (1/T1 − 1/T2)
Gibbs energyΔG° = −RT ln K = −2.303RT log K ; ΔG = ΔG° + RT ln Q
Two solids, common gas CpC = √(Kp1 + Kp2) ; PT = 2pC

Key Points :

  • Equilibrium is dynamic : rate of forward reaction = rate of backward reaction.
  • Pure solids and pure liquids are not written in K (activity = 1).
  • K depends only on temperature (and on how the equation is written), not on concentration, pressure, volume or catalyst.
  • Endothermic reaction : K increases with T ; exothermic reaction : K decreases with T.
  • A catalyst only helps reach equilibrium faster; it does not change K or the equilibrium yield.
  • Inert gas at constant volume : no effect. Inert gas at constant pressure : shift towards more gaseous moles.
  • Pressure increase shifts the equilibrium towards fewer gaseous moles (or towards the denser phase in physical equilibria).