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Chemical Equilibrium — Introduction
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).
| Equilibrium | Kc |
|---|---|
| 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
| Equilibrium | Kc | Kp |
|---|---|---|
| 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
2A ⇌ 2B ; K2 = [B]²[A]² = K1²
Knew = (Kold)multiplying factor
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Factors Affecting K — Reversing the Reaction
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
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
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 > 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):
| Time | PCl5 | PCl3 | Cl2 |
|---|---|---|---|
| t=0 | a | 0 | 0 |
| t=teq | a-x | x | x |
- 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
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
| Species | An | nA |
|---|---|---|
| Initial moles | a | 0 |
| 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
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]
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
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
26. Intercepts and Slopes
| Equation | Slope | Intercept |
|---|---|---|
| ln K = -ΔH°RT + ΔS°R | -ΔH°R | ΔS°R |
| log K = -ΔH°2.303R(1T) + ΔS°2.303R | -ΔH°2.303R | ΔS°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]
Q = [CO2]
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Effect of Pressure Change
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 :
| ΔH | Temperature Change | Keq Change | Shift |
|---|---|---|---|
| (+)ve | T ↑ | K ↑ | Forward |
| (+)ve | T ↓ | K ↓ | Backward |
| (−)ve | T ↑ | K ↓ | Backward |
| (−)ve | T ↓ | 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
| Quantity | Formula |
|---|---|
| Law of mass action | aA + bB ⇌ cC + dD : Kc = [C]c[D]d / [A]a[B]b |
| Kc from rate constants | Kc = kf / kb |
| Kp – Kc | Kp = Kc(RT)Δng ; Δng = gaseous moles (P) − gaseous moles (R) |
| Kx | Kp = Kx(PT)Δng |
| Manipulating K | Reversed : 1/K ; multiplied by n : Kn ; reactions added : K1 × K2 |
| Direction | Q < K : forward ; Q = K : equilibrium ; Q > K : backward |
| A ⇌ B + C (α, PT) | Kp = α2PT / (1 − α2) |
| N2O4 ⇌ 2NO2 | Kp = 4α2PT / (1 − α2) |
| Vapour density method | α = (D − d) / [d(n − 1)] ; D = initial VD, d = equilibrium VD, n = moles of products from 1 mol |
| Van't Hoff equation | log(K2/K1) = ΔH° / 2.303R × (1/T1 − 1/T2) |
| Gibbs energy | ΔG° = −RT ln K = −2.303RT log K ; ΔG = ΔG° + RT ln Q |
| Two solids, common gas C | pC = √(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).