Chemical Equilibrium, Equilibrium Constants (Kc, Kp) & Le Chatelier's Principle
Chemical reactions in closed systems do not proceed unidirectionally to absolute completion; instead, they approach a state of dynamic equilibrium where the forward and backward reaction rates become equal. At this macroscopic standstill, concentrations of reactants and products remain invariant over time despite persistent microscopic conversions.
1. Dynamic Nature of Equilibrium and Law of Chemical Equilibrium
Consider a general reversible homogeneous gaseous reaction:
a A(g) + b B(g) ⇔ c C(g) + d D(g)
By Guldberg and Waage's Law of Mass Action, the rate of forward reaction is rf = kf [A]a [B]b and the backward rate is rb = kb [C]c [D]d. At dynamic equilibrium, rf = rb:
Kc = kf / kb = [C]c [D]d / ([A]a [B]b)
Where concentrations are expressed in moles per liter (mol/L or M) at equilibrium.
2. Partial Pressure Formulation (Kp) & Relation to Kc
For gaseous systems, using Dalton's law of partial pressures and ideal gas law pi = (ni/V) R T = ci R T, the equilibrium constant can be formulated in terms of equilibrium partial pressures:
Substituting pi = [i] R T leads to the fundamental identity:
Kp = Kc (R T)Δng
Where Δng = (c + d) − (a + b) is the change in stoichiometric number of gaseous moles. Critical consequences:
- If Δng = 0 (e.g., H2(g) + I2(g) ⇔ 2HI(g)): Kp = Kc. Both constants are dimensionless, and pressure has no effect on equilibrium yield.
- If Δng > 0 (e.g., PCl5(g) ⇔ PCl3(g) + Cl2(g), Δng = +1): Kp > Kc (at RT > 1). Units of Kp = atm, Kc = mol L−1.
- If Δng < 0 (e.g., N2(g) + 3H2(g) ⇔ 2NH3(g), Δng = −2): Kp < Kc (at RT > 1).
Reaction Quotient Q vs Equilibrium Constant K & Directionality
Q < K Q = K Q > K
[Reactants High] [Dynamic Equilibrium] [Products High]
Forward Shift → No Net Shift ← Backward Shift
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Q = [Products]^st / [Reactants]^st at ANY arbitrary point in time
K = [Products]^st / [Reactants]^st EXCLUSIVELY at chemical equilibrium
Thermodynamic Link:
ΔG = ΔG° + R T ln Q
At Equilibrium: ΔG = 0, Q = K ==> ΔG° = -R T ln K ==> K = exp(-ΔG° / RT)
3. Heterogeneous Equilibria & Rules of Manipulating K
In heterogeneous equilibria involving pure solids or pure liquids, the molar concentration (density divided by molar mass) remains constant. Consequently, their active masses are taken as unity (1) and omitted from K expressions:
- CaCO3(s) ⇔ CaO(s) + CO2(g) ⇒ Kc = [CO2], Kp = pCO2
- NH4HS(s) ⇔ NH3(g) + H2S(g) ⇒ Kp = pNH3 · pH2S = (Ptotal / 2)2 = P2 / 4
Mathematical Operations on Equilibrium Constants
| Operation on Equation | Original Equation & Constant | New Equation & New Constant (K') |
|---|---|---|
| Reversing the reaction | A ⇔ B (K) | B ⇔ A ⇒ K' = 1 / K |
| Multiplying by factor n | A ⇔ B (K) | nA ⇔ nB ⇒ K' = Kn |
| Dividing by factor n (multiplying by 1/n) | A ⇔ B (K) | (1/n)A ⇔ (1/n)B ⇒ K' = K1/n = n√K |
| Adding two reactions | A ⇔ B (K1), B ⇔ C (K2) | A ⇔ C ⇒ K' = K1 · K2 |
4. Le Chatelier's Principle & Perturbation Effects
Le Chatelier's Principle: If a system at equilibrium is subjected to a disturbance in concentration, pressure, volume, or temperature, the equilibrium shifts in that direction which tends to annul or counteract the effect of the applied change.
| Disturbance Applied | Direction of Equilibrium Shift | Effect on Equilibrium Constant (K) |
|---|---|---|
| Increase [Reactant] or Remove Product | Shifts Forward (→) | No change in K |
| Increase [Product] or Remove Reactant | Shifts Backward (←) | No change in K |
| Increase Pressure / Decrease Volume | Shifts towards fewer gaseous moles (smaller Δng) | No change in K |
| Decrease Pressure / Increase Volume | Shifts towards more gaseous moles (larger Δng) | No change in K |
| Increase Temperature (ΔH > 0, Endothermic) | Shifts Forward (→) (absorbs added heat) | K increases |
| Increase Temperature (ΔH < 0, Exothermic) | Shifts Backward (←) (absorbs added heat) | K decreases |
| Addition of Catalyst | No shift (accelerates forward & reverse equally) | No change in K (reaches equilibrium faster) |
| Inert Gas Added at Constant Volume | No shift (partial pressures & molar concentrations unchanged) | No change in K |
| Inert Gas Added at Constant Pressure | Shifts towards larger number of gaseous moles (Δng > 0) | No change in K |
5. Temperature Dependence of K: Van 't Hoff Equation
The equilibrium constant K depends strictly on temperature and is independent of concentrations, pressures, or catalysts. Quantitatively, the temperature dependence is given by the differential Van 't Hoff equation:
Integrated form: log10(K2 / K1) = [ΔH° / (2.303 R)] · [(T2 − T1) / (T1 T2)]
6. Solved Practice Examples (JEE Main / Advanced / NEET)
1. Find Δng: Δng = 2 − (1 + 3) = 2 − 4 = −2.
2. Apply relation: Kp = Kc (R T)Δng = Kc (R T)−2 ⇒ Kc = Kp (R T)2.
3. Substitute values: R T = 0.0821 × 673 ≈ 55.25.
4. Kc = (1.6 × 10−4) × (55.25)2 = (1.6 × 10−4) × 3052.5 ≈ 0.488 ≈ 0.49 mol⁻² L².
1. Initially, solid decomposes into equal moles of NH3 and H2S:
pNH3 = pH2S = Ptotal / 2 = 0.84 / 2 = 0.42 atm.
2. Calculate Kp = pNH3 · pH2S = (0.42) × (0.42) = 0.1764 atm2.
3. Since temperature is constant, Kp remains 0.1764 atm2.
4. At new equilibrium, p'NH3 = 0.70 atm:
Kp = p'NH3 · p'H2S ⇒ 0.1764 = (0.70) · p'H2S
p'H2S = 0.1764 / 0.70 = 0.252 atm.
7. Frequently Asked Questions (FAQs)
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