Law of Chemical Equilibrium & Equilibrium Constant - Chemistry Class 11

Dynamic Equilibrium: Static and Dynamic Equilibrium

When two opposing processes operate simultaneously and independently between two states (state 1 and state 2) with continuous interchange between them, the system may reach a condition called dynamic equilibrium. Dynamic equilibrium is represented by a double-headed arrow in a reversible process: A ⇌ B. Depending on the nature of the states involved, equilibria are classified as physical or chemical.

What is Dynamic Equilibrium?

Dynamic equilibrium is the state of a system in which a reversible process continues to occur in both directions, but the macroscopic composition (concentrations, pressures, etc.) no longer changes with time because the forward and reverse rates are equal. Although there is continuous microscopic movement of particles between reactants and products, there is no net change in amounts. Systems in dynamic equilibrium are examples of steady states.

Static versus Dynamic Equilibrium

Static equilibrium refers to a condition in which all motion and change have ceased; forces or fluxes are balanced so that there is no microscopic exchange between states. In contrast, in dynamic equilibrium microscopic exchanges continue but produce no net macroscopic change because the opposing rates are equal. In both cases the resultant force or net rate is zero and no visible change is observed.

                                                    

Examples of Dynamic Equilibrium

• An unopened bottle of an aerated drink: on opening and partial removal of liquid, CO₂ leaves the liquid until a new equilibrium is reached; the rate of CO₂ leaving the liquid equals the rate of CO₂ dissolving back.
• Acetic acid dissociation in a single phase, described by CH₃COOH ⇌ CH₃COO^- + H⁺.
• Dimerization of nitrogen dioxide in the gas phase: 2NO₂ ⇌ N₂O₄.
• Henry's law applies to the first example: equilibrium concentration of CO₂ in the liquid is proportional to the partial pressure of CO₂ gas above it.
• Industrial synthesis of ammonia (Haber process): N₂(g) + 3H₂(g) ⇌ 2NH₃(g), where production and decomposition proceed simultaneously at equilibrium.

Chemical Equilibrium

• When a reversible reaction is carried out in a closed vessel, the system attains a stage where the rate of the forward reaction equals the rate of the backward reaction; this stage corresponds to chemical equilibrium.
• At equilibrium: Rate of forward reaction = Rate of backward reaction.

Characteristics of Chemical Equilibrium

• An equilibrium can be attained starting either from reactants or from products.
• Equilibrium is dynamic: reactions continue microscopically though macroscopic concentrations remain constant.
• At equilibrium, concentrations (or partial pressures) of species remain constant with time.
• The presence of a catalyst does not change the position of equilibrium; it only helps the system attain equilibrium faster.
• Observable physical properties (pressure, density, colour, etc.) become constant at equilibrium.

Law of Mass Action

• Guldberg and Waage states that the rate of a chemical reaction is directly proportional to the product of the active masses of the reacting substances. 
• For a general reaction,
  
  where k_f and k_b are rate constants.
• Δn_g = no. of moles of gaseous products - no. of moles of gaseous reactants
  i.e. Δn_g = ((c + d) - (a + b))
• In heterogeneous equilibrium, the active mass of pure solids and liquids are taken as unity.

Law of Chemical Equilibrium (Equilibrium Expression)

For the general reaction aA + bB ⇌ cC + dD at equilibrium, the quotient of the product of concentrations of products (each raised to its stoichiometric exponent) to the product of concentrations of reactants (each raised to its exponent) is constant at a given temperature. This constant is the equilibrium constant expressed in concentration terms and denoted by K_c:

Equilibrium Constant

• At equilibrium: rate of forward reaction = rate of backward reaction.
• The concentration-based equilibrium constant is denoted K_c.
• Unit of K_c = (mol L^-1)^Δn_g, where Δn_g = moles of gaseous products - moles of gaseous reactants.

Use of Partial Pressures: K_p

For gaseous reactions, it is often convenient to use partial pressures instead of concentrations because at a fixed temperature the partial pressure of a gas is proportional to its concentration (by the ideal gas law). For the gaseous reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), the pressure-based equilibrium constant is denoted K_p and defined as:

Unit of K_p = (pressure unit)^Δn_g, commonly atm^Δn_g or bar^Δn_g.

Example (Ammonia synthesis)

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g),

Units of Equilibrium Constants and Dimensionless Form

• The value of equilibrium constant K_c can be calculated by substituting the concentration terms in mol/L and for K_p partial pressure is substituted in Pa, kPa, bar, or atm. 
• This results in units of equilibrium constant based on molarity or pressure unless the exponents of both the numerator and denominator are the same.
• For the reactions,
  H₂(g) + I₂(g) 2HI, K_c and K_p have no unit.
  N₂O₄(g) 2NO₂ (g), K_c has unit mol/L and K_p has a unit bar or atm
• Equilibrium constants can also be expressed as dimensionless quantities if the standard state of reactants and products are specified. 
• For a pure gas, the standard state is 1 bar. Therefore a pressure of 4 bar in the standard state can be expressed as 4 bar/1 bar = 4, which is a dimensionless number. 
• The standard state (C₀) for a solute is 1 molar solution and all concentrations can be measured with respect to it. The numerical value of the equilibrium constant depends on the standard state chosen. 
• Thus in this system, both K_p and K_c are dimensionless quantities and represented as K_p° & K_C° respectively

Relation between K_c and K_p

For the reaction aA + bB ⇌ cC + dD:

The relation between K_p and K_c is:

K_p = K_c(RT)^Δn_g

where Δn_g = (c + d) - (a + b), R is the gas constant (0.08206 L·atm·K^-1·mol^-1 if pressure in atm and concentration in mol L^-1), and T is absolute temperature in K.

Relation between K_c and K_p for different types of reactions:
(i) If Δn_g = 0, then K_p = K_c.
(ii) If Δn_g > 0, then typically K_p > K_c (for given R and T).
(iii) If Δn_g < 0, then typically K_p < K_c.

Characteristics of the Equilibrium Constant (K_c or K_p)

• At a given temperature, K has a definite value for a particular chemical reaction.
• The magnitude of K indicates the extent to which reactants are converted to products at equilibrium: K ≪ 1 (equilibrium lies to the left, reactants dominate); K ≫ 1 (equilibrium lies to the right, products dominate); K ≈ 1 (significant amounts of both present).
• If a reaction is expressed as the sum of two or more reactions, the equilibrium constant of the overall reaction is the product of the equilibrium constants of the component reactions.
                                                                   A + B + E + F ⇌ C + D + G
  
• The value of K is independent of the initial concentrations of reactants and products.
• The value of K is independent of the presence of a catalyst (catalyst only speeds attainment of equilibrium).
• K for the reverse reaction is the reciprocal of K for the forward reaction.
  
  
  
• If every stoichiometric coefficient in the balanced equation is multiplied by n, the new equilibrium constant becomes Kⁿ. If divided by n, the new constant becomes the nth root of K.
  
• Using activities (dimensionless) in K expressions removes unit ambiguity; then K is dimensionless.

Significance of the Equilibrium Constant

1. Predicting Relative Concentrations

• The numerical value of the equilibrium constant provides information about the relative amounts of reactants and products at equilibrium:
  • K ≪ 1: equilibrium lies to the left (mostly reactants).
  • K ≫ 1: equilibrium lies to the right (mostly products).
  • K = 1: appreciable amounts of both reactants and products are present.

2. Calculating Equilibrium Concentrations (Worked Example)

Example: Phosgene (COCl₂) dissociates at high temperature into CO and Cl₂. At 600 K, K_p = 0.0041 atm for the reaction COCl₂(g) ⇌ CO(g) + 1/2 Cl₂(g). Initially 0.124 atm of COCl₂ is placed in the container and allowed to reach equilibrium. Find the equilibrium composition.

Sol.

P_CO=x=0.00769 atm
PCl2=x/2=0.00385 atm
P_COCl2=0.124-x=0.124-0.00769=0.1163 atm

3. Reaction Quotient (Q)

• At any instant during a reaction, a quantity analogous to the equilibrium expression can be written using the instantaneous concentrations or pressures; this is the reaction quotient, Q. For aA + bB ⇌ cC + dD:

                                                                                                                                     

• Comparison of Q and K predicts the reaction direction:
  • If Q > K, the reaction shifts toward reactants (backward) until equilibrium is reached.
  • If Q < K, the reaction shifts toward products (forward) until equilibrium is reached.
  • If Q = K, the system is at equilibrium.

Example. For the reaction NOBr(g) ⇌ NO(g) + 1/2 Br₂(g), K_p = 0.15 atm at 90°C. If partial pressures are NOBr = 0.5 atm, NO = 0.4 atm and Br₂ = 2.0 atm at this temperature, will Br₂ be consumed or formed?

Sol.

Compute Q using the instantaneous partial pressures and compare with K_p. Here Q > K_p, so the reaction shifts backward; therefore Br₂ will be consumed.

4. Degree of Dissociation & Vapour Density

Equilibrium constants are often related to the degree of dissociation (α) of weak electrolytes or molecular species. Using mole balance and the relation between average molecular mass (vapour density) and composition, α can be determined from experimental vapour density or equilibrium constant data.

Example. The vapour density of a mixture of PCl₅, PCl₃ and Cl₂ is 92. Find the degree of dissociation (α) of PCl₅ under the conditions of the measurement.

Sol.

Initial and equilibrium mole relations for PCl₅ ⇌ PCl₃ + Cl₂ are used to express the average molecular mass and relate it to the observed vapour density.

From the algebra one obtains the molecular mass value and hence

so α = 0.13.

Solved Examples and Practice

Q.1. Given the following equilibrium constants:

(1) CaCO₃(s) → Ca²⁺(aq) + CO₃^2-(aq); K₁ = 10^-8.4

(2) HCO₃^-(aq) → H⁺(aq) + CO₃^2-(aq); K₂ = 10^-10.3

Calculate the value of K for the reaction CaCO₃(s) + H⁺(aq) → Ca²⁺(aq) + HCO₃^-(aq)

Sol.

The net reaction is the sum of reaction (1) and the reverse of reaction (2):

Reaction (1): CaCO₃(s) → Ca²⁺(aq) + CO₃^2-(aq); K₁ = 10^-8.4

Reverse of reaction (2): H⁺(aq) + CO₃^2-(aq) → HCO₃^-(aq); K_-2 = 1 / K₂ = 10^+10.3

Net Reaction: CaCO₃(s) + H⁺(aq) → Ca²⁺(aq) + HCO₃^-(aq)

K = K₁ × K_-2 = 10^-8.4 × 10^+10.3 = 10^+1.9

Comment: This net reaction describes the dissolution of limestone by acid. The small intrinsic tendency of CaCO₃ to dissolve is driven by removal of CO₃^2- via protonation to HCO₃^-. Such coupled reactions are common, especially in biochemistry, where one reaction "pulls" another forward.

Q.2. The synthesis of HBr from hydrogen and liquid bromine has an equilibrium constant K_p = 4.5 × 10¹⁵ at 25°C. Given that the vapour pressure of liquid bromine is 0.28 atm, find K_p for the homogeneous gas-phase reaction at the same temperature.

Sol.

The heterogeneous synthesis (involving liquid Br₂) and the vaporization equilibrium of Br₂(l) ⇌ Br₂(g) can be combined to give the homogeneous gas-phase reaction. Use equilibrium algebra (sum reactions and multiply corresponding equilibrium constants) to obtain the gas-phase K_p. The vapour pressure of liquid bromine (0.28 atm) serves to relate the heterogeneous and homogeneous K values via multiplication or division of equilibrium expressions as appropriate.