Chemical equilibrium is a foundational concept in chemistry that explains how reversible reactions behave under closed system conditions. Mastering it is critical not just for scoring well in academic exams but also for understanding how real-world chemical processes operate from biological systems to industrial production.
What is Chemical Equilibrium?
Chemical equilibrium refers to the state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products. It is a dynamic condition, not a static one. This means molecules are constantly reacting in both directions, but the overall concentrations stay the same.
Real-Life Analogy:
Imagine a busy escalator with people going up and down. If people go up and down at the same rate, the number of people on each floor stays constant, even though movement continues. That’s chemical equilibrium: continuous activity with no overall change.
Key Terms:
|
Term |
Definition |
|
Dynamic Equilibrium |
Condition where reactions continue but concentrations remain unchanged |
|
Equilibrium Constant (K) |
Numerical value expressing the ratio of product and reactant concentrations |
|
Le Chatelier's Principle |
Rule that explains how equilibrium shifts when external conditions change |
Why Does Equilibrium Occur?
Equilibrium occurs because chemical reactions are often reversible. As products form, they can also revert back into reactants. Over time, these two opposing reactions balance out. This state of balance allows the system to reach minimum Gibbs free energy, a thermodynamically stable point.
The Gibbs Free Energy (∆G) becomes zero at equilibrium:
- ∆G < 0: Forward reaction is favored.
- ∆G > 0: Reverse reaction is favored.
- ∆G = 0: Equilibrium.
This condition helps biological systems maintain homeostasis and allows chemists to control product yields in industrial settings.
How Does Chemical Equilibrium Work?
Consider a generic reversible reaction:
aA + bB ⇌ cC + dD
The equilibrium constant (Kc) is expressed as:
Kc = [C]^c [D]^d / [A]^a [B]^b
This formula tells us the ratio of product to reactant concentrations at equilibrium. If K is large, the reaction favors products. If K is small, it favors reactants. The constant is only valid at a specific temperature.
Chemical equilibrium also responds to external changes. According to Le Chatelier’s Principle:
- Adding reactants shifts the reaction forward.
- Removing products also shifts it forward.
- Increasing temperature favors the endothermic direction.
These principles help in designing chemical processes like the Haber process for ammonia production.
Difference Between Kc and Kp
- Kc is based on concentration (mol/L).
- Kp is based on partial pressures (atm).
They're related by:
Kp = Kc(RT)^Δn
Where:
- R = Gas constant (0.0821 L·atm/mol·K)
- T = Temperature in Kelvin
- Δn = moles of gaseous products - moles of gaseous reactants
Understanding both is crucial for tackling problems involving gases in equilibrium.
Applications and Importance
Chemical equilibrium is not just a textbook concept. It has real-life implications:
- Biological Systems: Oxygen binding to hemoglobin is an equilibrium reaction sensitive to pH and CO2 levels.
- Industrial Chemistry: Equilibrium control is critical in maximizing yields.
- Environmental Science: Atmospheric reactions, such as ozone formation and breakdown, are governed by equilibrium principles.
Common Misconceptions
- Equilibrium doesn’t mean equal concentration: It means equal rates of forward and reverse reactions.
- The reaction hasn’t stopped: It continues dynamically.
- K is not always 1: Its value depends on the reaction and conditions.
In a chemical reaction, reactants are consumed to form products. It has been observed that reactants are not consumed cent percent and the resulting mixture contains both reactants and products.
On the basis of the above observation, chemical reactions are broadly classified into two types:
1. Irreversible reactions
The reactions which move in forward direction only are called irreversible reactions. For example:
BaCl₂ + Na₂SO₄ → BaSO₄ ↓ + 2NaCl
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2. Reversible reactions
The chemical reactions which take place in both directions under similar conditions are called reversible reactions. For example:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
N₂O₄(g) ⇌ 2NO₂(g)
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
EQUILIBRIUM
Equilibrium is the state of a reversible reaction in which the rate of forward reaction is equal to the rate of backward reaction.
Characteristics of Equilibrium
(i) Equilibrium can be attained from either direction.
(ii) At equilibrium, the concentration of reactants and products remain constant.
(iii) It is dynamic in nature. The reaction does not stop at equilibrium but proceed in both directions with same rate.
(iv) Equilibrium can be established if none of the product is allowed to escape out.
(v) A catalyst does not change the state of equilibrium, but it decreases time to reach the equilibrium.
Types of equilibria
(i) Physical equilibria: It involves physical changes.
(a) Solid ⇌ Liquid
e.g. ice ⇌ water
(b) Liquid ⇌ Gas
e.g. H₂O(ℓ) ⇌ H₂O(vapours)
(c) Solid ⇌ Gas
e.g. I₂(s) ⇌ I₂(vapours)
e.g. sublimation
Ice ⇌ Water ⇌ Vapours
[Note: The temperature and pressure at which all the three states exist together is called triple point. Triple point of H₂O is 0.0098°C and 4.58 mm.]
(ii) Chemical equilibria: It involves chemical reactions, e.g.
H₂(g) + I₂(g) ⇌ 2HI(g)
(iii) Homogeneous equilibria: When all the reactants and products are present in same phase, e.g.
PCl₅(g) ⇌ Cl₂(g) + PCl₃(g)
(iv) Heterogeneous equilibria: When reactants and products are not in same phase, e.g.
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
LAW OF MASS ACTION
It states that at constant temperature the rate of a chemical reaction is directly proportional to the product of active masses of the reactants raised to the power equal to stoichiometric coefficients. For dilute solutions, active mass is equal to the molar concentration.
For the reaction: aA + bB → Product
Rate ∝ [A]ᵃ [B]ᵇ and Rate = K [A]ᵃ [B]ᵇ
where, [A] and [B] are molar concentrations of reactants A and B respectively and K is the rate constant of the reaction.
Law of Chemical Equilibrium
For the reaction: aA + bB ⇌ cC + dD at equilibrium the ratio Kf/Kb = [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ = Kc is called equilibrium constant where Kf and Kb are the rate constants of forward and backward reactions respectively.
Concentrations of pure solid or pure liquid is taken as unity.
For the reaction: BaCl₂(s) ⇌ Ba²⁺(aq) + 2Cl⁻(aq)
K = [Ba²⁺][Cl⁻]²; ∵ [BaCl₂] = 1
Equilibrium constant in terms of pressure is given by Kp = (P𝐶ᶜ · P𝐷ᵈ)/(P𝐀ᵃ · P𝐁ᵇ)
Units of K
Unit of Kc = (mol L⁻¹)(c+d)-(a+b)
Units of Kp = (atm)(c+d)-(a+b)
[Note: When expressed in terms of activities, K becomes dimensionless.]
Relation between Kc and Kp
Kp = Kc(RT)^Δn
where, Δng = (nproduct - nreactants)gaseous
[Note: In terms of mole fractions K is given by: Kx = (XCc · XDd)/(XAa · XBb) and Kp = Kx(P)Δx where, Δx = (xp - xR)gaseous]
Ex.: The Kp value for a reaction A(g) ⇌ B(g) + C(g) is 0.9 atm at 300 K. The Kc value of the reaction is
(A) 0.9
(B) 0.0365
(C) 0.009
(D) can't be calculated
Solution: (B).
Kp/Kc = (RT)Δn and Δn = 1
Kc = Kp/RT = 0.9/(0.0821 × 300) = 0.0365
Ex.: The ratio Kp/Kc for the reaction CO(g) + ½O₂(g) ⇌ CO₂(g) is
(A) 1
(B) RT
(C) (RT)(-1/2)
(D) (RT)(-1/2)
Ex.: For which of the following reaction Kp = Kc?
(A) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
(B) N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
(C) H₂(g) + Cl₂(g) ⇌ 2HCl(g)
(D) PCl₃(g) + Cl₂(g) ⇌ PCl₅(g)
REACTION QUOTIENT
In a reaction at any stage other than equilibrium the ratio [C]c[D]d/[A]a[B]b = Q is called reaction quotient.
(i) If Q = K, the reaction is at equilibrium.
(ii) If Q < K, the reaction proceeds in forward direction.
(iii) If Q > K, the reaction proceeds in backward direction.
Characteristics of Equilibrium Constant
(i) It is independent of original concentration of reactants.
(ii) It is constant for a particular reaction and depends upon temperature only.
(iii) If the reaction is written in reversed direction, then Kb = 1/Kf.
(iv) It is independent of the presence of catalyst.
(v) If the reaction is multiplied by 2, equilibrium constant becomes square of the original value.
e.g. ½N₂ + ³⁄₂H₂ ⇌ NH₃; K₁
N₂ + 3H₂ ⇌ 2NH₃; K₂
then K₂ = (K₁)²
Ex.: A ten-fold increase in pressure on the reaction, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at equilibrium results ………… in Kp,
(A) ten-fold increase
(B) ten-fold decrease
(C) no change
(D) 4/9 fold decrease
Solution: (C).
Kp for a reaction is constant at constant temperature.
Exercise:
(1) If K₁ and K₂ are the equilibrium constants of the equilibrium (i) and (ii) respectively, what is the relationship between the two constants?
(i) SO₂(g) + ½O₂(g) ⇌ SO₃(g); K₁
(ii) 2SO₃(g) ⇌ 2SO₂(g) + O₂(g); K₂
(A) (K₁)² = 1/K₂
(B) K₂ = (K₁)²
(C) K₁ = 1/K₂
(D) K₁ = K₂
(2) Eight mole of a gas AB₃ attain equilibrium in a closed container of volume 1 dm³ as,2AB₃ ⇌ A₂(g) + 3B₂(g). If at equilibrium, 2 mole of A₂ are present then, equilibrium constant is
(A) 72 mol²L⁻²
(B) 36 mol²L⁻²
(C) 3 mol²L⁻²
(D) 27 mol²L⁻²
Effect of Temperature on K
Effect of temperature on K can be explained with the help of van't Hoff equation, which is given by
d ln Kp/dT = ΔH°/RT²
In integrated form: log K₂/K₁ = ΔH°/2.303R [T₂-T₁/T₁T₂], where, ΔH° = Enthalpy of reaction and T₂ > T₁
[Note: If ΔH = 0, log K₂/K₁ = 0 or K₁ = K₂. Equilibrium constant does not change with temperature.]
van't Hoff Reaction Isotherm
It gives the relation between standard free energy change of a reaction (when concentration of all the reactants and product is 1 molL⁻¹) and its equilibrium constant. The equation is given by:
ΔG° = -RT ln K = -2.303 RT log K
where, K may be Kc or Kp, R is the gas constant.
LE CHATELIER'S PRINCIPLE
If a system at equilibrium is subjected to a change of temperature, pressure or concentration, the equilibrium shifts in such a direction so as to reduce or neutralize the effect of the change applied.
Effect of Temperature
For exothermic reactions (ΔH = -ve), increase in temperature shifts the equilibrium in reverse direction.
For endothermic reactions (ΔH = +ve), increase in temperature shifts the equilibrium in forward direction.
Effect of Pressure
Increase in pressure shifts the equilibrium in the direction having lesser number of moles of gaseous species and vice-versa.
Effect of Concentration
Increase in the concentration of reactants shift the equilibrium in forward direction and vice-versa.
Effect of Adding Inert Gas
(i) Addition of inert gas at constant volume does not change the state of equilibrium.
(ii) Addition of inert gas at constant pressure will shift the equilibrium to the direction where number of moles of gases are greater.
[Note: If np = nR, no effect of addition of inert gas]
Calculation of degree of dissociation from vapour density measurements
Degree of dissociation (α) is defined as the fraction of the total number of molecules dissociated at the particular temperature. For the reaction:
A ⇌ nB
α = (D-d)/(n-1)d or α = (Mi-M)/(n-1)M
where, D = Theoretical vapour density
d = Observed vapour density
Mi = Initial molecular mass
M = Molecular mass at equilibrium or observed molecular mass
e.g. PCl₅ ⇌ PCl₃ + Cl₂
α = (D-d)/d = (D-d)/d
Ex.: The vapour density of N₂O₄ at a certain temperature is 30. The percentage dissociation of N₂O₄ at this temperature is
(A) 53.3
(B) 26.7
(C) 0.533
(D) 2.67
Solution: (A).
N₂O₄ ⇌ 2NO₂
1 0
1-α 2α
Average molar mass = 2α × 46 + (1-α) × 92 / (1+α)
92α + 92 - 92α / (1+α) = 92/(1+α)
30 × 2 = 92/(1+α)
∴ α = 0.533
Exercise
(1) For 3A + 2B ⇌ 2C + D, initial mole of A is double of B. At equilibrium moles of A and D are equal. Hence percentage dissociation of A is
(A) 50%
(B) 25%
(C) 75%
(D) none of the above
(2) In the reaction X ⇌ nY, the initial vapour density of X is D and vapour density at equilibrium is d. If degree of dissociation of X is α, then α is equal to
(A) D-d/d(n-1)
(B) D-d/(n-1)D
(C) D-d/nd-1
(D) d(n-1)/(D-d)
Applications of Le Chatelier's Principle
1. The suitable conditions for getting maximum yield of different chemicals in the industry can be predicted.
(i)N₂(g) + 3H₂(g) ⇌ 2NH₃, ΔH = -92.5 kJ
The favourable conditions for increasing the yield of ammonia are: (a) high conc. of H₂ and N₂
(b) high pressure and
(c) low temperature
(ii)2SO₂(g) + O₂(g) ⇌ 2SO₃(g); ΔH = -193.2 kJ
The favourable conditions for increasing the yield of SO₃ are: (a) high conc. of reactants
(b) low temperature and
(c) high pressure
(iii) Bosch process for the manufacture of H₂.
H₂(g) + CO(g) + H₂O(g) ⇌ 2H₂(g) + CO₂(g); ΔH = +42 kJ Water gas
The favourable conditions for increasing the yield of hydrogen are: (a) high temperature
(b) high concentration of reactants
(c) no effect of pressure as number of moles of gaseous reactants and products are equal.
2. To predict the equilibria of physical changes.
(i) Melting of ice:
Ice + Heat ⇌ Water (more volume) (less volume)
Favourable conditions for melting of ice are: (a) high temperature and
(b) high pressure
Solids whose volume increases on melting require low pressure, because high pressure results in solidification.
(ii) Solubility of gases in liquids: When a gas dissolves in a liquid there is decrease in volume, hence solubility of gas increases with increase of pressure.
[Note: Dissolution of a gas is a spontaneous process. It is accompanied by decrease in randomness (ΔS = -ve) hence from the equation, ΔG = ΔH - TΔS, ΔG can be negative only when ΔH is negative. Hence dissolution of gas in liquid is always exothermic.]
Solved Examples
1. A 10 litre box contains O₃ and O₂ at equilibrium at 2000 K. Kp = 4.17 × 10¹⁴ for 2O₃ ⇌ 3O₂. Assume that PO₂ >> PO₃ and if total pressure is 7.33 atm, then partial pressure of O₃ will be
(A) 9.71 × 10⁻⁵ atm
(B) 9.71 × 10⁻⁷ atm
(C) 9.71 × 10⁻⁶ atm
(D) 9.71 × 10⁻² atm
Sol. (B).
2O₃ ⇌ 3O₂
Initially P° 0
At eq. P° - 2P 3P
Total pressure at equilibrium = PO₂ + PO₃ = 7.33
Since PO₂ >> PO₃
Hence PO₂ = 7.33
3P = 7.33
P = 2.443
Now, Kp = (PO₂)³/(PO₃)²
PO₃ = ∛[(3P)³/Kp] = ∛[27 × (2.443)³/7.17 × 10¹⁴] = 9.71 × 10⁻⁷ atm
2. The Kp of a reaction is 10 atm at 300 K and 4 atm at 400 K. The incorrect statement about the reaction is
(A) the reaction is exothermic.
(B) the Ea of forward reaction is more than that of backward reaction.
(C) the rate of backward reaction increases more than that of forward reaction with increase of temperature.
(D) the difference between heat of reaction at constant pressure and that at constant volume is ΔnRT.
Sol. (B).Kp decreases with temperature. Hence, ΔH must be -ve as evident from the d ln Kp/dT = ΔH/RT² equation.
ΔH = Ea(FR) - Ea(BR)
∴ Ea(FR) < Ea(BR)