CHEMICAL EQUILIBRIUM

Chemical equilibrium is a state where there is no observable change in the properties of a system concerning time. An equilibrium system may be static or dynamic. A static equilibrium is one in which the system is stationary, while a dynamic equilibrium is the one in which forward and backward reactions occur at the same rate, thereby producing no net change in the concentration of the reactants or products. Examples of a reaction in dynamic equilibrium are:

3Fe(s) + 4H2O(g)     →            Fe3O4(s) + 4H2(g)

NH4Cl(s) heat       →          NH3(g) + HCl(g) 

LE-CHATELIER’S PRINCIPLE

The principle states that if a chemical system is in equilibrium and one of the factors involved in the equilibrium is altered, the equilibrium will shift so as to neutralize the effect of the change.

FACTORS AFFECTING EQUILIBRIUM

1. TEMPERATURE: An exothermic reaction is favoured by lowering the temperature while an endothermic reaction is favoured by raising the temperature of the system. The equilibrium position shifts to the side that if favoured. For example; 2H2 (g) + O2 (g)      →           2H2O(g); ΔH= – ve

Lowering the temperature favours the production of steam (forward reaction), the equilibrium position will therefore shift to the right. On the other hand, raising the temperature favours the production of hydrogen and oxygen (backward reaction), hence the equilibrium position will shift to the left.

2. PRESSURE: The side of the system with higher number of moles or volumes of gas is favoured by lowering the pressure while the side with lesser number of moles or volume of gas is favoured by raising the pressure of the system e.g.

3H2(g) + N2(g)        →        2HN3(g)

2SO2(g) + O2(g)     →       2SO3(g)

The forward reactions in the above reactions are favoured by increasing the pressure of the system, while the backward reactions are favoured by lowering the pressure.

3Fe(s) + 4H2O(g)      →    Fe3O4(s) + 4H2(g)

Pressure does not affect the reaction above because the volume of gases at both sides of the reaction is the same.

3. CONCENTRATION: If the concentration of the reactant is increased, more products will be formed i.e., the forward reaction will be favoured and equilibrium position will shift to the right. The same thing also happens if part of the products formed is removed. However, if part of the reactant is removed or the concentration of the products is increased, the backward reaction will be favoured.

EQUILIBRIUM CONSTANT

Equilibrium constant is the measure of the ratio of the equilibrium concentration of the products of a reaction to the equilibrium concentration of the reactants with each concentration raised to the power corresponding to the coefficient in the balanced equation of the reaction. For example

aA + bB            cC + dD

Kc = [C]c[D]d or Kp = PCcPDd

[A]a[B]b PAaPBb

[ ] means concentration

C means concentration in mol/dm3

P means partial pressure (for gases only)

If Kc is positive, it means that the forward reaction is favoured at equilibrium but if it is negative, then the backward reaction is favoured. Generally, under the influence of temperature; whenever the equilibrium position shifts to the left (i.e. favouring the backward reaction), Kc decreases. But when the equilibrium position shifts to the right (i.e. favouring forward reaction), Kc increases. The Kc value for any reaction is constant only if the temperature remains constant.

CALCULATION

1. Calculate the equilibrium constant of the following reaction and the equilibrium constant of its backward reaction at 450C if the pressure of the hydrogen, iodine and hydrogen iodide gases are 0.065, 0.45 and 0.245 respectively at the equilibrium position. H2(g) + I2(g)            2HI(g)

Solution

Kc for the forward reaction

Kp = PHI2 (0.245)2 = 2.05

PH2PI2 (0.065)(0.45)

Kp for the backward reaction

Kp = 1/Kp i.e, inverse of the Kp for forward reaction

Kp = 1/2.05 = 0.49

EQUILIBRIUM CONSTANT, FREE ENERGY, AND ELECTRODE POTENTIAL

Reactions with ΔG less than zero occur spontaneously. Also, reaction with large Kc values goes almost to completion, while those with small Kc values do not occur to a reasonable extent. Hence, ΔG and Kc relationship at constant temperature is represented in the Van’t Hoff Isotherm equation

ΔGѳ =-RTlnK

Where lnK = -2.303log10K

ΔGѳ = -2.303RTlog10K

ΔGѳ = standard free energy

K = equilibrium constant

R = molar gas constant

T = temperature of the system in Kelvin

For an electrochemical cell

Eѳ = RTlnK

nF

where; Eѳ = standard electrode potential

n = number of moles of electron

F = Faraday’s constant

By substitution, ΔGѳ = -nFEѳ

INDUSTRIAL APPLICATION OF CHEMICAL EQUILIBRIUM

1. THE HABER PROCESS

3H2(g) + N2(g)     →         2NH3(g); ΔH = -46.1kJmol-1

Conditions favouring the yield of NH3 are:

– Low temperature of about 4500C

– A high pressure of about 200atm

– Catalyst – finely divided iron; it speeds up the rate of higher yield of ammonia at about 2500C.

2. THE CONTACT PROCESS (Production of H2SO4)

The first step in the process is

2SO2(g) + O2(g)      →      2SO3(g) ΔH= -395.7kJmol-1

The conditions required for the high yield of SO3 are:

– Low temperature of about 450 – 5000C

– High pressure should be used but in practice, atmospheric pressure of 1 atm gives a high yield of SO3.

An increase in the concentration of one of the reactants favours the production of SO3. The SO3 is removed from the equilibrium mixture by dissolving it in a fairly conc. H2SO4 to form oleum, which is then diluted to form H2SO4 of the required concentration. – Catalyst – vanadium (V) oxide, V2O5.

Assignment

1. Calculate the ΔH of a chemical system when the entropy is 218.2 JK-1mol-1 and the temperature is 1000C.

Tutorial questions

2. Consider the following reversible reaction, which occurred at a temperature of 298K:

N2(g) + 3H2(g)      →           2NH3(g); ΔH = -92.37kJ

List two factors that would increase the yield of NH3(g)

2. Calculate the standard free energy of a reaction at 25°C, when the enthalpy and the entropy of reaction are -190Jmol-1 and 20Jmol-1K-1 respectively.
3. State whether the reaction in 2 above is feasible or not. Give a reason for your answer.
4. State Le Chatelier’s Principle
5. Consider the equilibrium reaction represented by the equation below;

A2(g) + 3B2(g)      →          2AB3(g); H= +xkJmol-1

5. Explain briefly the effects of the following changes on the equilibrium composition

– Increase in concentration of B

– Decrease in pressure of the system

– Addition of a catalyst

6. The lattice energies of three sodium halides are as follows:

Compound NaF, NaBr, NaI

Lattice Energy (kJmol-1) 890 719 670

Explain briefly the trend