# Equilibrium

### Equilibrium in Physical Processes

- Equilibrium is possible only in a closed system at a given temperature.
- Both the opposing forces occur at the same rate and there is a dynamic but stable condition.
- All measureable properties of the system remain constant.

Equilibrium is characterized by constant value of one of its parameters at a given temperature. The magnitude of such quantities at any stage indicates the extent to which the physical process has proceeded before reaching equilibrium.

Process | Conclusion |
---|---|

Liquid ⇌ Vapor H _{2}O (s) ⇌ H_{2}O (l) | P_{H2O} constant at given temperature |

Solute (s0 ⇌ Solute (solution) | Concentration of solute in solution is constant at given temperature. |

Gas (g) ⇌ Gas (aq) | [gas(aq)]/[gas(g)] is constant at given temperature. |

## Equilibrium in Chemical Processes

When the rates of forward and reverse reactions become equal, concentrations of reactants and product remain constant. This is the stage of chemical equilibrium. Let us consider following reversible reaction.

A + B ⇌ C + D

When the reaction begins, concentration of C and D is zero. In due course of time, concentration of C and D increases while that of A and B decreases. With this, the rate of forward reaction decreases, the rate of reverse reaction increases. Finally, a time comes when both reactions occur at the same rate, i.e. the system attains equilibrium.

#### Law of Chemical Equilibrium and Equilibrium Constant

The concentrations in an equilibrium mixture are related by equilibrium equation which is as follows. *(Guldberg and Waage 1864)*

`K_c=([C][D])/([A][B])`

Where, K_{c} is equilibrium constant and the expression on RHS is called the equilibrium constant expression.

**Equilibrium Law:** At a given temperature, the product of concentrations of products raised to the respective stoichiometric coefficient in balanced chemical equation divided by the product of concentrations of reactants raised to their individual stoichiometric coefficients has a constant value.

If a general reaction is written as follows:

aA + bB ⇌ cC + dD

Its equilibrium constant is written as follows:

`K_c=([C]_c[D]_d)/([A]_a[B]_b)`

Equilibrium constant for reverse reaction is the inverse of the equilibrium constant for forward reaction.

### Homogeneous Equilibria

All the reactants and products are in the same phase in a homogeneous system. Following are some examples.

N_{2} (g) + 3H_{2} (g) ⇌ 2NH_{3} (g)

CH_{3}COOC_{2}H_{5} (l) + H_{2}O (l) ⇌ CH_{3}COOH (aq) + C_{2}H_{5}OH (aq)

#### Equilibrium Constant in Gaseous System

Equilibrium constant is expressed in terms of molar concentrations of reactants and products. But for reactions involving gases, it is more convenient to express the equilibrium constant in terms of partial pressure.

We know that the ideal gas equation is

`pV=nRT`

Or, `p=n/V\RT`

In this equation, n is number of moles and V is volume. We know that mol per unit volume is concentration so `n/V=c`

So, above equation can be written as follows:

`p=cR\T`

We can also write p = [gas] RT

Where R = 0.0831 bar litre/mole K

*At constant temperature, pressure of gas is proportional to its concentration. (Boyle’s Law)*

Let us now consider following reaction.

H_{2} (g) + I_{2} (g) ⇌ 2HI (g)

Equilibrium constant can be written as follows:

`K_c=([HI (g)]_2)/([H_2(g)][I_2(g)])`

Or, `K_c=((p_(HI))^2)/((p_(H_2))(p_(I_2)))`

Since p_{HI} = [HI (g)]RT

p_{I2} = [I_{2} (g)]RT

p_{H2} = [H_{2} (g)]RT

So, above equation can be written as follows:

`K_p=([HI(g)]^2[RT]^2)/([H_2(g)]RT[I_2(g)]RT)`

Or, `K_p=([HI(g)]^2)/([H_2(g)][I_2(g)])=K_c`

In this example, K_{p} = K_{c}, but it is not always the case.

### Heterogeneous Equilibria

Equilibrium in a system with more than one phase is called heterogeneous equilibrium. Such equilibria generally involve pure solids or liquids. In such a case, equilibrium expressions can be simplified because molar concentration of pure solid or liquid is constant. But when a gas is involved, there will be variations. Let us consider following example.

CaCO_{3} (s) ⇌ CaO (s) + CO_{2} (g)

For this equation,

`K_c=([CaO(s)][CO_2(s)])/([CaCO_3(s)])`

Since molar concentrations of CaCO_{3} and CaO are constant, so modified equilibrium constant for this reaction will be

K'_{c} = [CO_{2} (g)]

Or, K_{p} = p_{CO3}