# Law of Mass Action and Equilibrium Constant

On the basis of observations of many equilibrium reactions, two Norwegian chemists Cato Goldberg and Peter Waage (1864) suggested a qualitative relationship between rates of reaction and the concentration of the reacting species. The relationship is known as Law of Mass Action. This law states that:

At constant temperature the rate of a chemical reaction is directly proportional to the product of the molar concentrations of reacting species with each concentration term raised to the power equal to the stoichiometric co-efficient of that species in the chemical equation.

For any general reaction,

aA + bB  à products

The law of mass action may be written as

where [A] and [B] are the molar concentration of the reactants A and B respectively, k is a constant of proportionality called rate constant

APPLICATION OF LAW OF MASS ACTION TO CHEMICAL EQUILIBRIUM

By applying law of mass action to a reversible reaction in equilibrium it is possible to derive a simple mathematical expression known as law of chemical equilibrium.

Let us consider a simple reversible reaction,

in which an equilibrium exists between the reactants A and B and the products C and D. According to law of mass action:

Rate of forward reaction= k f [A]a [B]b

where k f is the rate constant for the forward reaction. [A] and [B] are molar concentrations of reactants A and B respectively.

Similarly, the rate of backward reaction

= kb[C]C[D]d

where kb is the rate constant for the backward reaction and [C] and [D] are molar concentrations of products C and D respectively.

At equilibrium the rates of two opposing reactions become equal. Therefore, at equilibrium,

Rate of forward reaction = Rate of backward reaction

K f / kb = [C]C [D]d / [A]a [B]b / [A]a [B]b

Since k f and kb are constants, therefore, the ratio k f / kb is also constant and is represented by Kc.

K c = K f / kb = [C]C [D]d / [A]a [B]b

Thus, the expression of equilibrium constant can be written as:

K c= [C]C [D]d / [A]a [B]b

The constant K c is called equilibrium constant. The subscript ‘c’ indicates that K c is expressed in terms of concentrations (mol dm- 3) of reactants and products. The concentrations in expression (21.1) are the equilibrium concentrations.

The equilibrium constant, K c, may be defined as the ratio of product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric co-efficient of the substance in the balanced chemical equation.

Law of chemical equilibrium may be stated as:

At a given temperature, the ratio of product of equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the respective stoichiometric coefficients in the balanced chemical has a constant value.

The concentration ratio, [C]C [D]d/ [A]a [B]b is called concentration quotient and is denoted by Qc. According to law of chemical equilibrium,

Qc = K c at equilibrium.

EQUILIBRIUM CONSTANT IN GASEOUS SYSTEMS

If an equilibrium involves gaseous species, then the concentrations in the concentration quotient may be replaced by partial pressures because at a given temperature the partial pressure of a gaseous component is proportional to its concentration. If the above mentioned reaction has all the gaseous species, then

Q p = (pc)c (P D)d / (PA)a (PB)b = KP

The equilibrium constant, ~· defined in terms of partial pressures is not the same as the equilibrium constant K c, defined in terms of concentrations.

RELATIONSHIP BETWEEN KP AND K c

For ideal gas, PV = nRT

P = n / V RT = CRT

PA = CA RT

PB = CB RT

PC = CC RT

PD = CD RT

where C A‘ CB CC and CD are the molar concentrations of A, B, C and D respectively.

KP = (CC RT)C (CD RT)d / (CA RT)a (CB RT)b

KP = (CC RT) Δ n

Where            Δ n = (sum of the exponents in / the numerator of Qc)

– ( sum of the exponents in the denominator of Qc )

It is necessary that while calculating the value of KP units of R must be appropriate.

• If partial pressures are expressed in atm, the value of R should be 0.0821 atm dm3K-1 mol-l.

• If partial pressures are expressed in kPa, the value of R should be 8.314 kPa dm3 K-1 mol-1.

• If partial pressures are expressed in bar, the value of R should be 0.083 bar dm3 K-1 mol-1.

HOMOGENEOUS AND HETEROGENEOUS EQUILIBRIA

The equilibrium in which all the substances are present in the same phase is known as homogeneous equilibrium. For example,

The equilibrium in which the substances involved are present in different phases is called heterogeneous equilibrium.

For example,

EQUILIBRIUM EXPRESSIONS OF SOME REACTIONS

While writing equilibrium expressions certain conventions are followed. Let us illustrate these conventions with the help of some examples.

1. In case some solid is involved in the equilibrium, its concentration remains constant no matter how much of it is present. Therefore, by convention, the concentrations of all        solids are taken as unity, i.e., [Solid] = 1. For example

K = [CaO(s)] [CO2 (g)] / CaCO3 (s) ]

= [ CO2 (g) ] [ CaO(s) ] = [CaCO3 (s)] = 1

or        k p = pco2

The above expression shows that at a particular temperature, there is a constant pressure of CO2 in equilibrium with CaO(s) and CaCO3(s).

It may be emphasized here that in heterogeneous equilibria pure solids or liquids must be present for the equilibrium to exist, but their concentrations or partial pressures do not appear in the expression of the equilibrium constant. ·

2. In case some pure liquid is in equilibrium with some gas or gases, the concentration of pure liquid is taken as unity. For example,

K = [H2O(g))] / [H2O(l) ]

= [H2O(l)

K p = OH2O

3. For the equilibria in aqueous medium if H2O is also involved in the equilibrium, its concentration will not change appreciably because it is present in a large quantity (= 55.5 mol dm-3) and hence, [H2O] is taken as constant. For example,

K = [NH4 + (aq) ] [ OH ( aq) / [NH3 (aq) ] [ H2O(l)]

or                     K x [ H2O] = NH4 + (aq) ] [ OH (aq) / [NH3 (aq)

or                                     k = [NH4 + (aq) [OH (aq)] / [NH3 (aq)]

CHARACTERISTICS OF EQUILIBRIUM CONSTANT (K)

Some important characteristics of equilibrium constant are as follows:

1. The equilibrium constant has a definite value for every chemical reaction at a particular temperature   value of equilibrium constant is independen1of initial concentrations of reacting species.

For example, the equilibrium constant for the reaction

K = [ feSCN2+ ] / [Fe3+] [ SCN] = 138.0

Whatever may be the initial concentrations of the reactants, Fe3+ and SCN-1 ions, the value of K comes out to be 138.0 at 298 K.

2. The value of equilibrium constant, K changes with change in temperature.

However, it may be noted that the value of K may increase or decrease with increase in temperature depending upon whether the reaction is endothermic or exothermic in nature respectively. The values of KP at different temperatures for an endothermic reaction and for anexothermic reaction are given in Table 21.1.

Table 21.1. Variation of PK with Temperature for Some Reactions

3. For a reversible reaction, the equilibrium constant for the backward reaction is inverse of the equilibrium constant for the forward reaction.

For example, equilibrium constant, K, for the reaction of combination between hydrogen and iodine at 717 K is 48.

The equilibrium constant for the decomposition of hydrogen iodide at the same temperature is the inverse of the above equilibrium constant

4. The equilibrium constant is independent of the presence of catalyst. .

5. If equilibrium constant is expressed in terms of concentration, it has different units for different reactions.

This can be illustrated by following examples:

(a) If the number of moles of the product is same as the number of moles of reactants, K has no units (i.e., it is dimensionless). Thus, in the reaction between hydrogen and iodine to form hydrogen iodide, K has no units as illustrated below:

K =[ HI]2 / [H2] [ I2] = (mol dm-3)2 / (mol dm-3 ) ( mol -3)

(b) If the number of moles of the products is not the same as that of reactants, K will have certain units depending upon the change in the number of moles. Thus, for the reaction between nitrogen and hydrogen to form ammonia, K would have the units dm6 mol-2 as shown below:

K =[ NH]2 / [N2] [ H2]3 = (mol dm-3)2 / (mol dm-3 ) ( mol -3)

Thus, K will have the units dm6 mot2.