Law of Mass Action was first formulated by two Norwegian chemists, Guldberg and Waage in 1864.
The basis of their formulation is the observation of a huge deposit of Sodium Carbonate on Egyptian take shore. A large amount of NaCl intake water and CaCO₃ on the take shore made the reverse reaction possible.
The basis of their formulation is the observation of a huge deposit of Sodium Carbonate on Egyptian take shore. A large amount of NaCl intake water and CaCO₃ on the take shore made the reverse reaction possible.

CaCl2(aq) + Na2CO3(aq)
↓
CaCO3(s) + 2NaCl(aq)

The forward reaction occurs spontaneously in the laboratory. They States the Low as
The rate of chemical reaction at a constant temperature is directly proportional to the active mass of the reactants.
The active mass is thermodynamic quantity. We assumed active mass as,
 Molar Concentration (moles/lit) When the solution is dilute, that is when the system behaves ideally.
 Partial pressure in atmosphere unit for gaseous system and when the pressure of the system is very low.
 For pure solid and pure liquid, active mass is assumed to be unity since their mass does not affect the rate of reaction.
Mathematical expression of Law of Mass Action

Let us Consider a reaction,
A + B ⇆ C + D

Let the reacting system contains reactants only and [A] and [B] are their Concentrations in molar units.

According to the Law of mass action, the rate of the forward reaction,

R_{f} ∝ [A] × [B]
∴ R_{f} = K_{f} × [A] × [B]

Though the rate equation is an experimental quantity and powers of concentration terms are determined experimentally, here we consider a special type of rate equation where the stoichiometric coefficients are raised in the of the concentration term(this is true for the elementary or one step reaction).

As the reaction proceeds in the forward direction, concentration of A and B decreases and R_{f} also decreases.

When the products are getting accumulated in the system, the backward reaction also starts and the rate of the backward reaction according to the law of mass action,

R_{b} ∝ [C] × [D]
∴ R_{b} = K_{b} × [C] × [D]

Here K_{f} and K_{b} are the rate constants of the forward and backward reaction and they do not depend on concentration at a given temperature.

As the reactions proceed in the forward reactions R_{f} is decreasing but R_{b} is increasing. A state is then attained when they are equal.
This state is called the chemical equilibrium. There will be no further change in the composition of the system.

Thus at equilibrium,
R_{f} = R_{b}

or, K_{f} × [A] × [B] = K_{b} × [C] × [D]
∴ K_{f}/K_{b} = [C] × [D]/[A] × [B]

[A], [B], [C] and [D] are the equilibrium concentration of A, B, C and D.

Again, K_{f}/K_{b} = K_{c}
Where K_{c} is the concentration equilibrium constant of the reaction.

At a given temperature for a reaction, K_{c} is constant does not depend on the concentration of the reacting components.
K_{c} = [C] × [D]/[A] × [B] 
Equilibrium Constant from the law of mass action
 Concentration equilibrium constant

If we write the equation as,

Æ”₁A₁ + Æ”₂A₂ ⇆ Æ”₃A₃ + Æ”₄A₄
Where Æ”₁, Æ”₂, Æ”₃, and Æ”₄ are stoichiometric coefficient.
∴ Kc = [A₃]^{Æ”3} × [A₄]^{Æ”4}/[A₁]^{Æ”1} × [A₂]^{Æ”2} 

Where Kc is called concentration equilibrium constant of the reaction and [A₁], [A₂], [A₃], and [A₄] is the equilibrium concentration of A, B, C, and D.

However, the values of the equilibrium constant of a chemical reaction depend on the mode of writing its Stoichiometric (balanced) equation.

Thus, for the reaction of formation of NH3 from N2 and H2, we can write the equation as,

N2 + 3 H2 ⇆ 2 NH3

The equilibrium constant can be written as,
Kc = [NH3]²/[N2] [H2]²

But if the equation is written as,
½ N2 + 3/2 H2 ⇆ NH3

Then, K՛c = [NH3]/[N2]^{3/2}[H2]^{½}

It is clear then Kc and K՛c are not Equal in magnitude.
Thus, Kc = (K՛c)^{½}

The general rule is that if the equation is multiplied by n then the relation of Equilibrium concentration n will be raised to the power.
Kc = (K՛c)^{n} 
 Pressure Equilibrium Constant:

When all the reactants and products are gases (that is, gas  phase reacting system), the expression of the equilibrium constant for the equation,

Æ”_{1}A_{1} + Æ”_{2}A_{2} ⇆ Æ”_{3}A_{3} + Æ”_{4}A_{4}

Where Æ”_{1}, Æ”_{2}, Æ”_{3, }and Æ”_{4} are stoichiometric coefficient.
∴ K_{p} = (P_{3}^{Æ”3} × P_{4}^{Æ”4})/(P_{1}^{Æ”1} × P_{2}^{Æ”2}) 

Where K_{p} is called Pressure equilibrium constant of the reaction and P_{1}, P_{2}, P_{3} and P_{4} are the equilibrium partial pressure of reacting components.
 Problem 1:

For the dissociation N_{2}O_{4} ⇆ 2 NO_{2}, obtain an expression for the fraction of original N_{2}O_{4} dissociated at equilibrium in terms of K_{p} and total pressure.
 Answer:

The reaction is, N_{2}O_{4} ⇆ 2 NO_{2}

Let a mole of N_{2}O_{4} is taken initially and x moles of N_{2}O_{4} is dissociated at equilibrium then mole number of N_{2}O_{4} and NO_{2} at equilibrium is (a  x) and 2x.
Total moles number at equilibrium = (a x + 2x) = (a + x).
Equilibrium partial pressure of the components is {(a  x)/(a + x)}P and {2x/(a + x)}P where P = Total pressure of the reacting system.
The expression of the equilibrium constant,
K_{p} = (P_{NO2})^{2}/P_{N2O4}
= {2x/(a + x)}P/{(a  x)/(a + x)}P
∴ K_{p} = (4x^{2}P)/(a^{2}  x^{2})
K_{p} = (P_{NO2})^{2}/P_{N2O4}
= {2x/(a + x)}P/{(a  x)/(a + x)}P
∴ K_{p} = (4x^{2}P)/(a^{2}  x^{2})

The fraction of the original N_{2}O_{4} dissociated at equilibrium É‘ = x/a.
Replacing (x/a) by É‘, we have,
K_{p} = (4Î±^{2}P)/(1  Î±^{2}) 
 Mole Fraction Equilibrium Constant:

If we write the equation as,

Æ”_{1}A_{1} + Æ”_{2}A_{2} ⇆ Æ”_{3}A_{3} + Æ”_{4}A_{4}

Where Æ”_{1}, Æ”_{2}, Æ”_{3} and Æ”_{4} are the stoichiometric coefficient
∴ K_{x} = (x_{3}^{Æ”3} × x_{4}^{Æ”4})/(x_{1}^{Æ”1} × x_{2}^{Æ”2}) 

Where K_{x} is called Mole fraction Equilibrium Constant of the reaction and x_{1}, x_{2}, x_{3} and x_{4} are the equilibrium mole fraction of reacting components.
Equilibrium Constant 
Relation Between K_{p} and K_{c}:
The interrelations of these Equilibrium constants are as follows,
K_{p} = (P_{3}^{Æ”3} × P_{4}^{Æ”4})/(P_{1}^{Æ”1} × P_{2}^{Æ”2})
K_{p} = (P_{3}^{Æ”3} × P_{4}^{Æ”4})/(P_{1}^{Æ”1} × P_{2}^{Æ”2})

The ideal gas Equation, PV = nRT, may be written as,
P = (n/V)RT = CRT
Where C is the concentration of gas expressed as amount per unit volume.

= {(C_{3}RT)^{Æ”3} × (C_{4}RT)^{Æ”4}}/{(C_{1}RT)^{Æ”1} × (C_{2}RT)^{Æ”2}}
∴ K_{p} = K_{c}(RT)^{Î”Æ”} Î”Æ” = (Æ”_{3} + Æ”_{4})  (Æ”_{1} + Æ”_{2}) 
 For the reaction in which total number of reactant molecules and of resultant molecules are same:
 For the reactions in which the number of molecules of reactants differ from that of the resultant:
H_{2} (g) + I_{2} (g) ⇆ 2HI
KP = (P_{HI})^{2}/{(P_{H2})(P_{I2})}
= (C_{HI}RT)^{2}/(C_{H2}RT) (C_{I2}RT)
= [(C_{HI})^{2}/{(C_{H2}) (C_{I2})}] × [(RT)^{2}/(RT)(RT)]
= K_{c}
Thus when (Æ”_{3} + Æ”_{4}) = (Æ”_{1} + Æ”_{2}),
K_{p} = K_{c}
K_{p} = K_{c}
2SO_{2}(g) + O_{2}(g) ⇆ SO_{3}(g)
Here, K_{p} = K_{c}×(RT)^{{1  (2+1)}}
= K_{c}×(RT)^{2}
Thus when, (Æ”_{3} + Æ”_{4}) ≠ (Æ”_{1} + Æ”_{2}),
K_{p} ≠ K_{c}
K_{p} ≠ K_{c}
Reaction (Gaseous System) 
Relation between K_{p} and K_{c} 
H_{2} + Cl_{2} ⇆ 2HCl  K_{p} = K_{c} 
CO + H_{2}O ⇆ 2H_{2 }+ CO_{2}  K_{p} = K_{c} 
CO + NO_{2} ⇆ NO + CO_{2}  K_{p} = K_{c} 
PCl_{5} ⇆ PCl_{3} + Cl_{2}  K_{p} = K_{c} RT 
2H_{2} + O_{2} ⇆ 2H_{2}O  K_{p} = K_{c} RT^{ 1} 
2CO + O_{2} ⇆ 2CO_{2}  K_{p} = K_{c} RT^{ 1} 
N_{2} + 3H_{2} ⇆ 2NH_{3}  K_{p} = K_{c} RT^{ 2} 
 Problem 2:

Calculate the K_{c} value of the reaction N_{2} + 3H_{2} ⇆ 2NH_{3} at 400^{0}C,
Given Kp at the same temperature 1.64 × 10^{4}.
 Solution:

K_{p} = 0.5.
For Solution see,
Chemical Equilibrium Questions and Answers
 Problem 3:

At 100^{0}C the vapour density of N_{2}O_{4} is 25 at 1 atm. Show that K_{p} = 9.6.
 Solution:

N_{2}O_{4} (g) ⇆ 2NO_{2} (g)

Let 1 mole of N_{2}O_{4} is taken initially (t = 0) and x mole of N_{2}O_{4} has reacted at equilibrium.

So the mole number of each component is (1x) and 2x and total moles at equilibrium,

(1x+2x) = (1+x).

So total moles has increased from 1 to (1+x).
Let Volume increases from V_{1} to V_{2}.
So, (1+x) = V_{2}/V_{1}

As density and hence vapour density is inversely proportional to volume so vapour density will decrease from d_{1} to d_{2}.
Hence, (1+x) = V_{2}/V_{1}= d_{1}/d_{2}.
Molecular weight of N_{2}O_{4} is 92 and vapour density,
= 92/2
= 46
Due to dissociation it is = 25.
∴ 1+x = 46/25
or, x = 0.84
Now partial pressure are,
P_{NO2}= {2x/(1+x)}P
= (2×0.84)/1.84
= 0.913 atm
P_{N2O4} = {(1x)/(1+x)}P
= 0.16/1.84= 0.087
∴ K_{p} = (PNO2)2/PN2O4
= (0.913)2/0.087
≃ 9.6
Molecular weight of N_{2}O_{4} is 92 and vapour density,
= 92/2
= 46
Due to dissociation it is = 25.
∴ 1+x = 46/25
or, x = 0.84
Now partial pressure are,
P_{NO2}= {2x/(1+x)}P
= (2×0.84)/1.84
= 0.913 atm
P_{N2O4} = {(1x)/(1+x)}P
= 0.16/1.84= 0.087
∴ K_{p} = (PNO2)2/PN2O4
= (0.913)2/0.087
≃ 9.6
Relation Between Kp and Kx:
∴ K_{p} = K_{x}(P)^{Î”Æ”} Î”Æ” = (Æ”_{3} + Æ”_{4})  (Æ”_{1} + Æ”_{2}) 
 Problem 4:

Show that the equilibrium condition for any chemical reaction is given by Î”G = 0.
 Answer:
Vant Hoff reaction isotherm is Î”G =  RT lnK_{a} + RT lnQ_{a}But when the reaction attain equilibrium,Qa = Ka
Thus, Î”G = 0.