Consider that 'a' moles of PCl , is taken in container of volume 'V'
Let x moles of PCl₅ be dissociated into x moles of PCl₃ and x moles of Cl₂

Applying law of mass action
\(\\ \\ { K }_{ C }=\cfrac { \left[ { PCl }_{ 3 } \right] \left[ { Cl }_{ 2 } \right] }{ \left[ { PCl }_{ 5 } \right] } =\cfrac { \left( \frac { x }{ V } \right) \left( \frac { x }{ V } \right) }{ \frac { a-x }{ V } } =\cfrac { { x }^{ 2 } }{ \left( a-x \right) V } \) 
K_p, calculation: K_P= K_C . RT^Δng
Δn_g = 2-1 = 1
We know that = PV = nRT
\(RT=\cfrac { PV }{ n } \) 
Where 'n' is the total number of moles at equilibrium
n = a - x + x + x = a + x
\({ LK }_{ P }=\cfrac { { x }^{ 2 } }{ \left( a-x \right) V } .\cfrac { PV }{ n } \)
\({ K }_{ P }=\cfrac { { x }^{ 2 }\times PV }{ \left( a-x \right) V(a+x) } \)
\({ K }_{ P }=\cfrac { { x }^{ 2 }P }{ \left( a-x \right) \left( a+x \right) } \)

Let us consider the formation of HI in which, 'a' moles of hydrogen and 'b' moles of iodine gas are allowed to react in a container of volume V. Let 'x' moles of each of H₂ and I₂ react together to form 2x moles of HI.
H₂(g) + I₂(g) ⇌ 2HI(g)

Applying law of mass action
\({ K }_{ C }=\cfrac { \left[ HI \right] ^{ 2 } }{ \left[ { H }_{ 2 } \right] \left[ I_{ 2 } \right] } \) 
\({ K }_{ C }=\cfrac { \left( \cfrac { 2x }{ V } \right) ^{ 2 } }{ \cfrac { \left( a-x \right) }{ V } \cfrac { \left( b-x \right) }{ V } } =\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \)
The equilibrium constant K_p can also be calculated as follows:
We know the relationship between the K_c and K_p
K_P, = K_C . RT^Δng
Here the
Δn_g = n_p - n_r = 2 - 2 = 0
Hence, K_P = K_C
\(\\ \\ \\ { K }_{ P }=\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \) 

Let us consider a reversible reaction
x_A + y_B ⇌ IC + mD
where A, B are the reactants C and D are the product and x, y, I and m are the stoichiometric coefficients of A, B, C and D respectively.
Applying the law of mass action the rate of forward reaction.
r_f∝ [A]^x [B]^y or r_f = K_f[A]^x [B]^y
Similarly the rate of backward reaction
r_b ∝ [C]^l [D]m or
r_b,= K_b [C]^l [D]^m
where K_f and K_b are proportionality constants
At equilibrium, Rate of forward reaction (r_f) = Rate of backward reaction (r_b)
\(\therefore\) K_f[A]^x. [B]^y = K_b[C]^l [D]^m
or \(\cfrac { { k }_{ f } }{ k_{ b } } =\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] '\left[ B \right] ^{ y } } ={ K }_{ C }\)
where K_C is the equilibrium constant in terms of concentration.
At a given temperature, the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants is a constant known as equilibrium constant.
If the reactants and products of the above reaction are in gas phase, then the equilibrium constant can be written in terms of partial pressures.
\({ k }_{ P }=\cfrac { { p }_{ C }^{ 1 }\times { p }_{ D }^{ m } }{ { p }_{ A }^{ x }\times { p }_{ B }^{ Y } } \) 
where p_A, p_B, p_C and P_D are the partial pressure of gases A, B, C and D respectively.

The equilibrium constant Kc for the above reaction can be written as:
\({ K }_{ C }=\cfrac { \left[ { PCl }_{ 3 } \right] \left[ { Cl }_{ 2 } \right] }{ \left[ { Pcl }_{ 5 } \right] } =\cfrac { \left( 1.59 \right) ^{ 2 } }{ 1.4 } =1.79\)