1. If we reverse an equation K_{P} or _{ }K_{C} is reversed.
If A + B ⇔ C + D _{ }K_{C} = 10
then for C + D ⇔ A + B _{ }K_{C }= 10^{1}
2. If we multiply each of the coefficient in a balanced equation by a factor n,then equilibrium constant is raised to the same factor.
 ½ N_{2} + ½O_{2 }⇔ NO _{ }K_{C} = 5
N_{2} + O_{2 }⇔ 2NO K_{C} = 5^{2} = 25_{ }
3. If we divide each of the coefficients in a balanced equation by a factor n, then new equilibrium constant is the nth root of the previous value.
2SO_{2} + O_{2} ⇔ 2SO_{3 }_{ }K_{C} =25
SO_{2} + 1/2 O_{2} ⇔ SO_{3}B _{ }K_{C} = (25)^{1/2}
4. When we combine individual equation, we have to multiply their equilibrium constant for net reaction.
If K_{1 }, K_{2} and K_{3 } are stepwise equilibrium constant for A⇔B , B⇔C , C⇔D .
Then for A⇔D , the equilibrium constant is K= K_{1}K_{2}K_{3 }
Significance of the magnitude of equilibrium constant:

If a large value of K_{P} or _{ }K_{C} signifies that the forward reaction goes to completion or very nearly so.

A large value of K_{P} or K_{C} signifies that the forward reaction doesn’t occur to any significant extent.

A reaction is most likely to reach a state of equilibrium in which both reactant and product are present if the numerical value of K_{P} or _{ }K_{C} is neither very large nor very small
Units of the equilibrium constant:
For the sake of simplicity in the units of Kc or Kp the relative molarity of pressure of reactants and products are used with respect to standard condition. (For solution standard state 1 mol/liter and for gas standard pressure=1 atm).Now resulting equilibrium constant becomes unitless by using relative molarity and pressure. But the numerical value may change in the value of standard states molarity and pressure.