Relation between Kp and Kc and Law of Mass of action: Equilibrium constants:

LAW OF MASS ACTION: Law of mass action is applicable for only reversible chemical reactions and it is an imperial law.

The law state that “ At a fixed temperature the rate of a chemical reaction is directly proportional to the product of reactive mass of reactants raised to the their respective Stoichiometric coefficients ” The law of mass action is by Guldberg and Waage.

(1) Equilibrium Constants in term of concentration ( K_C):

At the constant temperature, let us consider the following reversible reaction

According to law of mass action - The rate of forward reaction

The rate of reverse reaction-

Where K_f and K_b  is the rate constant of the forward reaction and backward reaction respectively

We know at equilibrium, the two rates of forward as well as backward are equal. ie

Rate of reaction = Rate of forward reaction – Rate of backward reaction = 0

Kc=Kf/Kb

Unit of Kc= (Conc)ng

(2) Equilibrium Constants in term of Pressure( Kp):

Consider the general chemical  reaction taking place at constant temperature.

From law of mass action- rate forward reaction is directly proportional  to product of active mass of reactants and rate backward reaction is directly proportional  to product of active mass of products.

For an ideal gas PV=nRT

Where 

P= Pressure in atm

V=Volume in liters

n=Number of gaseous moles

R=Gas constant  

  =  0.0821 L atm/mol/K or 1/12 L atm /mole/K

T=Temperature in kelvin

                                      = total number of moles of gaseous products -total number of moles of gaseous reactants

Illustrated examples:

(1) The chemical equilibrium constant Kc for the decomposition of PCl5 is 0.0625 mole L^-1 at 300^C. What will be the value of Kp is ?

(2) The value of Kc for the reaction (formation of ammonia by haber process) is 0.50 at 400º C.What will be the value of Kp at 400ºC when concentration are expressed in mole litre.1 and pressure in atmosphere ?

(3)For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) The Kp is 4.3 × 10^-4 at 375°C. Calculate Kc for the reaction. ?

(4) Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g) ⇌ CH3OH(g) The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of Kp at this temperature ?

(5) At 700 K, the equilibrium constant Kp, for the reaction, dissociation of Sulphur trioxide to Sulphur dioxide is 1.8 × 10^3 kPa What is the numerical value of Kc for this reaction at the same temperature