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Question

Consider the following gaseous equilibria given below:

  1. N2+3H22NH3; Eqm. Constant = K1
  2. N2+O22NO; Eqm. constant = K2
  3. H2+12O2H2O; Eqm. constant= K3

The Equilibrium constant for the reaction: 2NH3+52O22NO+3H2O, In Terms of K1, K2 and K3will be _________.


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Solution

Step-1: Calculate the Equilibrium constant

  • In a chemical equilibrium, the equilibrium constant is written as the ratio of the concentration of product to the concentration of reactant raised to the power of their respective stoichiometric coefficient.

N2+3H22NH3

K1=[NH3]2[N2][H2]3---------------------------- (i)

N2+O22NO

K2=[NO]2[N2][O2]--------------------------------- (ii)

H2+12O2H2O

K3=[H2O][H2][O2]12 --------------------------- (iii)

Step-2: Equilibrium constant of 2NH3+52O22NO+3H2O

K=[NO2]2[H2O]3[NH3]2[O2]52----------------------(iv)

Step-3: Calculate K in terms of K1,K2,andK3

Cubing equation (iii)

K33=[H2O]3[H2]3[O2]32----------- (v)

multiplying eq(v) and eq(ii)

K33×K2=[NO]2[H2O]3[N2][H2]3[O2]52----------- (vi)

Divide eq(vi) by eq(I)

K33×K2K1=[NO]2[H2O]3[N2][H2]3[N2][H2]3[O2]52[NH3]2=[NO2]2[H2O]3[NH3]2[O2]52=K

Hence, The Equilibrium constant K in terms of K1, K2 and K3 is K=K33×K2K1


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