Question

Question 7.12 A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this
temperature, the equilibrium constant, Kc for the reaction

N2(g) + 3H2(g) <---> 2NH3(g) is 1.7 x 10^2

Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Class XI Equilibrium Page 225

Answer 1

N2(g) +3H2(g) <=> 2NH3(g)
so, reaction quotient ( Qc) = [NH3]²/[H2]³[N2]

Given ,
[ NH3 ] = 8.13 mol/20L { concentration = mole/volume }
= 0.4065 M

[N2] = 1.57mol/20L = 0.0785 M
[H2] = 1.92 mol/20L = 0.096 M

Qc = (0.4065M)²/(0.096M)³ (0.0785M)
= 2.379 × 10³ /M²

Here, you can see that , Kc ≠ Qc , so the reaction is not in equilibrium.
because Qc > Kc it means that the reaction will proceed in backward direction or in the direction of reactants .

Answer 2

Hi

Here is your answer,

N₂(g) + 2H₂(g) ⇆ 2NH₃
  
                     Qc = [ NH₃]²/[N₂] [H₂]³

Given,

                 [NH₃] = 8.13/20 M = 0.4065 M ; [N₂] = 1.57/20 M = 0.0785 M

                     [H₂] = 1.92/20 M = 0.096

           Qc = [0.4065 M ]²/[0.0785] [0.096 M ]³ 

                              = 2.379 × 10³ M⁻²


∴ Qc ≠ Kc so the reaction mixture is not in equilibrium. Qc > Kc, it indicates that the reaction will proceed in the direction of reactants.



Hope it helps you !