Question #a83c1

1 Answer
Aug 16, 2016

The reaction was initiated by volimetric ratio of Hydrigen and Nitrogen as 3:1 and at equilibrium the ratio remained more or less same 36%:13%3:1 (Within experimental error.)
So the volume of NH3 in the equilibrium mixture may be taken as (1003613)%=51%

If the pressure of the reaction mixture at equilibrium be P atm then the partial pressures of the constituents of the reaction mixture will be as follows.

Partial presssure of Hydrogen
pH2=36%ofP=0.36P atm

Partial presssure of Nitrogen
pN2=13%ofP=0.13P atm

Partial presssure of Ammonia
pNH3=51%ofP=0.51P atm

The equation of the gaseous reaction:

3H2(g)+N2(g)2NH3(g)

Now the equlibrium constant of the gaseous reaction in respect of pressure is given by

Kp=(pNH3)2pN2×(pH2)3

Kp=(0.51P)20.13P×(0.36P)3 atm2

42.88P2 atm2

If the value of P is known then Kp can be calculated out.