Question #a83c1

1 Answer
P dilip_k
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 #NH_3# in the equilibrium mixture may be taken as # (100-36-13)%=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
#p_(H_2)=36%ofP=0.36P" atm"#

Partial presssure of Nitrogen
#p_(N_2)=13%ofP=0.13P" atm"#

Partial presssure of Ammonia
#p_(NH_3)=51%ofP=0.51P" atm"#

The equation of the gaseous reaction:

#3H_2(g)+N_2(g)rightleftharpoons2NH_3(g)#

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

#K_p=(p_(NH_3))^2/(p_(N_2)xx(p_(H_2))^3)#

#=>K_p=(0.51P)^2/(0.13Pxx(0.36P)^3)" atm"^-2#

#~~42.88P^-2" atm"^-2#

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

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