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Question

Ammonia is heated at 15 atm from 27oC to 347oC assuming volume constant. The new pressure becomes 50 atm at equilibrium. Calculate % of NH3 actually decomposed.

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Solution

Pressure of NH3 at 27oC=15atm
Pressure of NH3 at 347oC=Patm
P620=15300
P=31atm
Let a moles of ammonia be present. Total pressure at equilibrium =50atm
2NH3N2(g)+H2(g)
At equi, (a2x) x 3x
total moles a2x+x+3x=a+2x
Initialno.ofmolesMolesatequilibrium=InitialpressureEquilibriumpressure
aa+2x=3150
x=1962a
Amount of ammonia decomposed 2x=2×1962a=1931a
% of ammonia decomposed19×a31×a×100
=61.3

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