A flask containing 0.50 atm of ammonia contains some solid NH4HS which undergoes dissociation according to NH4HS(s)⇌NH3(g)+H2S(g). Calculate the pressure of NH3 and H2S at equilibrium KP=0.11). Also calculate the total pressure.
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Solution
The correct option is APNH3 = 0.6653 atm PH2S = 0.1653 atm PTotal = 0.8306 atm NH4HS(s)⇌NH3(g)+H2S(g)Pressure before dissociationx0.50Pressure after dissociation(x−P)(0.5+P)P
Let pressure equivalent to P atm is developed by NH3 and H2S on dissociation of NH4HS, when 0.5 atm of NH3 is already present. ∴Kp=PNH3×PH2S=(0.5+P)(P)
or 0.11=(0.5+P)(P) ∴P=0.1653 ∴PNH3=0.5+0.1653=0.6653atm ∴PH2S=0.1653atm PTotal =PNH3+PH2S PTotal = 0.1653atm+0.6653atm PTotal =0.8306atm