At 25°C in a closed system, ammonium hydrogen sulfide exists as the following eq

Question

At 25°C in a closed system, ammonium hydrogen sulfide exists as the following equilibrium 3. NH,HS(s) NH3(g)+ H2S(g) When a sample of pure NH4HS(s) is placed in an evacuated reaction vessel and allowed to come to equilibrium at 25°C, the total pressure is 0.660 atm. What is the value of K? a) To this system, sufficient H2S(g) is injected until the pressure of H2S is three times that of the ammonia at equilibrium. What are the partial pressures of NH3 and H2S at equilibrium? b) c) In a different experiment, 0.750 atm of NH3 and 0.500 atm of H2S are introduced into a 1.00-L vessel at 25°C. How many grams of NH4HS are present when equilibrium is established?

Explanation / Answer

for the reaction NH4HS(s) --------->NH3(g)+ H2S(g), at equilibrium, PNH3= PH2S=0.66 atm, the activity of NH4HS is 1,

K= Equilibrium constant = PNH3*PH2S/[NH4HS] becomes K= PNH3*PH2S= 0.66*0.66=0.4356

2, At Equilibrium, PH2S= 3*PNH3

from the relation of K,

hence K=0.4356 = PNH3*( 3PNH3), PNH3= 0.38 atm and PH2S= 3*0.38= 1.14 atm

3. Q= reaction coefficient = PH2*PNH3= 0.75*0.5= 0.375<K, so the reaction proceeds in the forward direction.

let x= partial pressure change of NH3= partial pressure of H2S to reach equilibrium

At Equilibrium, PNH3=0.75+x and PH2S=0.5+x

K= 0.4356= (0.75+x)*(0.5+x)

when sovled ,x = 0.047, at Equilibrium, partial pressure of NH3=0.75+0.047=0.797 atm

partial pressure of H2S( the limiting reactant)= 0.5+0.047=0.547

Partial pressure is the pressure exerted if the gas along occupies the entire volume

P= 0.546 atm, V=1L, T= 25 deg.c= 25+273= 298K , R= gas constant =0.0821 L.atm/mole.K

n= no of moles of H2S= PV/RT= 0.546*1/(0.0821*298)= 0.022 moles of H2S

1 mole of H2S requires 1 mole of NH4HS= 0.022 moles of NH4HS

molar mass of NH4HS= 51g/mole, moles of NH4HS at equilibrium= 0.022*51 gm=1.122 gm

no of moles of NH3=

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