Remember to ignore that 55.0 g NH4HS since it is a solid and doesn't enter into Kc (or Kp). For every mole of NH4HS that decomposes there will be x mols NH3 and x mols H2S.
......NH4HS --> NH3 + H2S
I.......solid....0.....0
C........-x......x.....x
E......solid-x...x.....x
Kc = 0.11 = x*x
Substitute and solve for x (in mols), then substitute into PV = nRT and calculate P in atmospheres.
Ammonium hydrogrn sulfide decomposes according to the reaction.
NH4HS(s) <----> NH3(g) + H2S(g)
If 55.0g of solid NH4HS is placed in a sealed 5.0L container, what is the partial pressure of NH3 and H2S at equilibrium ?
Kc = 0.11 at 250°C
1 answer
1 answer