When solid NH4HS and 0.28 mol NH3(g) were placed in a 2 L vessel at 24◦C, the equilibrium NH4HS(s)⇀↽NH3(g) + H2S(g)for which Kc= 0.00016, was reached. What is the equilibrium concentration of NH3?Answer in units of mol/L

Answer 1

In a 2 L vessel the (NH3) initially is 0.28 mols so it is 0.28/2 = 0.14 M.
.................NH4Hs(s) ==> NH3(g) + H2S(g)
I................solid..................0.14.............0
C...............solid..................+x..............+x
E...............solid...............0.14+x.............+x

Kc = 0.00016 = (NH3)(H2S)/(NH4HS)(s)
Plug in the E line, solve for x, evaluate 0.14+x and you will obtain (NH3) in mols/L @ 24 C. Note: By definition, (NH4HS) = 1 since it is a solid. Post your work if you get stuck.

Answer 2

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