For the reaction #NH_4HS_((s)) rightleftharpoons NH_(3(g)) + H_2S_((g))#, #K_c = 1.2*10^(-4)# at 491K. What concentration of #NH_3# will be present at equilibrium if a sample of #NH_4HS# is placed in a vessel at 491K?
The equilibrium concentration of ammonia will be equqal to
Even before doing any calculations, you can look at the magnitude of the equilibrium constant,
The first important thing to notice is that ammonium hydrosulfide is actually a solid. This means that its concentration is assumed to be constant during the reaction.
This is why the problem doesn't provide an initial concentration for the ammonium hydrosulfide.
Use an ICE table to help you determine what the equilibrium concentrations of the two products will be
The expression of the eequilibrium constant will look like this
This means that
Rounded to two sig figs, the equilibrium concentration of ammonia will thus be