Question

The equilibrium constant, K, for the following reaction is 1.80×10-4 at 298 K. NH4HS(s) NH3(g) + H2S(g) An equilibrium mixture of the solid and the two gases in a 1.00 L flask at 298 K contains 0.231 mol NH4HS, 1.34×10-2 M NH3 and 1.34×10-2 M H2S. If the concentration of H2S(g) is suddenly increased to 2.08×10-2 M, what will be the concentrations of the two gases once equilibrium has been reestablished? [NH3] = M [H2S] = M please finish it to the end!

Answer 1

              NH4HS(s) ---> NH3(g) + H2S(g)

first equil    0.231 M      0.0134 M 0.0134 M

added                                  0.0208 M
                           
change           +x            -x      -x

re established 0.231+x        0.0134-x 0.0208-x

   K = [H2S][NH3]/[NH4HS]

(1.80*10^-4) = (0.0134-X)(0.0208-X)/(0.231+X)

X = 0.0096

[NH4SH] = 0.231+0.0096 = 0.2406 M

[NH3] =0.134-0.0096 = 0.1244 M

[H2S] = 0.0208-0.0096 = 0.0112 M