Question

The squares in the equation are equilibrium signs.

The equilibrium constant, Ky, for the following reaction is 0.110 at 298 K: NH_HS(s) NH3(g) + H2S(g) Calculate the equilibrium partial pressure of H2S when 0.416 moles of NH_HS(s) is introduced into a 1.00 L vessel at 298 K. Phys= c atm

Answer 1

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

I         y                                   0          0

C     -x                                    +x         +x

E     y-x                                   +x         +x

at equilibrium PNH3 = PH2S = x

Kp   = [NH3][H2S]

0.11    = x*x

x^2 = 0.11

x = 0.3316

PH2S   = x   = 0.3316atm

------ PCl5(g) -----------> PCl3(g)    + Cl2(g)

I ----- 1.42   -------------- 0 --------    0

C-----   -x   --------------- +x ----------- +x

E -----1.42-x ------------- +x --------- +x

            Kp   = PPCl3*PCl2/PPCl5

           0.497   = x*x/(1.42-x)

          0.497*(1.42-x) = x^2

       x = 0.63

     PPCl5 = 1.42-x   = 1.42-0.63   = 0.79atm

     PPCl3   = 0.63atm

     PCl2      = 0.63atm