Question

The equilibrium constant (Kp) for the reaction below is 70.0 at 32 °C: NH4HS(s) <---> NH3(g) + H2S(g)

A sample of NH4HS(s) is placed in an evacuated container and allowed to come to equilibrium. (You should calculate what the P is of all gases at this point) The partial pressure of H2S is then increased by the addition of 1.00 bar of H2S, and the system responds accordingly( shifts?). Calculate the partial pressure of every gas after such shift.

Answer 1

Kp=p(NH3)*p(H2S)

As we can see partial pressure will be similar for both products,

We will call it as p1.

Kp=p1*p1 = p1^2

70 = p1^2

8.366 = p1

p(NH3) = p(H2S) = 8.366

The partial pressure of H2S is then increased by the addition of 1.00 bar of H2S.

Initial pressure NH3 = Initial pressure H2S = 8.366 bar

After the addition of 1 bar of H2S the equilibrium will shift on the left

p NH3 = 8.366-x

p H2S = 9.366 –x

70 = (8.366-x)( 9.366 –x)

Solving by quadratic equations,

X = 0.484 bar

pNH3 = 8.366 – 0.484 = 7.882

pH2S = 9.366 – 0.484 = 8.882