Question

At 22°C, Kp = 0.070 for the equilibrium:

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

a) A sample of solid NH4HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (atm) of ammonia, assuming that some solid NH4HS remains.

b) After the equilibrium in part a) was established, the reaction flask is then charged with an additional 0.59 atm of NH3. When the system has equilibrated, what is the partial pressure of hydrogen sulfide?

Answer 1

a)

NH4HS --> NH3 + H2S

Kp = P-NH3 * P-H2S

Kp = 0.07

in equilibrium:

0.07 =  P-NH3 * P-H2S

assume 1 mol of NH3 = 1 mol of H2S so

P = sqrt(0.07 ) = 0.26457

so

P-NH3 = 0.26457 atm

P-H2S = 0.26457 atm

b)

extra 0.59 atm of NH3

initially

P-H2S = 0.26457 atm

P-NH3 = 0.26457+ 0.59 = 0.85457 atm

in equilibrium

P-H2S = 0.26457 - x

P-NH3 = 0.85457 - x

substitute in Kp

Kp = P-NH3 * P-H2S

0.07 = (0.26457 - x)( 0.85457 - x)

0.07 = 0.26457*0.85457 - (0.85457 +0.26457 )x + x^2

x^2 - 1.11914x + 0.26457*0.85457-0.07 = 0

x^2 - 1.1191x + 0.15609 = 0

x = 0.1633

P-H2S = 0.26457 - 0.1633 = 0.10127 atm

P-NH3 = 0.85457 - 0.1633 = 0.69127 atm