What are the partial pressures of NH3 and H2S at equilibrium, that is, what are the values of PNH3 and PH2S, respectively?Enter the partial pressure of ammonia followed by the partial pressure of hydrogen sulfide numerically in atmospheres separated by a comma.

Ammonium bisulfide, NH4HS , forms ammonia, NH3 , and hydrogen sulfide, H2S , through the reaction NH4HS(s)⇄NH3(g)+H2S(g) This reaction has a Kp value of 0.120 at 25°C .

An empty 5.00-L flask is charged with 0.300 g of pure H2S(g) , at 25°C

Partial pressure of NH3 = 0.325 atm

Partial pressure of H2S  = 0.368 atm

Explanation:

Temperature = 25 °C = 298 K

Volume = 5 L

Mass of H2S = 0.30 g

Moles of H2S = 0.30 / 34.08

= 0.0088

Using PV = nRT

Initial pressure of H2S = 0.0088 * 0.0821 * 298 / 5

= 0.043 atm

Kp = 0.120

(a). NH4HS(s) ⇄ NH3(g) + H2S(g)

Initial - 0.0 0.043

Change - + x + x

Equilibrium - x 0.043 + x

Kp = P(NH3) * P(H2S)

0.120 = x (0.043 + x)

x = 0.325 atm

Hence, at equilibrium:

Partial pressure of NH3 = 0.325 atm

Partial pressure of H2S = 0.043 + 0.325 = 0.368 atm

fe 55.85 g/mol and h2o 18.02 g/mol.

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.