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

b.final

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :