Respuesta :
Answer:
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
The pressures of NH3.
As per the question, the partial pressure of the NH3 and H2S at the equilibrium is of the value PNH3 and PH2S. As per the equilibrium of the partial pressure of ammonia that is followed by the partial pressure in hydrogen sulfate numerically the air is separated. The amount will be as:
Answer pressure of NH3 is equal to 0.325 atm and the pressure of H2S equal to 0.368 atm.
As per the ammonium bisulfite, the NH4HS forms ammonia, and the hydrogen sulfide by the reaction NH4HS(s)⇄NH3(g)+H2S(g)
This reaction if the value of K p-value is 0.120 at 25°C. The partial pressure value of NH3 is 0.325 with that of H2S of 0.368 atm.
Find more information about the pressure of NH3.
brainly.com/question/6960398.
