The decomposition of ammonium hydrogen sulfide: NH4HS (s) ⬄ NH3 (g) + H2S (g) is an endothermic process. A 6.1589 g sample of the solid is placed in an evacuated 4.00L vessel at 24oC. After equilibrium is established, the total pressure inside is 0.709 atm. Some of the solid remains at equilibrium.
a. Determine the Kp for the equilibrium.
b. If more of the ammonium hydrogen sulfide had been added after equilibrium was established, how would this affect the total pressure inside the vessel? Briefly explain.
Equilibrium Question?
2019-02-12 11:42 am
回答 (1)
2019-02-12 12:21 pm
a.
NH₄HS(s) ⇌ NH₃(g) + H₂S(g) …… Kp
Total pressure = P(NH₃) + P(H₂S) = 0.709 atm
But P(NH₃) = P(H₂S)
Hence, P(NH₃) = P(H₂S) = 0.709/2 atm
As NH₄HS is solid, it is not shown in the expression of Kp.
Kp = P(NH₃) × P(H₂S) = (0.709/2)² = 0.126 (to 2 sig. fig.)
b.
As ammonium hydrogen sulfide is a solid and its effective concentration (its density) is constant. The addition of ammonium hydrogen sulfide does not affect its effective concentration. Hence, the equilibrium position is unchanged and the total pressure inside the vessel is unchanged.
NH₄HS(s) ⇌ NH₃(g) + H₂S(g) …… Kp
Total pressure = P(NH₃) + P(H₂S) = 0.709 atm
But P(NH₃) = P(H₂S)
Hence, P(NH₃) = P(H₂S) = 0.709/2 atm
As NH₄HS is solid, it is not shown in the expression of Kp.
Kp = P(NH₃) × P(H₂S) = (0.709/2)² = 0.126 (to 2 sig. fig.)
b.
As ammonium hydrogen sulfide is a solid and its effective concentration (its density) is constant. The addition of ammonium hydrogen sulfide does not affect its effective concentration. Hence, the equilibrium position is unchanged and the total pressure inside the vessel is unchanged.
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