At a certain temperature,the Kp for the decomposition of H2S is 0.784 H2S(g)-->H2(g)+S(g)?
2016-02-29 9:24 pm
Initially,only H2S is present at a pressure of 0.170 atm in a closed container. What is the total pressure in the container at equilibrium?
回答 (1)
2016-02-29 9:37 pm
H₂S(g) ⇌ H₂(g) + S(g) ...... Kp = 0.784
Initial :
P(H₂S) = 0.170 atm
P(H₂) = P(S) = 0 atm
At eqm :
P(H₂) = P(S) = y atm
P(H₂S) = (0.170 - y) atm
Kp = P(H₂) × P(S) / P(H₂S)
y² / (0.170 - y) = 0.784
y² + 0.784y - 0.13328 = 0
y = 0.144 or y = -0.928 (rejected)
Total pressure = (0.170 - 0.144) + 0.144 + 0.144 = 0.314 atm
Initial :
P(H₂S) = 0.170 atm
P(H₂) = P(S) = 0 atm
At eqm :
P(H₂) = P(S) = y atm
P(H₂S) = (0.170 - y) atm
Kp = P(H₂) × P(S) / P(H₂S)
y² / (0.170 - y) = 0.784
y² + 0.784y - 0.13328 = 0
y = 0.144 or y = -0.928 (rejected)
Total pressure = (0.170 - 0.144) + 0.144 + 0.144 = 0.314 atm
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160229132400AA21Ler