At a certain temperature, the Kp for the decomposition of H2S is 0.814 atm. EASY 10 PTS?

2019-03-13 5:04 am

H2S (g) yields H2(g) + S(g)

Initially, only H2S is present at a pressure of 0.271 atm in a closed container. What is the total pressure in the container at equilibrium?

I am having trouble setting up the first part

I set it up as 0.814=(x)(x)/0.271-x and I know I have to rearrange to solve it as a quadratic, but how do I get c for the formula?

Thanks

回答 (1)

胡雪8°

2019-03-13 6:47 am

✔ 最佳答案

________________ H₂S(g) ____ ⇌ ____ H₂(g) ____ + ____ S(g) ____ Kp = 0.814 atm
Initial: _________ 0.271 atm ________ 0 atm _________ 0 atm
Change: _________ - y atm ________ +y atm _________ +y atm
Equilibrium: __ (0.271 - y) atm ______ y atm __________ y atm

At equilibrium: Kp = P(H₂) P(S) / P(H₂S)
0.814 = y² / (0.271 - y)
y² + 0.814y - 0.220594 = 0
y = [-0.814 ± √(0.814² + 4×0.220594)] / 2
y = 0.214 or y = -1.03 (rejected)

Total pressure at equilibrium
= P(H₂S) + P(H₂) + P(S)
= [(0.271 - y) + y + y] atm
= (0.271 + y) atm
= (0.271 + 0.214) atm
= 0.485 atm

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