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At a certain temperature, the K p for the decomposition of H 2 S is 0.739 . H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g ) Initially, only H 2 S is present at a pressure of 0.215 atm in a closed container. What is the total pressure in the container at equilibrium?

Respuesta :

Answer:

The total pressure in the container is 0.389 atm

Explanation:

Step 1: Data given

Kp = 0.739

The initial pressure of H2S = 0.215 atm

Step 2: The balanced equation

H2S(g) ⇆ H2(g) + S(g)

Step 3: The initial pressures

pH2S = 0.215 atm

pH2 = 0 atm

pS = 0 atm

Step 4: The pressures at the equilibrium

pH2S = 0.215 - X atm

pH2 = X atm

pS = X atm

Step 5:

Kp = 0.739 = (pS)*(pH2) / (pH2S)

0.739 = X*X / (0.215 - X)

0.739 = X² / (0.215 - X)

X² = 0.739*(0.215-X)

X² = 0.1589 - 0.739X

X² +0.739X - 0.1589 = 0

X = 0.174

pH2S = 0.215 - 0.174 atm = 0.041 atm

pH2 = 0.174 atm

pS = 0.174 atm

Step 6: Calculate the total pressure in the container

Total pressure = 0.041 atm + 0.174 atm + 0.174 atm

Total pressure = 0.389 atm

The total pressure in the container is 0.389 atm

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