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?

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