At a certain temperature, the kp for the decomposition of h2s is 0.795.
h2s (g) > h2(g) + s(g)
initially, only h2s is present at a pressure of 0.203 atm in a closed container. what is the total pressure in the container at equilibrium?

The total pressure in the container at equilibrium is 0.37065 atm

Explanation:

Step 1: Data given

Kp for the decomposition of H2S is 0.795

Only H2S is present at a pressure of 0.203 atm

Step 2: The balanced equation

H2S (g) → H2(g) + S(g)

Step 3: Calculate partial pressure of H2 and S

Initially, there is no H2 and S, they have no pressure

There will react X

The pressure at the equilibrium we have:

H2: X

S: X

H2S: 0.203 - X

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

0.795 = (x)(x)/(0.203 - X)

X = 0.16765

Step 4: Calculate total pressure at equilibrium:

Total pressure = the partial pressure of H2 + partial pressure of S + partial pressure of H2S

P = X + X + (0.203 - X)

P = 0.16765 + 0.16765 + (0.203 - 0.16765)

P =  0.37065 atm

The total pressure in the container at equilibrium is 0.37065 atm

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