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At a certain temperature, the Kp for the decomposition of H2S is 0.759. H2S (g) <----->...

At a certain temperature, the Kp for the decomposition of H2S is 0.759.

H2S (g) <-----> H2(g) + S(g)

Initially, only H2S is present at a pressure of 0.164 atm in a closed container. What is the total pressure(atm) in the container at equilibrium?

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Answer #1

                         H2S (g) <-----> H2(g) + S(g)

Initial                0.164                0         0

Final               (0.164-x)             x         x

Pressure in the chamber after equilibrium = 0.164 + x = 0.164 + 0.1385 = 0.3025 atm

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