At a certain temperature, the K p for the decomposition of H 2 S is 0.859 . H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g ) Initially, only H 2 S is present at a pressure of 0.245 atm in a closed container. What is the total pressure in the container at equilibrium?

P = 0.444 atm

Explanation:

The reaction of decomposition of H₂S is:

                    H₂S(g)   ⇄  H₂(g)  + S(g)      

Initial            0.245          0           0

Equilibrium  0.245-x       x            x

The equilibrium constant (Kp) of the above reaction is the following:

    (1)

Knowing that initally, only H₂S is present at P = 0.245, the Kp of equation (1) at equilibrium is:

Using the quadratic formula we get x₁ = 0.199 and x₂ = -1.058.

Taking only the possitive number we have that:

Therefore, the total pressure in the container at equilibrium is:

I hope it helps you!

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