Question

In: Chemistry

At a certain temperature, the Kp for the decompositon of H2S is 0.871. H2S(g) <----> H2(g)...

At a certain temperature, the Kp for the decompositon of H2S is 0.871. H2S(g) <----> H2(g) + S(g) Initally, only H2S is present at a pressure of 0.226 bar in a closed container. What is the total pressure in the container at equilibrium?

Expert Solution

Solution

Write reaction and using ICE find final equilibrium pressures

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

I 0.226 0 0

C -x + x +x

E (0.226-x) x x

Kp = P (H2) P (S) / P(H2S)

0.871 = x²/ (0.226-x)

Lets solve quadratic equation.

(0.226-x) * 0.871 = x ²

x = 0.1862

Equilibrium pressure of H2S = 0.226 – x = 0.226 - 0.1862 = 0.0398 bar

Equilibrium pressure of H2 = x = 0.1862 bar

Equilibrium pressure of S = x = 0.1862 bar

Total pressure = 0.0398 + 0.1862 + 0.1862

= 0.412 bar

Total pressure is 0.412 bar

queen_honey_blossom answered 2 years ago

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