At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a

Question

8 days ago

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t 800 K initially contains 3.00 atm of H2S. If the reaction is allowed to equilibrate, what is the equilibrium pressure of S2?

1 answer:

Tems11 [2.7K] · Answer 1 · 2026-04-08T11:32:21+03:00

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

$Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2$

Where P represents the pressure of each component at equilibrium.

Starting with an initial pressure of H2S at 3.00 atm, the equilibrium concentrations are:

H2S = 3.00 atm - 2X

H2 = 2X

S2 = X

Substituting these values into the equation gives:

$3.2x10^{-7} = (2X)^2X / (3-2X)^2$

$3.2x10^{-7} = 4X^3 / 9- 6X+4X^2$

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Calculating X yields:

X = 0.008945 atm

In equilibrium, the pressure of S2 is X, so the pressure stands at 0.008945 atm