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2H2S(g)⇌2H2(g)+S2(g),Kc=1.67×10−7 at 800∘C is carried out at 800 ∘C with the following initial concentrations: [H2S]=0.100M, [H2]=0.100M, and [S2]=0.00 M. Find the equilibrium concentration of S2.

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Answer:

Equilibrium concentration of [S₂] is 1.67x10⁻⁷M

Explanation:

Based in the reaction:

2H₂S(g ) ⇌ 2H₂(g) + S₂(g)

Kc = 1.67x10⁻⁷ = [S₂] [H₂]² / [H₂S]²

The reaction is in equilibrium when [S₂] [H₂]² / [H₂S]² = Kc

With initial concentrations, the equilibrium concentrations must be:

[H₂S] = 0.100M - 2X

[S₂] = X

[H₂] = 0.100M + 2X

Replacing these values in Kc:

1.67x10⁻⁷ = [X] [0.100M + 2X]² / [0.100M - 2X]²

1.67x10⁻⁷ = 4X³ + 0.4X² + 0.01X / 4X² - 0.4X + 0.01

6.68x10⁻⁷X² - 6.68x10⁻⁸X + 1.67x10⁻⁹ = 4X³ + 0.4X² + 0.01X

0 =  4X³ + 0.4X² + 0.01X - 1.67x10⁻⁹

Solving for X:

X = 1.67x10⁻⁷M

As [S₂] = X, equilibrium concentration of [S₂] is 1.67x10⁻⁷M

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