Question

Consider the reaction for the decomposition of hydrogen disulfide: 2H2S(g)⇌2H2(g)+S2(g), Kc = 1.67×10−7 at 800∘C A 0.500 L reaction vessel initially contains 0.163 mol of H2S and 5.00×10−2 mol of H2 at 800∘C. Find the equilibrium concentration of [S2].

Answer 1

Answer:

Molar concentration of S₂ is 1.77×10⁻⁶M

Explanation:

For the reaction:

2H₂S(g) ⇄ 2H₂(g) + S₂(g)

The equilibirum constant, K, is defined as:

$K = \frac{[S_2][H_2]^2}{[H_2S]^2}$(1)

Concentrations in equilibirum are:

[H₂S] : 0,163/0.500L - X

[H₂] : 0,0500/0.500L + X

[S₂] : X

Replacing the concentrations and the equilibrium value in (1):

$K = \frac{[X][0.1+X]^2}{[0326-X]^2}$

1.67x10⁻⁷ = X (X² + 0.2X + 0.01) / (X² -0.652X + 0.106)

1.67x10⁻⁷X² - 1.09x10⁻⁷X + 1.77x10⁻⁸ = X³ + 0.2X² + 0.01X

0 =  X³ + 0.2X² + 0.01X - 1.77x10⁻⁸

Solving for X:

X = 1.77×10⁻⁶

As [S₂] = X, molar concentration of S₂ is 1.77×10⁻⁶M

I hope it helps!