Question

Question 6 1.75 pts The following reaction 2H2S(g)=2H2(g)+S2(g), Kc=1.625x10-7 at 800°C is carried out at the same temperature with the following initial concentrations: [H,S]=0.162M, [H2]=0.184 M, and [S2]=0.00 M. Find the equilibrium concentration of S2. Express the molarity to three significant figures. Answer in units of nM.

Answer 1

Answer:

[S₂] = 1.27×10⁻⁷ M

Explanation:

2 H₂S(g) ⇄ 2 H₂(g) + S₂(g), Kc=1,625x10⁻⁷

The equation of this reaction is:

1,625x10⁻⁷ = $\frac{[H_2]^2[S_2]}{[H_{2}S]^2}$

The equilibrium concentrations are:

[H₂S] = 0,162 - 2x

[H₂] = 0,184 + 2x

[S₂] = x

Replacing:

1,625x10⁻⁷ = $\frac{[0,184+2x]^2[x]}{[0,162-2x]^2}$

Solving:

4x³ + 0,736x² + 0,033856x - 4,3x10⁻⁹

x = 1.27×10⁻⁷

Thus, concentration of S₂ is:

[S₂] = 1.27×10⁻⁷ M