Question

The following reaction

2H₂S(g)⇌2H₂(g)+S₂(g),K_c=1.654×10⁻⁷ at 800∘C

is carried out at the same temperature with the following initial concentrations: [H₂S]=0.132M, [H₂]=0.165 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:

Given reaction is

2H₂S(g)⇌2H₂(g)+S₂(g), K_c=1.654×10⁻⁷

Initial concentrations: [H₂S]=0.132M, [H₂]=0.165 M, and [S₂]=0.00 M

2H₂S(g) ⇌ 2H₂(g)+S₂(g)

Initial 0.132 0.165 0

Change -2x +2x +x

Equilibrium 0.132-2x 0.165+2x x

Kc=[H₂]²[S₂]/H₂S]²

(1.654×10⁻⁷) = (0.165+2x)²(x)/(0.132-2x)²

Since Kc is very small then 0.165+2x ~ 0.165 and 0.132-2x ~ 0.132.

Therefore (1.654×10⁻⁷) (0.132)²/(0.165)²=x

x=1.058×10⁻⁷

Therefore the equilibrium concentration of S2, [S2]=x=1.058×10⁻⁷ M

Since 1nM= 1 x 10^-9 M

The equilibrium concentration of S2, [S2]=105.8 nM ~ 106 nM (3 significant figures)

Please let me know if you have any doubt. Thanks.