Question

Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30 10-8 at 700°C. 2 H2S(g) 2 H2(g) + S2(g) If 0.31 mol H2S is placed in a 4.1 L container, what is the equilibrium concentration of H2(g) at 700°C?

Answer 1

Answer:

The equilibrium concentration of hydrogen gas is 0.0010 M.

Explanation:

The equilibrium constant of the reaction = $K_c=9.30\times 10^{-8$}

Moles of hydrogen sulfide = 0.31 mol

Volume of the container = 4.1 L

$[concentration]=\frac{moles}{volume (L)}$

$[H_2S]=\frac{0.31 mol}{4.1 L}=0.076 M$

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

Initially

0.076 M

At equilibrium

(0.076-2x)                         2x     x

The expression of an equilibrium constant :

$K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

$9.30\times 10^{-8}=\frac{(2x)^2\times x}{(0.076-x)^2}$

Solving for x:

x = 0.00051

The equilibrium concentration of hydrogen gas:

$[H_2]=2x=2\times 0.00051 M=0.0010 M$