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?

The equilibrium concentration of hydrogen gas is 0.0010 M.

Explanation:

The equilibrium constant of the reaction = }

Moles of hydrogen sulfide = 0.31 mol

Volume of the container = 4.1 L

Initially

0.076 M

At equilibrium

(0.076-2x)                         2x     x

The expression of an equilibrium constant :

Solving for x:

x = 0.00051

The equilibrium concentration of hydrogen gas:

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b