Question

Hydrogen sulfide decomposes according to the following reaction: 2H2S(g) ⇋ 2H2(g) + S2(g) Kc=9.30x10-8 at 700.°C.If 0.45 mol of H2S is placed in a 3.0 L container, what is the equilibrium concentration of H2(g) at 700.°C?

Answer 1

Answer:

[H₂] = 1.61x10⁻³ M

Explanation:

2H₂S(g) ⇋ 2H₂(g) + S₂(g)

Kc = 9.30x10⁻⁸ = $\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}$

First we calculate the initial concentration:

0.45 molH₂S / 3.0L = 0.15 M

The concentrations at equilibrium would be:

[H₂S] = 0.15 - 2x

[H₂] = 2x

[S₂] = x

We put the data in the Kc expression and solve for x:

$\frac{(2x^2) * x}{(0.15-2x)^2}=9.30x10^{-8}$

$\frac{4x^3}{0.0225-4x^2}=9.30*10^{-8}$

We make a simplification because x<<< 0.0225:

$\frac{4x^3}{0.0225} =9.30*10^{-8}$

x = 8.058x10⁻⁴

[H₂] = 2*x = 1.61x10⁻³ M

Answer 2

Answer:

Equilibrium concentration of H2(g)

= 0.0016mol/l

Explanation:

For a gas equilibrium constant, kc;

Kc = [product]/[reactant] all in equilibrium gaseous state.

2H2S(g) --> 2H2(g) + S2(g)

Concentration of the reactant, H2S = 0.45/3 = 0.15mol/l

Initial Concentrations;

H2S = 0.15mol/l

H2 = 0

S2 = 0

Change in Concentrations;

H2S = -2y

H2 = 2y

S2 = y

Kc = 9.30 x 10^-8

Concentrations at equilibrium;

H2S = 0.15 - 2y

H2 = 2y

S2 = y

Inputting the Concentrations at equilibrium into the gas equilibrium constant;

(2y)^2* y/(0.15 - 2y)^2 = 9.3 x 10^-8

Solving for y;

y = 0.0008mol/l

Equilibrium concentration of H2(g) = 2y

= 2 * 0.0080mol/l

= 0.0016mol/l