At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel at 800 K initially contains 3.00 atm of H2S. If the reaction is allowed to equilibrate, what is the equilibrium pressure of S2?

Question

contestada

Respuesta :

jufsanabriasa jufsanabriasa · Answer 1 · 2020-04-09T03:45:02+02:00

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

[tex]Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2[/tex]

Where P is the pressure of each compound in equilibrium.

If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:

H2S = 3.00 atm - 2X

H2 = 2X

S2: = X

Replacing:

[tex]3.2x10^{-7} = (2X)^2X / (3-2X)^2[/tex]

[tex]3.2x10^{-7} = 4X^3 / 9- 6X+4X^2[/tex]

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Solving for X:

X = 0.008945 atm

As in equilibrium, pressure of S2 is X, pressure is 0.008945 atm

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-12-02T15:09:11+01:00

The pressure of the H2S at equilibrium is 0.0089 atm.

The ICE table is set up as follows;

H2S(g) ⇌ 2 H2(g) + S2(g)

I 3 0 0

C -x + x +x

E 3 - x x x

Given that;

Kp = P.H2^2 × P.S2/P.H2S^2

Kp = 3.2 × 10–7

3.2 × 10–7 =4 x^3/ (3 - x)^2

3.2 × 10–7 = 4x^3/9 - 6x + x^2

4x³ - 1.28x10⁻⁶x² + 1.92x10⁻⁶x - 2.88x10⁻⁶ = 0

x = 0.0089 atm

Learn more: https://brainly.com/question/6284546