Question

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.76×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.22×10−19, what is the equilibrium constant Kfinal for the following reaction?

Answer 1

Answer:

$K=8.40x10^{25}$

Explanation:

Hello,

In this case, the equilibrium constant is asked for the following reaction:

S2−+2H+⇌H2S

Thus, by means of the given equilibrium constants and reactions, a suitable combination results by:

1. Inverting:

H2S(aq)⇌HS−(aq)+H+(aq)

as

HS−(aq)+H+(aq) ⇌ H2S(aq)

2. Inverting:

HS−(aq)⇌S2−(aq)+H+(aq)

as

S2−(aq)+H+(aq)⇌HS−(aq)

Thus, by adding them we obtain:

HS−(aq)+H+(aq)+S2−(aq)+H+(aq)⇌HS−(aq)+H2S(aq)

Resulting in:

S2−(aq)+2H+(aq)⇌H2S(aq)

Which is simplified but proposing it law of mass action it turns out:

$K=\frac{[H_2S]}{[HS^-][H^+]} *\frac{[HS^-]}{[S^{2-}][H^+]} = \frac{1}{K_1}*\frac{1}{K_2}=\frac{1}{9.76x10^{-8}} *\frac{1}{1.22x10^{-19}} =8.40x10^{25}$

Best regards.

Answer 2

Answer:

Kf  = 8.4 x 10²⁶

Explanation:

H₂S(aq) ⇌ HS⁻(aq) + H⁺(aq)

K1 = [HS⁻] [H⁺] / [H₂S]

K1 = 9.76×10−8

9.76×10−8 =  [HS⁻] [H⁺] / [H₂S]

HS⁻(aq) ⇌ S²⁻(aq) + H⁺(aq)

K2 = 1.22×10−19

K2 = [S²⁻] [H⁺] / [HS⁻]

S²⁻ (aq)  + 2H⁺(aq)  ⇌ H₂S(aq)

Kf = [H₂S] /[S²⁻] [H⁺]²

To obtain the value of Kf, the relationship between Kf, k1 and k2 have to be determined;

K1K2 = ( [HS⁻] [H⁺] / [H₂S]) * [S²⁻] [H⁺] / [HS⁻]

K1K2 = [H⁺]² [S²⁻] / [H₂S]

K1K2 = 1 / Kf

Kf = 1  / (K1K2)

Kf = 1 / (9.76×10⁻⁸ * 1.22×10⁻¹⁹)

Kf = 1 / (11.91 x 10-27)

Kf = 0.084 x 10²⁸

Kf  = 8.4 x 10²⁶