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?
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Ответ:
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
Ответ: