Respuesta :
Answer:
Therefore the equilibrium constant [tex]K_{final}[/tex] is 9.35× 10²⁵
Explanation:
Equilibrium constant:
Equilibrium constant is used to find out the ratio of the concentration of the product to that of reactant.
xA+yB→zC
The equilibrium constant K,
[tex]k=\frac{[C]^z}{[A]^x[B]^y}[/tex]
Here [A] is equilibrium concentration of A
[B] is equilibrium concentration of B
[C] is equilibrium concentration of C.
1. [tex]H_2S(aq)\rightleftharpoons HS^-(aq)+H^+(aq)[/tex] [tex]K_1= 9.06 \times 10^{-8}[/tex]
[tex]K_1=\frac{[HS^-][H^+]}{[H_2S]}[/tex]
2. [tex]HS^-(aq)\rightleftharpoons S^{2-}(aq)+H^+(aq)[/tex] [tex]K_2= 1.18 \times 10^{-19}[/tex]
[tex]K_2=\frac{[S^{2-}][H^+]}{[HS^-]}[/tex]
3.
[tex]S^{2-}(aq)+2H^+(aq)\rightleftharpoons H_2S(aq)[/tex]
[tex]K_3=\frac{[H_2S]}{[S^{2-}][H^+]^2}[/tex]
Therefore,
[tex]k_1k_2=\frac{[HS^-][H^+]}{[H_2S]}\frac{[S^{2-}][H^+]}{[HS^-]}[/tex]
[tex]\Rightarrow k_1k_2=\frac{[S^{2-}][H^+]^2}{[H_2S]}[/tex]
[tex]\Rightarrow k_1k_2=\frac{1}{K_3}[/tex]
[tex]\Rightarrow K_3=\frac{1}{K_1K_2}[/tex]
[tex]\Rightarrow K_3=\frac{1}{9.06\times 10^{-8}\times 1.18\times 10^{-19}}[/tex]
⇒K₃=9.35× 10²⁵
[tex]K_3=K_{final}[/tex]
Therefore the equilibrium constant [tex]K_{final}[/tex] is 9.35× 10²⁵
