Respuesta :
Answer: The value of [tex]K_c[/tex] for the final reaction is [tex]7.16\times 10^{25}[/tex]
Explanation:
The given chemical equations follows:
Equation 1: [tex]H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1[/tex]
Equation 2: [tex]HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2[/tex]
The net equation follows:
[tex]S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c[/tex]
As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.
The value of equilibrium constant for net reaction is:
[tex]K_c=\frac{1}{K_1}\times \frac{1}{K_2}[/tex]
We are given:
[tex]K_1=9.57\times 10^{-8}[/tex]
[tex]K_2=1.46\times 10^{-19}[/tex]
Putting values in above equation, we get:
[tex]K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}[/tex]
Hence, the value of [tex]K_c[/tex] for the final reaction is [tex]7.16\times 10^{25}[/tex]
The equilibrium constant for the reaction has been [tex]7.16\;\times\;10^2^5[/tex].
The equilibrium constant has been the value of the reaction quotient when the concentration of the product and reactant has been equal, i.e. the reaction has been at equilibrium.
Equilibrium constant
The given reaction has been:
[tex]\rm H_2S\;\leftrightharpoons\;HS^-\;+\;H^+ \;\;\;\;\;\;\;\;\;K_1=9.57\;\times\;10^-^8\\ HS^-\leftrightharpoons\;S_2^-\;+\;H^+\;\;\;\;\;\;\;\;\;\;\;\;K_2=1.46\;\times\;10^-^1^9[/tex]
The final reaction to be obtained has been:
[tex]\rm S_2^-\;+\;2\;H^+\;\leftrightharpoons\;H_2S[/tex]
The final reaction has been obtained by the reverse of the given two reactions. Thus, the equilibrium constant ([tex]K_c[/tex]) has been achieved by:
[tex]K_c=\dfrac{1}{K_1} \;\times\;\dfrac{1}{K_2} [/tex]
Substituting the values:
[tex]K_c=\dfrac{1}{9.57\;\times\;10^-^8} \;\times\;\dfrac{1}{1.46\;\times\;10^-^1^9} \\ K_c=7.16\;\times\;10^2^5[/tex]
Thus, the equilibrium constant for the reaction has been [tex]7.16\;\times\;10^2^5[/tex].
Learn more about equilibrium constant, here:
https://brainly.com/question/17960050
