Respuesta :
Answer : The value of [tex]K_c[/tex] for the final reaction is, 184.09
Explanation :
The equilibrium reactions in aqueous solution are :
(1) [tex]HNO_2(aq)\rightleftharpoons H^+(aq)+NO_2^-(aq)[/tex] [tex]K_{c_1}=4.5\times 10^{-4}[/tex]
(2) [tex]H_2SO_3(aq)\rightleftharpoons 2H^+(aq)+SO_3^{2-}(aq)[/tex] [tex]K_{c_2}=1.1\times 10^{-9}[/tex]
The final equilibrium reaction is :
[tex]2HNO_2(aq)+SO_3^{2-}(aq)\rightleftharpoons H_2SO_3(aq)+2NO_2^-(aq)[/tex] [tex]K_{c}=?[/tex]
Now we have to calculate the value of [tex]K_c[/tex] for the final reaction.
Now equation 1 is multiply by 2 and reverse the equation 2, we get the value of final equilibrium reaction and the expression of final equilibrium constant is:
[tex]K_c=\frac{(K_{c_1})^2}{K_{c_2}}[/tex]
Now put all the given values in this expression, we get :
[tex]K_c=\frac{(4.5\times 10^{-4})^2}{1.1\times 10^{-9}}=184.09[/tex]
Therefore, the value of [tex]K_c[/tex] for the final reaction is, 184.09
The value of Kc for the given reaction is 184.09.
What is equilibrium constant kc?
The equilibrium constant kc is the ratio between the concentration of reactant and product.
Given,
[tex]\bold{HNO_2(aq) = H+(aq) + NO_2^-(aq)}[/tex]
[tex]\bold{H_2SO_3(aq) = 2H+(aq) + SO_3^2^-(aq)}[/tex]
[tex]\bold{Kc1 = 4.5 \times10^-^4}[/tex]
[tex]\bold{Kc2 = 1.1 \times 10^-^9 }[/tex]
Equilibrium reaction:
[tex]\bold{ 2HNO_2(aq) + SO_3^2^-(aq) = H_2SO_3(aq) + 2NO_2^-(aq) }[/tex]
Now from the equilibrium constant kc:
[tex]\bold{Kc = \dfrac{(kc_1)^2}{kc_2} }[/tex]
[tex]\bold{Kc = \dfrac{(\bold{ 4.5 \times10^-^4})^2}{\bold{1.1 \times 10^-^9 }} =184.09 }[/tex]
Thus, the equilibrium constant kc for the reaction is 184.09.
Learn more about equilibrium constant, here:
https://brainly.com/the /17960050
