Respuesta :
Explanation:
For reaction (1). [tex]N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)[/tex]
So, expression for [tex]K_{C1}[/tex] of this reaction is as follows.
[tex]K_{C1}[/tex] = [tex]\frac{[NO]^{2}}{[N_{2}][O_{2}]}[/tex]
= [tex]2.5 \times 10^{-31}[/tex]
For reaction (2). [tex]N_{2}(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons N_{2}O(g)[/tex]
So, expression for [tex]K_{C2}[/tex] of this reaction is as follows.
[tex]K_{C2}[/tex] = [tex]\frac{[N_{2}O]}{[N_{2}][O_{2}]^{\frac{1}{2}}}[/tex]
= [tex]3.31 \times 10^{-24}[/tex]
For reaction (3). [tex]N_{2}O(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons 2NO(g)[/tex]
So, expression for [tex]K_{C3}[/tex] of this reaction is as follows.
[tex]K_{C3}[/tex] = [tex]\frac{[NO]^{2}}{[N_{2}O][O_{2}]^{\frac{1}{2}}}[/tex]
= [tex]\frac{[NO]^{2}}{[N_{2}][O_{2}]} \times \frac{[N_{2}][O_{2}]}{[N_{2}O][O_{2}]^{\frac{1}{2}}}[/tex]
= [tex]K_{C1} \times \frac{1}{K_{C2}}[/tex]
= [tex][tex]2.5 \times 10^{-31} \times \frac{1}{3.31 \times 10^{-24}}[/tex]
= [tex]7.55 \times 10^{-8}[/tex]
Thus, we can conclude that equilibrium constant for reaction (3) is [tex]7.55 \times 10^{-8}[/tex].
Equilibrium reactions are the depiction of the reactants and the products that do not undergo any changes. [tex]7.55\times 10^{-8}[/tex] is the equilibrium constant for (3).
What is equilibrium constant?
In a chemical reaction, the equilibrium constant is the depiction of the reaction quotient when the system is at an equilibrium chemical state.
The first reaction is shown as:
[tex]\rm N_{2}(g) + O_{2}(g)\rightleftharpoons 2 NO(g)[/tex]
The equilibrium constant for the first reaction can be shown as,
[tex]\begin{aligned}\rm K_{c1} &= \rm \dfrac{[NO]^{2}}{[N_{2}][O_{2}]}\\\\&= 2.5\times 10^{-31}\end{aligned}[/tex]
The second reaction is shown as:
[tex]\rm N_{2}(g) + \dfrac{1}{2} O_{2}(g)\rightleftharpoons N_{2}O(g)[/tex]
The equilibrium constant for the second reaction can be shown as,
[tex]\begin{aligned}\rm K_{c2} &= \rm\dfrac{[N_{2}O]}{[N_{2}][O_{2}]^{\frac{1}{2}}}\\\\&= 3.31 \times 10^{-24}\end{aligned}[/tex]
The third reaction is shown as:
[tex]\rm N_{2}O(g) + \dfrac{1}{2} O_{2}(g)\rightleftharpoons 2 NO(g)[/tex]
The equilibrium constant for the third reaction can be shown as,
[tex]\begin{aligned}\rm K_{c3} &= \rm \dfrac{[NO]^{2}}{[N_{2}O][O_{2}]^{\frac{1}{2}}}\\\\&= \rm K_{c1} \times \dfrac{1}{K_{c2}}\\\\&= 7.55\times 10^{-8}\end{aligned}[/tex]
Therefore, [tex]7.55\times 10^{-8}[/tex] is the equilibrium constant for the third reaction.
Learn more about the equilibrium constant here:
https://brainly.com/question/13601814
