Respuesta :
Answer:
At equilibrium, the concentration of [tex]NO_{3 (g)}[/tex] is 1.60 M
Explanation:
Initial values
We have the balanced chemical reaction is:
[tex]NO_{3 (g)}[/tex] + [tex]NO_{(g)}[/tex] ⇄2[tex]NO_{2(g)}[/tex]
[tex]NO_{3 (g)}[/tex] :
[tex]\frac{1.6 mol}{0.5000L} = 3.2 M[/tex]
[tex]NO_{(g)}[/tex] :
[tex]\frac{1.2 mol}{0.5000L} = 2.4 M[/tex]
Equilibrium values
[tex]NO_{3 (g)}[/tex] and [tex]NO_{(g)}[/tex] are going to consume, while [tex]NO_{2(g)}[/tex] is going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: [[tex]NO_{3 (g)}[/tex]]=3.2M ; [[tex]NO_{(g)}[/tex]]= 2.4M ; [[tex]NO_{2(g)}[/tex]]=0
C: [[tex]NO_{3 (g)}[/tex]]=-x ; [[tex]NO_{(g)}[/tex]]= -x ; [[tex]NO_{2(g)}[/tex]]=+2x
E: [[tex]NO_{3 (g)}[/tex]]=3.2-x ; [[tex]NO_{(g)}[/tex]]= 2.4-x ; [[tex]NO_{2(g)}[/tex]]=0+2x
Now we can use the constant information:
K=8.01
[tex]8.01=\frac{(2x)^2}{(3.2-x)(2.4-x)}[/tex]
Solving for x we have 1.6 and 9.6 as possible results. But 9.6 has no sense because the concentration left can not be negative. So x=1.6
It means that at equilibrium, the concentration of [tex]NO_{3 (g)}[/tex] is going to be 1.6 M
