Respuesta :
Answer:
At equilibrium, the concentration of [tex]N_{2 (g)}[/tex] is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
[tex]N_{2 (g)}[/tex] + [tex]O_{2 (g)}[/tex] ⇄ 2[tex]NO_{(g)}[/tex]
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no [tex]N_{2 (g)}[/tex] nor [tex]O_{2 (g)}[/tex] present. Immediately, [tex]N_{2 (g)}[/tex] and[tex]O_{2 (g)}[/tex] are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: [[tex]N_{2 (g)}[/tex]]=0 ; [[tex]O_{2 (g)}[/tex] ]= 0 ; [[tex]NO_{(g)}[/tex]]=0.60M
C: [[tex]N_{2 (g)}[/tex]]=+x ; [[tex]O_{2 (g)}[/tex] ]= +x ; [[tex]NO_{(g)}[/tex]]=-2x
E: [[tex]N_{2 (g)}[/tex]]=0+x ; [[tex]O_{2 (g)}[/tex] ]= 0+x ; [[tex]NO_{(g)}[/tex]]=0.60-2x
Now we can use the constant information:
[tex]K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }[/tex]
[tex]4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}[/tex]
[tex]4.10* 10^{-4}[/tex]= [tex]\frac{(0.60-2x)^{2}}{x^{2} }[/tex]
[tex]4.10* 10^{-4} * x^{2}[/tex]= [tex](0.60-2x)^{2}}[/tex]
[tex]\sqrt{4.10* 10^{-4} * x^{2}}[/tex]= [tex]\sqrt{(0.60-2x)^{2}}}[/tex]
[tex]0.0202 x =0.60 - 2x[/tex]
[tex]2x+0.0202x=0.60[/tex]
[tex]x=\frac{0.60}{2.0202}[/tex][tex]= 0.30[/tex]
At equilibrium, the concentration of [tex]N_{2 (g)}[/tex] is going to be 0.30M
