Question

If the concentration equilibrium constant for this reaction is Kc = 3.36 at reaction conditions, what will the equilibrium concentration of NO(g) be if a sealed reaction vessel is filled with gases at initial concentrations of 0.0 M of NO3(g), 1.8 M of NO(g), and 2.8 M of NO2(g)?

Answer 1

`[NO]_(eq)=0.329M`

Step-by-step explanation:

Hello,

The balanced chemical reaction must be stated as follows:

`NO(g)+NO_2(g)-->N_2O_3(g)`

Thus, the equilibrium constant, thanks to the law of mass action turns out being:

`Kc=([N_2O_3]_(eq))/([NO]_(eq)[NO_2]_(eq))`

Based on the equilibrium condition, one leave the law of mass action in terms of the change
`x` due to the chemical reaction:

`Kc=(x)/((1.8M-x)(2.8M-x))\\(5.04-4.6x+x^2)*3.36=x\\16.9344-15.456x-x+3.36x^2=0\\3.36x^2-16.456x+16.9344=0\\x_1=1.471M\\x_2=3.427M`

The adequate result is 1.471M since the other one turns out into a negative molarity for NO.

Finally, one conclude that the equilibrium concentration of NO is:

`[NO]_(eq)=1.8M-1.471M=0.329M`

Best regards.