Question

The reaction 2 NO(g) ⇌ N2(g) + O2(g) has a value of Keq = 2400 at a temperature of 2000 K. If 0.570 mol of NO(g) is initially placed in a 3.0 L container, calculate the equilibrium concentrations of each gas

Answer 1

`[N_2]_(eq)=0.094M`

`[O_2]_(eq)=0.094M`

`[NO]_(eq)=0.002M`

Step-by-step explanation:

Hello,

In this case, for the given chemical reaction, we write the following law of mass action:

`Keq=([N_2][O_2])/([NO]^2)`

That in terms of the change
`x` due to the reaction extent (ICE procedure) we can write:

`Keq=(x*x)/(([NO]_0-2x)^2)`

Thus, the initial concentration of nitrogen monoxide is:

`[NO]_0=(0.570mol)/(3.0L) =0.19M`

Thereby, we write:

`2400=(x*x)/((0.19-2x)^2)`

That we can solve by suing the quadratic equation formula or solver to obtain two roots:

`x_1=0.094M\\x_2=0.096M`

Nevertheless, the correct answer is 0.094 M since the other root will produce a negative concentration of nitrogen monoxide at equilibrium, therefore, the equilibrium concentrations turn out:

`[N_2]_(eq)=x=0.094M`

`[O_2]_(eq)=x=0.094M`

`[NO]_(eq)=0.19M-2x=0.19M-2(0.094M)=0.002M`

Best regards.