Question

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, the reaction below can proceed to a measurable extent. N2(g) + O2(g) ⇔ 2 NO(g) At 3000 K, the reaction above has Keq = 0.0153. If 0.3152 mol of pure NO is injected into an evacuated 2.0-L container and heated to 3000K, what will be the equilibrium concentration of NO?

Answer 1

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

`\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=(0.3152mol)/(2.0L)=0.1576M`

The given equilibrium reaction is,

`N_2(g)+O_2(g)\rightleftharpoons 2NO(g)`

Initially conc. 0 0 0.1576

At eqm. (x) (x) (0.1576-2x)

The expression of
`K_c` will be,

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

`0.0153=((0.1576-2x)^2)/((x)* (x))`

By solving the term, we get:

`x=0.0742,0.0839`

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.