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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?

Respuesta :

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

Solution :

First we have to calculate the concentration of NO.

[tex]\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M[/tex]

The given equilibrium reaction is,

                           [tex]N_2(g)+O_2(g)\rightleftharpoons 2NO(g)[/tex]

Initially conc.      0        0           0.1576

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

The expression of [tex]K_c[/tex] will be,

[tex]K_c=\frac{[NO]^2}{[N_2][O_2]}[/tex]

[tex]0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}[/tex]

By solving the term, we get:

[tex]x=0.0742,0.0839[/tex]

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.