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The reaction below is carried out at a different temperature at which Kc=0.055. This time, however, the reaction mixture starts with only the product, [NO]=0.0100M, and no reactants. Find the equilibrium concentrations of N2, O2, and NO at equilibrium.

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The reaction below is carried out at a different temperature at which Kc = 0.055. This time, however, the reaction mixture starts with only the product, [NO] = 0.0100 M, and no reactants. Find the equilibrium concentrations of N2, O2, and NO at equilibrium. The equation is N2(g) + O2(g) <--> 2NO(g)

Answer: Concentration of [tex]NO[/tex] at equilibrium = 0.001 M

Concentration of [tex]N_2[/tex] = 0.0045 M

Concentration of [tex]O_2[/tex]  = 0.0045 M

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stoichiometric coefficients.

Initial concentration of [tex]NO[/tex] = 0.0100 M

The given balanced equilibrium reaction is,

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

Initial conc.            0 M           0 M         0.0100 M

At eqm. conc.     (x) M          (x) M       (0.0100-2x) M

The expression for equilibrium constant for this reaction will be,

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

Now put all the given values in this expression, we get :

[tex]0.055=\frac{(0.0100-2x)^2}{(x)\times (x)}[/tex]

By solving the term 'x', we get :

x = 0.0045

Thus, the concentrations of [tex]N_2,O_2\text{ and }NO[/tex] at equilibrium are :

Concentration of [tex]NO[/tex] at equilibrium = (0.0100-2x) M  = [tex](0.0100-2\times 0.0045)=0.001M[/tex]

Concentration of [tex]N_2[/tex] = (x) M = 0.0045 M

Concentration of [tex]O_2[/tex] = (x) M = 0.0045 M

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