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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. N2(g)+O2(g)⇌2NO(g)

Respuesta :

Answer : The concentration of [tex]N_2,O_2\text{ and }NO[/tex] at equilibrium is, 0.0045 M, 0.0045 M and 0.001 M respectively.

Explanation :  Given,

Equilibrium constant = 0.055

[NO] = 0.0100 M

The given chemical reaction is:

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

Initial conc.       0        0           0.0100

At eqm.             x        x         (0.0100-2x)

The expression for equilibrium constant is:

[tex]K=\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]

x = 0.0045 and x = 0.0057

There are two values of 'x'. But we are neglecting the value of x = 0.0057 because the concentration at equilibrium can not be more than initial concentration.

So, x = 0.0045 M

The concentration of [tex]N_2[/tex] at equilibrium = x = 0.0045 M

The concentration of [tex]O_2[/tex] at equilibrium = x = 0.0045 M

The concentration of [tex]NO[/tex] at equilibrium = (0.0100-2x) = (0.0100-2(0.0045)) = 0.001 M

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