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)

Answer : The concentration of 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:

Initial conc.       0        0           0.0100

At eqm.             x        x         (0.0100-2x)

The expression for equilibrium constant is:

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

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 at equilibrium = x = 0.0045 M

The concentration of at equilibrium = x = 0.0045 M

The concentration of at equilibrium = (0.0100-2x) = (0.0100-2(0.0045)) = 0.001 M

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