The balanced equation for the dissociation of nitrogen monoxide (NO) is:
2NO(g) ⇌ N2(g) + O2(g)
According to the problem statement, 2 moles of NO were placed in a 1 dm^3 bottle and allowed to reach equilibrium, and at equilibrium, 1.2 moles of NO had dissociated. This means that the initial concentration of NO was:
[NO]initial = 2 mol / 1 dm^3 = 2 M
And the concentration of NO at equilibrium is:
[NO]equilibrium = (2 - 1.2) mol / 1 dm^3 = 0.8 M
Since the stoichiometry of the balanced equation is 2:1:1 for NO, N2, and O2, respectively, the equilibrium concentrations of N2 and O2 will also be 0.6 M.
The equilibrium constant (Kc) can be calculated using the equilibrium concentrations of the reactants and products, raised to the power of their stoichiometric coefficients. Therefore:
Kc = ([N2][O2]) / ([NO]^2)
Substituting the equilibrium concentrations into the equation, we get:
Kc = (0.6 M x 0.6 M) / (0.8 M x 0.8 M)
Kc = 0.5625
Therefore, the value of the equilibrium constant for the reaction at that temperature is 0.5625. Note that the units of Kc depend on the stoichiometry of the balanced equation. Since the stoichiometric coefficients are all 1, the units of Kc in this case are M^-1