One mole each of 2NO (g) and 2 4N O (g)were mixed in a 1 3dm flask and allowed toequilibrate. The following equilibrium wasestablished : 32 2 42NO (g) N O (g)If at equilibrium the reaction mixturecontained 0.75 moles of 2 4N O (g) , calculatethe value of equilibrium constant ck for thereaction

To calculate the equilibrium constant (Kc) for the given reaction, we need to use the balanced equation and the concentrations of the reactants and products at equilibrium.

The balanced equation for the reaction is: 2NO (g) + 2N2O4 (g) ⇌ 4NO2 (g)

At equilibrium, the reaction mixture contains 0.75 moles of N2O4 (g). Since the stoichiometric coefficient of N2O4 (g) is 2, the concentration of N2O4 (g) at equilibrium is 0.75/1.3 = 0.577 moles/dm^3.

Using the stoichiometry of the balanced equation, we can determine the concentration of NO (g) at equilibrium. Since the stoichiometric coefficient of NO (g) is also 2, the concentration of NO (g) at equilibrium is 2 * 0.577 = 1.154 moles/dm^3.

The concentration of NO2 (g) at equilibrium can be calculated by subtracting the concentration of N2O4 (g) from the initial concentration of NO2 (g). Since the initial concentration of N2O4 (g) is 0 moles/dm^3, the concentration of NO2 (g) at equilibrium is 2 * 0.577 = 1.154 moles/dm^3.

Now, we can substitute the equilibrium concentrations into the equilibrium expression to calculate the value of Kc: Kc = [NO2]^4 / ([NO]^2 * [N2O4]^2) = (1.154^4) / ((1.154^2) * (0.577^2)) = 2.56

Therefore, the value of the equilibrium constant (Kc) for the reaction is 2.56.