For the reaction equilibrium N2O4 (g) 2 NO2(g) the concentrations of N2O4 and NO2 at equilibrium are 4.8×10-2 and 1.2×10-2 mol L-1 respectively. The value of Kc for the reaction is

To determine the equilibrium constant \( K_c \) for the reaction \( N_2O_4 (g) \rightleftharpoons 2 NO_2 (g) \), we can use the concentrations of the reactants and products at equilibrium. The equilibrium constant expression for this reaction is given by the formula:

Equilibrium Constant Expression

The general form of the equilibrium constant \( K_c \) for a reaction of the type \( aA + bB \rightleftharpoons cC + dD \) is:

Kc = \frac{[C]^c[D]^d}{[A]^a[B]^b}

For our specific reaction, the expression becomes:

Kc = \frac{[NO_2]^2}{[N_2O_4]}

Substituting the Equilibrium Concentrations

From the problem, we know the equilibrium concentrations:

• [N2O4] = 4.8 × 10^-2 mol L^-1
• [NO2] = 1.2 × 10^-2 mol L^-1

Now, we can substitute these values into the equilibrium expression:

Kc = \frac{(1.2 × 10^{-2})^2}{(4.8 × 10^{-2})}

Calculating the Values

First, calculate the numerator:

(1.2 × 10^-2)² = 1.44 × 10^-4

Next, we can substitute this back into the equation for \( K_c \):

Kc = \frac{1.44 × 10^{-4}}{4.8 × 10^{-2}}

Final Calculation

Now, divide the two values:

Kc = 1.44 × 10^-4 ÷ 4.8 × 10^-2

Performing the division gives:

Kc = 3.0 × 10^-3

Conclusion

Thus, the equilibrium constant \( K_c \) for the reaction \( N_2O_4 (g) \rightleftharpoons 2 NO_2 (g) \) at the given concentrations is:

Kc = 3.0 × 10^-3

This value indicates the extent to which the reaction favors the formation of products (NO2) over reactants (N2O4) at equilibrium. A smaller \( K_c \) value suggests that, at equilibrium, the concentration of reactants is greater than that of the products.