Final answer:
The equilibrium concentration of NO2 in a 1.00 L solution of 0.129 mol N2O4 with chloroform as solvent is approximately 1.17 × 10^-3 M when the initial change in N2O4 concentration is considered negligible.
Step-by-step explanation:
To calculate the equilibrium concentration of both species in a 1.00 L solution prepared from 0.129 mol of N2O4 with chloroform as the solvent where the equilibrium constant Kc is 1.07 × 10-5 in chloroform, we begin by setting up the initial concentrations. The initial concentration of N2O4 is 0.129 M, while the initial concentration of NO2 is 0 M since no reaction has occurred yet. After some of the N2O4 dissociates into 2 NO2, the equilibrium expression can be written as:
Kc = [NO2]^2 / [N2O4]
Considering the change in concentration of N2O4 is negligible, we can solve for the equilibrium concentration of NO2 by taking the square root of the product of Kc and the initial concentration of N2O4:
[NO2] = sqrt(Kc × [N2O4_initial])
Substituting the values:
[NO2] = sqrt(1.07 × 10-5 × 0.129 M) = sqrt(1.3803 × 10-6 M) ≈ 1.17 × 10-3 M
Thus, the equilibrium concentration of NO2 is approximately 1.17 × 10-3 M, confirming that the given options in the question are indeed correct.