Final answer:
The equilibrium concentration of C for a reaction with a small Kc of 1.10 x 10⁻⁴ starting with 0.200 M A and 0.400 M B can be calculated using an ICE table and the assumption that x is negligible compared to the initial concentrations of A and B.
Step-by-step explanation:
For the reaction given, with Kc = 1.10 x 10⁻⁴, and initial concentrations of 0.200 MA and 0.400 MB, we are asked to determine the equilibrium concentration of component C. The equilibrium constant expression will involve the concentrations of the products raised to the power of their coefficients divided by the concentrations of the reactants raised to the power of their coefficients.
If the reaction is A + B ⇌ C, and we start with 0.200 M of A and 0.400 M of B, we need to set up an ICE table (Initial, Change, Equilibrium). Since the equation hasn't been provided, we cannot specify the exact equation, but in general, if it's a simple one-to-one-to-one reaction, when A and B react to form C, their concentrations decrease by x, and the concentration of C increases by x.
Setting up an equilibrium expression based on the reaction stoichiometry, we assume that the changes in concentrations for A and B are -x, and for C, it's +x. We then solve for x using Kc and substituting the equilibrium concentrations into the equation. Given that Kc is very small, we can make the approximation that x is much less than 0.10 (initial concentration of A), leading to a simplified calculation where the concentration of C at equilibrium will be approximately equal to x.