Final answer:
To find the equilibrium concentrations of H2O and Cl2O, we set up an equilibrium table using the reaction's Kc value and solve for x. The calculated equilibrium concentrations are approximately [H2O] = 0.70 M, [Cl2O] = 0.70 M, and [HOCl] = 0.60 M, but these are not an exact match to any provided options.
Step-by-step explanation:
To determine the equilibrium concentrations of a reaction where [H₂O] and [Cl₂O] both start at 1.00 M, we can use the reaction equation and the equilibrium constant (Kc) provided. The reaction in question is:
H₂O(g) + Cl₂O(g) → 2HOCl(g), with Kc = 0.0900.
Let's set up a table to track the changes in concentrations as the reaction reaches equilibrium:
- Initial concentrations: [H₂O] = [Cl₂O] = 1.00 M, [HOCl] = 0 M
- Change in concentrations: Since the reaction produces 2 moles of HOCl for every mole of H₂O and Cl₂O that reacts, we'll denote the change in concentration of H₂O and Cl₂O as -x and the change in HOCl as +2x.
- Equilibrium concentrations: [H₂O] = [Cl₂O] = 1.00 - x M, [HOCl] = 2x M
We can now write the equilibrium expression:
Kc = [HOCl]² / ([H₂O] * [Cl₂O])
Plugging in the equilibrium concentrations, we get:
0.0900 = (2x)² / (1.00 - x)²
Solving for x, assuming x is small relative to the initial concentrations due to the small Kc value, yields x ≈ 0.30. Thus, the equilibrium concentrations are approximately:
- [H₂O] = [Cl₂O] = 1.00 - 0.30 ≈ 0.70 M
- [HOCl] = 2 * 0.30 ≈ 0.60 M
However, neither of these options is listed in the problem statement, which could indicate an error in the question or possible misinterpretation of the correct answer. The closest option to the calculated values would be option b, but it is still not exact.