At a certain temperature, the equilibrium constant, Kc, equals 0.11 for the reaction: 2 ICl(g) ⇌ I2(g) + Cl2(g). What is the equilibrium concentration of ICl if 0.45 mol of I2 and 0.45 mol of Cl2 are initially mixed in a 2.0-L flask?

• A. 0.15 M
• B. 0.60 M
• C. 0.30 M
• D. 0.45 M

Answer: (C)

Verified step by step guidance

1. Identify the initial concentrations of I2 and Cl2. Since 0.45 mol of each is placed in a 2.0 L flask, calculate their initial concentrations using the formula: \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \).
2. Set up an ICE (Initial, Change, Equilibrium) table to track the concentrations of ICl, I2, and Cl2. Initially, the concentration of ICl is 0 M, and the concentrations of I2 and Cl2 are both 0.225 M.
3. Define the change in concentration for the reaction. Let x be the change in concentration of I2 and Cl2 that reacts to form ICl. Therefore, the change for ICl will be +2x, and for I2 and Cl2, it will be -x.
4. Write the expression for the equilibrium concentrations: \([\text{ICl}] = 2x\), \([\text{I2}] = 0.225 - x\), \([\text{Cl2}] = 0.225 - x\).
5. Substitute these expressions into the equilibrium constant expression \( K_c = \frac{[\text{I2}][\text{Cl2}]}{[\text{ICl}]^2} \) and solve for x, which will allow you to find the equilibrium concentration of ICl.