Suppose a 250. mL flask is filled with 2.0 mol of Cl,, 1.4 mol of CHCI, and 1.0 mol of HCl. The following reaction becomes possible: C1, (2) + CHCI, (3) = HCI(g) +CCl,(g) The equilibrium constant K for this reaction is 0.510 at the temperature of the flask. Calculate the equilibrium molarity of Cl,. Round your answer to two decimal places.

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**Chemical Equilibrium Calculation**

**Problem Statement:**

Suppose a 250. mL flask is filled with 2.0 mol of Cl₂, 1.4 mol of CHCl₃, and 1.0 mol of HCl. The following reaction becomes possible:

\[ \text{Cl}_2(g) + \text{CHCl}_3(g) \rightleftharpoons \text{HCl}(g) + \text{CCl}_4(g) \]

The equilibrium constant \( K \) for this reaction is 0.510 at the temperature of the flask.

Calculate the equilibrium molarity of Cl₂. Round your answer to two decimal places.

**Solution Approach:**

Here is how to approach the problem:

1. **Initial Concentrations:**
   - Initially, we have 2.0 moles of Cl₂ in 250 mL, 1.4 moles of CHCl₃, and 1.0 mole of HCl.
   - Convert the volumes to liters: \( \text{Volumes} = 0.250 \text{ L} \).

2. **Calculations:**
   - Find the initial molarities:
       - Molarity of Cl₂: \[ \frac{2.0 \text{ moles}}{0.25 \text{ L}} = 8.0 \text{ M} \]
       - Molarity of CHCl₃: \[ \frac{1.4 \text{ moles}}{0.25 \text{ L}} = 5.6 \text{ M} \]
       - Molarity of HCl: \[ \frac{1.0 \text{ moles}}{0.25 \text{ L}} = 4.0 \text{ M} \]

3. **Set up the ICE Table (Initial, Change, Equilibrium) for the reaction:**
   - Let \( x \) be the change in concentration at equilibrium:
       - Initial: [Cl₂] = 8.0 M, [CHCl₃] = 5.6 M, [HCl] = 4.0 M
       - Change: [Cl₂] = -x, [CHCl₃] = -x, [HCl] = +x, [CCl₄] = +x
       -
Transcribed Image Text:**Chemical Equilibrium Calculation** **Problem Statement:** Suppose a 250. mL flask is filled with 2.0 mol of Cl₂, 1.4 mol of CHCl₃, and 1.0 mol of HCl. The following reaction becomes possible: \[ \text{Cl}_2(g) + \text{CHCl}_3(g) \rightleftharpoons \text{HCl}(g) + \text{CCl}_4(g) \] The equilibrium constant \( K \) for this reaction is 0.510 at the temperature of the flask. Calculate the equilibrium molarity of Cl₂. Round your answer to two decimal places. **Solution Approach:** Here is how to approach the problem: 1. **Initial Concentrations:** - Initially, we have 2.0 moles of Cl₂ in 250 mL, 1.4 moles of CHCl₃, and 1.0 mole of HCl. - Convert the volumes to liters: \( \text{Volumes} = 0.250 \text{ L} \). 2. **Calculations:** - Find the initial molarities: - Molarity of Cl₂: \[ \frac{2.0 \text{ moles}}{0.25 \text{ L}} = 8.0 \text{ M} \] - Molarity of CHCl₃: \[ \frac{1.4 \text{ moles}}{0.25 \text{ L}} = 5.6 \text{ M} \] - Molarity of HCl: \[ \frac{1.0 \text{ moles}}{0.25 \text{ L}} = 4.0 \text{ M} \] 3. **Set up the ICE Table (Initial, Change, Equilibrium) for the reaction:** - Let \( x \) be the change in concentration at equilibrium: - Initial: [Cl₂] = 8.0 M, [CHCl₃] = 5.6 M, [HCl] = 4.0 M - Change: [Cl₂] = -x, [CHCl₃] = -x, [HCl] = +x, [CCl₄] = +x -
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