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.
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.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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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
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8c591db-e344-457f-9816-5d949c4e5036%2F0c704505-5ce5-44bb-9c89-eb4d747d6944%2Fd324ndh_processed.png&w=3840&q=75)
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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