Suppose a 500. mL flask is filled with 0.80 mol of Br₂, 1.2 mol of OC1₂ and 1.7 mol of BrOCI. The following reaction becomes possible: Br₂(g) + OC1₂(g) → BrOC1 (g) + BrC1 (g) The equilibrium constant K for this reaction is 1.75 at the temperature of the flask. Calculate the equilibrium molarity of BrOC1. Round your answer to two decimal places. M X ?
Suppose a 500. mL flask is filled with 0.80 mol of Br₂, 1.2 mol of OC1₂ and 1.7 mol of BrOCI. The following reaction becomes possible: Br₂(g) + OC1₂(g) → BrOC1 (g) + BrC1 (g) The equilibrium constant K for this reaction is 1.75 at the temperature of the flask. Calculate the equilibrium molarity of BrOC1. Round your answer to two decimal places. M X ?
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Chapter1: Chemical Foundations
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![### Chemistry Equilibrium Calculation
#### Problem Statement
Suppose a 500. mL flask is filled with 0.80 mol of Br\(_2\), 1.2 mol of OCl\(_2\), and 1.7 mol of BrOCl. The following reaction becomes possible:
\[ \text{Br}_2 (g) + \text{OCl}_2 (g) \rightleftharpoons \text{BrOCl} (g) + \text{BrCl} (g) \]
The equilibrium constant \(K\) for this reaction is 1.75 at the temperature of the flask.
Calculate the equilibrium molarity of BrOCl. Round your answer to two decimal places.
#### Instructions for Calculation
1. **Initial Concentrations**:
- Determine the initial concentrations of the reactants and products by dividing the number of moles by the volume of the flask.
- Flask Volume = 500 mL = 0.500 L
For Br\(_2\):
\[ [Br_2]_0 = \frac{0.80 \text{ mol}}{0.500 \text{ L}} = 1.60 \text{ M} \]
For OCl\(_2\):
\[ [OCl_2]_0 = \frac{1.2 \text{ mol}}{0.500 \text{ L}} = 2.40 \text{ M} \]
For BrOCl:
\[ [BrOCl]_0 = \frac{1.7 \text{ mol}}{0.500 \text{ L}} = 3.40 \text{ M} \]
2. **Reaction Table**:
- Set up an ICE (Initial, Change, Equilibrium) table to determine the changes in concentration as the system achieves equilibrium.
3. **Equilibrium Expression**:
- Use the equilibrium constant expression for the balanced equation.
\[ K = \frac{[BrOCl][BrCl]}{[Br_2][OCl_2]} \]
4. **Solving for \(x\)**:
- Introduce the changes in concentration (\(-x\) for reactants and \(+x\) for products).
- Apply the equilibrium constant \(K\) to find the value of \(x\).
5. **Calculate](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf399169-ef14-4107-9621-c5ec1888471d%2Fa712d36b-e5e2-497c-b86f-357a95fc7613%2Fw7oa8q_processed.png&w=3840&q=75)
Transcribed Image Text:### Chemistry Equilibrium Calculation
#### Problem Statement
Suppose a 500. mL flask is filled with 0.80 mol of Br\(_2\), 1.2 mol of OCl\(_2\), and 1.7 mol of BrOCl. The following reaction becomes possible:
\[ \text{Br}_2 (g) + \text{OCl}_2 (g) \rightleftharpoons \text{BrOCl} (g) + \text{BrCl} (g) \]
The equilibrium constant \(K\) for this reaction is 1.75 at the temperature of the flask.
Calculate the equilibrium molarity of BrOCl. Round your answer to two decimal places.
#### Instructions for Calculation
1. **Initial Concentrations**:
- Determine the initial concentrations of the reactants and products by dividing the number of moles by the volume of the flask.
- Flask Volume = 500 mL = 0.500 L
For Br\(_2\):
\[ [Br_2]_0 = \frac{0.80 \text{ mol}}{0.500 \text{ L}} = 1.60 \text{ M} \]
For OCl\(_2\):
\[ [OCl_2]_0 = \frac{1.2 \text{ mol}}{0.500 \text{ L}} = 2.40 \text{ M} \]
For BrOCl:
\[ [BrOCl]_0 = \frac{1.7 \text{ mol}}{0.500 \text{ L}} = 3.40 \text{ M} \]
2. **Reaction Table**:
- Set up an ICE (Initial, Change, Equilibrium) table to determine the changes in concentration as the system achieves equilibrium.
3. **Equilibrium Expression**:
- Use the equilibrium constant expression for the balanced equation.
\[ K = \frac{[BrOCl][BrCl]}{[Br_2][OCl_2]} \]
4. **Solving for \(x\)**:
- Introduce the changes in concentration (\(-x\) for reactants and \(+x\) for products).
- Apply the equilibrium constant \(K\) to find the value of \(x\).
5. **Calculate
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