A student ran the following reaction in the laboratory at 566 K: CO(g) + Cl2(g) = COC2(g) When she introduced 0.655 moles of CO(g) and 0.680 moles of Cl,(g) into a 1.00 liter container, she found the equilibrium concentration of COCI,(g) to be 0.618 M. Calculate the equilibrium constant, K, she obtained for this reaction. Kc

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**Experiment: Determining the Equilibrium Constant (Kc) for the Reaction \( \text{CO(g)} + \text{Cl}_2\text{(g)} \rightleftharpoons \text{COCl}_2\text{(g)} \) at 566 K**

**Introduction:**
In this experiment, a student investigated the reaction between carbon monoxide (CO) and chlorine gas (Cl₂) to form phosgene (COCl₂) in the laboratory at a temperature of 566 K.

**Chemical Equation:**
\[ \text{CO(g)} + \text{Cl}_2\text{(g)} \rightleftharpoons \text{COCl}_2\text{(g)} \]

**Initial Conditions:**
- **CO(g):** 0.655 moles
- **Cl₂(g):** 0.680 moles
- Volume of container: 1.00 liter

**Equilibrium Concentration:**
- **COCl₂(g):** 0.618 M

**Calculation Objective:**
Determine the equilibrium constant, \( K_c \), for the reaction.

**Detailed Procedure:**
1. **Establish Initial and Equilibrium Concentrations:**

   - Initial concentration of CO(g) = \(\frac{0.655 moles}{1.00 L} = 0.655 M\)
   - Initial concentration of Cl₂(g) = \(\frac{0.680 moles}{1.00 L} = 0.680 M\)
   - Equilibrium concentration of COCl₂(g) = 0.618 M

2. **Calculate Change in Concentration:**

   The stoichiometry of the reaction informs us that the change in concentration of COCl₂(g) will be equivalent to the change in concentration for both CO(g) and Cl₂(g), but in the opposite direction.
   
   Let \( x \) represent the change in concentration for CO(g) and Cl₂(g):
   - \([ \text{COCl}_2]\) gained = +x
   - \([ \text{CO]}\) lost = -x
   - \([ \text{Cl}_2]\) lost = -x

   Since 0.618 M of COCl₂ is formed:
   \[ x = 0.618 M \]

   Therefore:
   - Change in CO
Transcribed Image Text:**Experiment: Determining the Equilibrium Constant (Kc) for the Reaction \( \text{CO(g)} + \text{Cl}_2\text{(g)} \rightleftharpoons \text{COCl}_2\text{(g)} \) at 566 K** **Introduction:** In this experiment, a student investigated the reaction between carbon monoxide (CO) and chlorine gas (Cl₂) to form phosgene (COCl₂) in the laboratory at a temperature of 566 K. **Chemical Equation:** \[ \text{CO(g)} + \text{Cl}_2\text{(g)} \rightleftharpoons \text{COCl}_2\text{(g)} \] **Initial Conditions:** - **CO(g):** 0.655 moles - **Cl₂(g):** 0.680 moles - Volume of container: 1.00 liter **Equilibrium Concentration:** - **COCl₂(g):** 0.618 M **Calculation Objective:** Determine the equilibrium constant, \( K_c \), for the reaction. **Detailed Procedure:** 1. **Establish Initial and Equilibrium Concentrations:** - Initial concentration of CO(g) = \(\frac{0.655 moles}{1.00 L} = 0.655 M\) - Initial concentration of Cl₂(g) = \(\frac{0.680 moles}{1.00 L} = 0.680 M\) - Equilibrium concentration of COCl₂(g) = 0.618 M 2. **Calculate Change in Concentration:** The stoichiometry of the reaction informs us that the change in concentration of COCl₂(g) will be equivalent to the change in concentration for both CO(g) and Cl₂(g), but in the opposite direction. Let \( x \) represent the change in concentration for CO(g) and Cl₂(g): - \([ \text{COCl}_2]\) gained = +x - \([ \text{CO]}\) lost = -x - \([ \text{Cl}_2]\) lost = -x Since 0.618 M of COCl₂ is formed: \[ x = 0.618 M \] Therefore: - Change in CO
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