In addition to filling in the blanks below, show all of your work for this problem on paper for later upload. The reaction C(s) + CO2(g) 2 2 CO(g) Was carried out with 1.50 mol CO, in a 20.0 L flask at 1100 K. At equilibrium, the concentration of CO was found to be 7.00' 102 M. Find the equilibrium concentration of CO, and determine the value of K, (not KA) at this temperature. Enter your value for [CO2} in the first box and an appropriate unit of measure in the second box. Enter your value for K, in the box.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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### Equilibrium Concentration and Kp Calculation

To solve the equilibrium problem provided, follow these steps carefully. Please show all your work when solving this on paper for later upload.

#### Problem Statement
The reaction \( \text{C(s) + CO}_2(g) \rightleftharpoons 2 \text{CO(g)} \) was conducted with 1.50 mol \( \text{CO}_2 \) in a 20.0 L flask at 1100 K. At equilibrium, the concentration of \( \text{CO} \) was found to be \( 7.00 \times 10^{-2} \) M. Find the **equilibrium concentration** of \( \text{CO}_2 \) and determine the value of \( K_p \) (not \( K_c \)) at this temperature.

#### Step-by-Step Solution:

1. **Initial Conditions**:
   - Initial moles of \( \text{CO}_2 \) = 1.50 mol
   - Volume of the flask = 20.0 L
   - Initial concentration of \( \text{CO}_2 \) = \(\frac{1.50 \text{ mol}}{20.0 \text{ L}} = 0.075 \) M

2. **Equilibrium Conditions**:
    - At equilibrium, concentration of \( \text{CO} \) = \( 7.00 \times 10^{-2} \) M
    - Let \( x \) be the change in concentration of \( \text{CO}_2 \) and \( \text{CO} \) during the reaction.
    - Change in concentration: 
      - \( \text{CO}_2 \) decreases by \( x \)
      - \( \text{CO} \) increases by \( 2x \)
    - Since \( \text{CO} \) concentration at equilibrium is \( 7.00 \times 10^{-2} \) M, then: 
      \[ 2x = 7.00 \times 10^{-2} \]
      \[ x = 3.50 \times 10^{-2} \]

3. **Equilibrium Concentration of \( \text{CO}_2 \)**:
    - Initial concentration of \( \text{CO}_2 \) was 0.075 M.
    - Change is \(
Transcribed Image Text:### Equilibrium Concentration and Kp Calculation To solve the equilibrium problem provided, follow these steps carefully. Please show all your work when solving this on paper for later upload. #### Problem Statement The reaction \( \text{C(s) + CO}_2(g) \rightleftharpoons 2 \text{CO(g)} \) was conducted with 1.50 mol \( \text{CO}_2 \) in a 20.0 L flask at 1100 K. At equilibrium, the concentration of \( \text{CO} \) was found to be \( 7.00 \times 10^{-2} \) M. Find the **equilibrium concentration** of \( \text{CO}_2 \) and determine the value of \( K_p \) (not \( K_c \)) at this temperature. #### Step-by-Step Solution: 1. **Initial Conditions**: - Initial moles of \( \text{CO}_2 \) = 1.50 mol - Volume of the flask = 20.0 L - Initial concentration of \( \text{CO}_2 \) = \(\frac{1.50 \text{ mol}}{20.0 \text{ L}} = 0.075 \) M 2. **Equilibrium Conditions**: - At equilibrium, concentration of \( \text{CO} \) = \( 7.00 \times 10^{-2} \) M - Let \( x \) be the change in concentration of \( \text{CO}_2 \) and \( \text{CO} \) during the reaction. - Change in concentration: - \( \text{CO}_2 \) decreases by \( x \) - \( \text{CO} \) increases by \( 2x \) - Since \( \text{CO} \) concentration at equilibrium is \( 7.00 \times 10^{-2} \) M, then: \[ 2x = 7.00 \times 10^{-2} \] \[ x = 3.50 \times 10^{-2} \] 3. **Equilibrium Concentration of \( \text{CO}_2 \)**: - Initial concentration of \( \text{CO}_2 \) was 0.075 M. - Change is \(
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