When 1.0 mol of water vapor and 3.0 mol of carbon monoxide is allowed to reach equilibrium at 986°C (Kp = 0.63), the total pressure at equilibrium is 2.0 atm. Co(g) + H20(g) = CO2(g) + H2(g) a. What is the partial pressure of each gas at equilibrium. b. How many moles of each species are present at equilibrium?

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### Chemical Equilibrium Problem

**Given Scenario:**

When 1.0 mol of water vapor and 3.0 mol of carbon monoxide is allowed to reach equilibrium at 986°C (\(K_p = 0.63\)), the total pressure at equilibrium is 2.0 atm.

The reaction is as follows:

\[ \text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)} \]

**Questions:**

a. What is the partial pressure of each gas at equilibrium?

b. How many moles of each species are present at equilibrium?

**Detailed Explanation for Students:**

1. **Identify Initial Conditions:**
   - Initial moles of \(\text{CO}\) = 3.0 mol
   - Initial moles of \(\text{H}_2\text{O}\) = 1.0 mol
   - Allowed to reach equilibrium at a total pressure of 2.0 atm

2. **Establish the Equilibrium Expression:**
   - For the reaction, the equilibrium constant \(K_p\) is given by:
     \[
     K_p = \frac{P_{\text{CO}_2} \cdot P_{\text{H}_2}}{P_{\text{CO}} \cdot P_{\text{H}_2\text{O}}} = 0.63
     \]

3. **Assume the Change in Moles:**
   - Let \(x\) be the change in moles of each gas as it reaches equilibrium.

   - Initial \(\text{CO}\) moles = 3.0 mol, equilibrium moles = 3.0 - \(x\) mol
   - Initial \(\text{H}_2\text{O}\) moles = 1.0 mol, equilibrium moles = 1.0 - \(x\) mol
   - \(\text{CO}_2\) and \(\text{H}_2\) are initially 0 mol, equilibrium moles = \(x\) mol each

4. **Express Total Pressure:**
   - Given total pressure at equilibrium = 2.0 atm

5. **Calculate Partial Pressures:**
   - Let's denote the
Transcribed Image Text:### Chemical Equilibrium Problem **Given Scenario:** When 1.0 mol of water vapor and 3.0 mol of carbon monoxide is allowed to reach equilibrium at 986°C (\(K_p = 0.63\)), the total pressure at equilibrium is 2.0 atm. The reaction is as follows: \[ \text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)} \] **Questions:** a. What is the partial pressure of each gas at equilibrium? b. How many moles of each species are present at equilibrium? **Detailed Explanation for Students:** 1. **Identify Initial Conditions:** - Initial moles of \(\text{CO}\) = 3.0 mol - Initial moles of \(\text{H}_2\text{O}\) = 1.0 mol - Allowed to reach equilibrium at a total pressure of 2.0 atm 2. **Establish the Equilibrium Expression:** - For the reaction, the equilibrium constant \(K_p\) is given by: \[ K_p = \frac{P_{\text{CO}_2} \cdot P_{\text{H}_2}}{P_{\text{CO}} \cdot P_{\text{H}_2\text{O}}} = 0.63 \] 3. **Assume the Change in Moles:** - Let \(x\) be the change in moles of each gas as it reaches equilibrium. - Initial \(\text{CO}\) moles = 3.0 mol, equilibrium moles = 3.0 - \(x\) mol - Initial \(\text{H}_2\text{O}\) moles = 1.0 mol, equilibrium moles = 1.0 - \(x\) mol - \(\text{CO}_2\) and \(\text{H}_2\) are initially 0 mol, equilibrium moles = \(x\) mol each 4. **Express Total Pressure:** - Given total pressure at equilibrium = 2.0 atm 5. **Calculate Partial Pressures:** - Let's denote the
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