2NH3(g) N2(g) + 3H2(g) If an equilibrium mixture of the three gases in a 14.3 L container at 723K contains NHz at a pressure of 0.743 atm and N2 at a pressure of 0.867 atm, the equilibrium partial pressure of H, is atm.

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**Equilibrium and Partial Pressures**  
**Understanding Chemical Equilibrium**

In this section, we will explore the concept of equilibrium constants (Kp) concerning a chemical reaction and how to use them.

**Example Reaction and Given Data:**

The equilibrium constant, \( K_p \), for the following reaction is \( 2.20 \times 10^4 \) at 723K.

\[ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) \]

If an equilibrium mixture of the three gases in a 14.3 L container at 723K contains NH\(_3\) at a pressure of 0.743 atm and N\(_2\) at a pressure of 0.867 atm, calculate the equilibrium partial pressure of H\(_2\).

**Steps to Calculation:**

1. **Identify the Reaction Equation and Kp Expression:**
   \[ K_p = \frac{P_{\text{N}_2} \cdot (P_{\text{H}_2})^3}{(P_{\text{NH}_3})^2} \]

2. **Plug in the Known Values:**
   \[ P_{\text{NH}_3} = 0.743 \text{ atm} \]
   \[ P_{\text{N}_2} = 0.867 \text{ atm} \]
   \[ K_p = 2.20 \times 10^4 \]

3. **Solve for the Unknown Partial Pressure \( P_{\text{H}_2} \):**

   Rearrange the equation to solve for \( P_{\text{H}_2} \):
   \[ 2.20 \times 10^4 = \frac{0.867 \cdot (P_{\text{H}_2})^3}{(0.743)^2} \]

4. **Detailed Algebraic Steps:**
   Multiply both sides by \((0.743)^2\):
   \[ 2.20 \times 10^4 \cdot (0.743)^2 = 0.867 \cdot (P_{\text{H}_2})^3 \]
   \[ 2.20 \times 10^4 \cdot 0.552049 = 0.867 \
Transcribed Image Text:**Equilibrium and Partial Pressures** **Understanding Chemical Equilibrium** In this section, we will explore the concept of equilibrium constants (Kp) concerning a chemical reaction and how to use them. **Example Reaction and Given Data:** The equilibrium constant, \( K_p \), for the following reaction is \( 2.20 \times 10^4 \) at 723K. \[ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) \] If an equilibrium mixture of the three gases in a 14.3 L container at 723K contains NH\(_3\) at a pressure of 0.743 atm and N\(_2\) at a pressure of 0.867 atm, calculate the equilibrium partial pressure of H\(_2\). **Steps to Calculation:** 1. **Identify the Reaction Equation and Kp Expression:** \[ K_p = \frac{P_{\text{N}_2} \cdot (P_{\text{H}_2})^3}{(P_{\text{NH}_3})^2} \] 2. **Plug in the Known Values:** \[ P_{\text{NH}_3} = 0.743 \text{ atm} \] \[ P_{\text{N}_2} = 0.867 \text{ atm} \] \[ K_p = 2.20 \times 10^4 \] 3. **Solve for the Unknown Partial Pressure \( P_{\text{H}_2} \):** Rearrange the equation to solve for \( P_{\text{H}_2} \): \[ 2.20 \times 10^4 = \frac{0.867 \cdot (P_{\text{H}_2})^3}{(0.743)^2} \] 4. **Detailed Algebraic Steps:** Multiply both sides by \((0.743)^2\): \[ 2.20 \times 10^4 \cdot (0.743)^2 = 0.867 \cdot (P_{\text{H}_2})^3 \] \[ 2.20 \times 10^4 \cdot 0.552049 = 0.867 \
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