The equilibrium constant, Ke, for the following reaction is 6.30 at 723K. 2NH3(g) N2(g) + 3H2(g) If an equilibrium mixture of the three gases in a 14.3 L container at 723K contains 0.318 mol of NH3(g) and 0.364 mol of N2, the equilibrium concentration of H, is М.

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### Chemical Equilibrium Calculation

#### Problem Statement
The equilibrium constant, \( K_c \), for the following reaction is **6.30 at 723 K**.

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

If an equilibrium mixture of the three gases in a **14.3 L container at 723 K** contains **0.318 mol of NH\(_3\)(g)** and **0.364 mol of N\(_2\)**, the equilibrium concentration of H\(_2\) is \_\_\_\_\_\_\_ M.

#### Explanation
To solve this problem:
1. **Determine the concentrations** of NH\(_3\) and N\(_2\) at equilibrium:
   - Concentration of NH\(_3\): \(\frac{0.318 \text{ mol}}{14.3 \text{ L}}\)
   - Concentration of N\(_2\): \(\frac{0.364 \text{ mol}}{14.3 \text{ L}}\)

2. **Set up the expression** for the equilibrium constant \( K_c \):

\[
K_c = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} = 6.30
\]

3. **Plug in** the known concentrations and solve for the concentration of H\(_2\).
   
#### Visual Explanation
The problem does not contain graphical elements, but solving it involves a straightforward calculation using the given formulae.

This approach ensures that the ratios and equilibrium relations are calculated correctly to find the missing concentration of H\(_2\).
Transcribed Image Text:### Chemical Equilibrium Calculation #### Problem Statement The equilibrium constant, \( K_c \), for the following reaction is **6.30 at 723 K**. \[ 2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g) \] If an equilibrium mixture of the three gases in a **14.3 L container at 723 K** contains **0.318 mol of NH\(_3\)(g)** and **0.364 mol of N\(_2\)**, the equilibrium concentration of H\(_2\) is \_\_\_\_\_\_\_ M. #### Explanation To solve this problem: 1. **Determine the concentrations** of NH\(_3\) and N\(_2\) at equilibrium: - Concentration of NH\(_3\): \(\frac{0.318 \text{ mol}}{14.3 \text{ L}}\) - Concentration of N\(_2\): \(\frac{0.364 \text{ mol}}{14.3 \text{ L}}\) 2. **Set up the expression** for the equilibrium constant \( K_c \): \[ K_c = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} = 6.30 \] 3. **Plug in** the known concentrations and solve for the concentration of H\(_2\). #### Visual Explanation The problem does not contain graphical elements, but solving it involves a straightforward calculation using the given formulae. This approach ensures that the ratios and equilibrium relations are calculated correctly to find the missing concentration of H\(_2\).
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