mmonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 50.0 L tank with 2.8 mol of mmonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 0.84 mol. alculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 gnificant digits. K = D x10 S ?

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### Calculating the Equilibrium Constant for Ammonia Decomposition

**Problem Statement:**
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 50.0 L tank with 2.8 mol of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 0.84 mol. 

Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

**Reaction:**
Ammonia decomposes as follows:
\[ 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \]

**Solution Steps:**

1. **Determine the change in moles of NH₃:**

Since 0.84 mol of N₂ is produced and the stoichiometric ratio between NH₃ and N₂ is 2:1, the moles of NH₃ that decompose:
\[ 2 \times 0.84 \text{ mol} = 1.68 \text{ mol} \]

2. **Calculate the moles of NH₃ at equilibrium:**

Initial moles of NH₃ – moles that decomposed:
\[ 2.8 \text{ mol} - 1.68 \text{ mol} = 1.12 \text{ mol} \]

3. **Calculate moles of H₂ produced:**

The stoichiometric ratio between NH₃ and H₂ is 2:3.
\[ 3 \times 0.84 \text{ mol} = 2.52 \text{ mol} \]

4. **Determine the concentrations at equilibrium:**

For the 50.0 L tank:
\[ [\text{NH}_3] = \frac{1.12 \text{ mol}}{50.0 \text{ L}} = 0.0224 \text{ M} \]
\[ [\text{N}_2] = \frac{0.84 \text{ mol}}{50.0 \text{ L}} = 0.0168 \text{ M} \]
\[ [\text{H}_2] = \frac{2.52 \text{ mol}}{50.0 \text{ L}} = 0.0504 \text{ M} \]
Transcribed Image Text:### Calculating the Equilibrium Constant for Ammonia Decomposition **Problem Statement:** Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 50.0 L tank with 2.8 mol of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 0.84 mol. Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits. **Reaction:** Ammonia decomposes as follows: \[ 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \] **Solution Steps:** 1. **Determine the change in moles of NH₃:** Since 0.84 mol of N₂ is produced and the stoichiometric ratio between NH₃ and N₂ is 2:1, the moles of NH₃ that decompose: \[ 2 \times 0.84 \text{ mol} = 1.68 \text{ mol} \] 2. **Calculate the moles of NH₃ at equilibrium:** Initial moles of NH₃ – moles that decomposed: \[ 2.8 \text{ mol} - 1.68 \text{ mol} = 1.12 \text{ mol} \] 3. **Calculate moles of H₂ produced:** The stoichiometric ratio between NH₃ and H₂ is 2:3. \[ 3 \times 0.84 \text{ mol} = 2.52 \text{ mol} \] 4. **Determine the concentrations at equilibrium:** For the 50.0 L tank: \[ [\text{NH}_3] = \frac{1.12 \text{ mol}}{50.0 \text{ L}} = 0.0224 \text{ M} \] \[ [\text{N}_2] = \frac{0.84 \text{ mol}}{50.0 \text{ L}} = 0.0168 \text{ M} \] \[ [\text{H}_2] = \frac{2.52 \text{ mol}}{50.0 \text{ L}} = 0.0504 \text{ M} \]
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