Consider the apparatus shown below. The valve is closed. the left bulb contains nitrogen gas and the right bulb hydrogen gas. N₂ H₂ 300 K 3.00 L 300 K 9.00 L 2.00 atm 2.00 atm When the valve is opened the two gases mix and react completely to form ammonia gas (NH3). The temperature remains constant. What is the pressure in the apparatus after the chemical reaction? 2.00 atm none of the answers are correct O 1.00 atm

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### Understanding Gas Reactions: Ammonia Formation

Consider the apparatus shown below. The valve is closed. The left bulb contains nitrogen gas (N₂) and the right bulb contains hydrogen gas (H₂).

![Apparatus Diagram](#)
- **Left Bulb (N₂):**
  - Temperature: 300 K
  - Volume: 3.00 L
  - Pressure: 2.00 atm

- **Right Bulb (H₂):**
  - Temperature: 300 K
  - Volume: 9.00 L
  - Pressure: 2.00 atm

When the valve is opened, the two gases mix and react completely to form ammonia gas (NH₃). The temperature remains constant. What is the pressure in the apparatus after the chemical reaction?

**Options:**
1. 2.00 atm
2. None of the answers are correct
3. 1.00 atm

**Explanation:**

1. The initial state of the gases in separate bulbs indicates that both gases are at the same pressure (2.00 atm) and temperature (300 K) but occupy different volumes.
2. According to the balanced chemical equation for the formation of ammonia:
   \[ N₂ + 3 H₂ \rightarrow 2 NH₃ \]
3. For every 1 mole of nitrogen, 3 moles of hydrogen react. Therefore, the limiting reagent must be determined:
    - Moles of N₂: \( n(N₂) = \frac{P \cdot V}{R \cdot T} = \frac{2.00 \cdot 3}{0.0821 \cdot 300} = \frac{6}{24.63} \approx 0.244 \text{ moles} \)
    - Moles of H₂: \( n(H₂) = \frac{2.00 \cdot 9}{0.0821 \cdot 300} = \frac{18}{24.63} \approx 0.73 \text{ moles} \)

Since 1 mole of N₂ requires 3 moles of H₂, to react with 0.244 moles of N₂, we need \(3 \cdot 0.244 = 0.732 \text { moles of H₂}\). Therefore, N₂ is the limiting reagent.

4. Using gas
Transcribed Image Text:### Understanding Gas Reactions: Ammonia Formation Consider the apparatus shown below. The valve is closed. The left bulb contains nitrogen gas (N₂) and the right bulb contains hydrogen gas (H₂). ![Apparatus Diagram](#) - **Left Bulb (N₂):** - Temperature: 300 K - Volume: 3.00 L - Pressure: 2.00 atm - **Right Bulb (H₂):** - Temperature: 300 K - Volume: 9.00 L - Pressure: 2.00 atm When the valve is opened, the two gases mix and react completely to form ammonia gas (NH₃). The temperature remains constant. What is the pressure in the apparatus after the chemical reaction? **Options:** 1. 2.00 atm 2. None of the answers are correct 3. 1.00 atm **Explanation:** 1. The initial state of the gases in separate bulbs indicates that both gases are at the same pressure (2.00 atm) and temperature (300 K) but occupy different volumes. 2. According to the balanced chemical equation for the formation of ammonia: \[ N₂ + 3 H₂ \rightarrow 2 NH₃ \] 3. For every 1 mole of nitrogen, 3 moles of hydrogen react. Therefore, the limiting reagent must be determined: - Moles of N₂: \( n(N₂) = \frac{P \cdot V}{R \cdot T} = \frac{2.00 \cdot 3}{0.0821 \cdot 300} = \frac{6}{24.63} \approx 0.244 \text{ moles} \) - Moles of H₂: \( n(H₂) = \frac{2.00 \cdot 9}{0.0821 \cdot 300} = \frac{18}{24.63} \approx 0.73 \text{ moles} \) Since 1 mole of N₂ requires 3 moles of H₂, to react with 0.244 moles of N₂, we need \(3 \cdot 0.244 = 0.732 \text { moles of H₂}\). Therefore, N₂ is the limiting reagent. 4. Using gas
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