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

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
icon
Related questions
icon
Concept explainers
Question
100%
### 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
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

Blurred answer
Knowledge Booster
Ideal and Real Gases
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY