At a certain temperature, 0.920 mol SO3 is placed in a 4.50 L container. 2 SO3(g) 2 SO₂(g) + O₂(g) At equilibrium, 0.140 mol O₂ is present. Calculate K.
At a certain temperature, 0.920 mol SO3 is placed in a 4.50 L container. 2 SO3(g) 2 SO₂(g) + O₂(g) At equilibrium, 0.140 mol O₂ is present. Calculate K.
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...
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At a certain temperature, 0.920 mol SO3 is placed in a 4.50 L container.
2SO3(g) ↽−−⇀ 2SO2(g) + O2(g)
At equilibrium, 0.140 mol O2 is present. Calculate ?c.
Can you please show your work so I can understand where I am going wrong? Thank you!
![### Chemistry Equilibrium Problem Explanation
**Problem Statement**
At a certain temperature, 0.920 moles of sulfur trioxide (\(SO_3\)) is placed in a 4.50 L container.
**Reaction:**
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
At equilibrium, 0.140 moles of oxygen (\(O_2\)) is present. Calculate the equilibrium constant \(K_c\).
**Steps to Solve:**
1. **Initial Concentration Calculation:**
- Calculate the initial concentration of \(SO_3\).
- \[ \text{Initial Concentration of } SO_3 = \frac{0.920 \, \text{mol}}{4.50 \, \text{L}} = 0.2044 \, \text{M} \]
2. **Change in Concentration:**
- Let the change in concentration of \(SO_3\) be \( -2x \) since 2 moles of \(SO_3\) produce 2 moles of \(SO_2\) and 1 mole of \(O_2\).
- At equilibrium, \(O_2\) concentration is given as 0.140 M, so \( x = 0.140 \, \text{M} \).
3. **Equilibrium Concentrations:**
- \( SO_3 \) = Initial concentration - \( 2x \)
- \[ [SO_3] = 0.2044 - 2(0.140) = 0.2044 - 0.280 \]
- \[ [SO_3] = -0.0756 \, \text{M} \rightarrow (\text{Not possible, needs re-evaluation}) \]
4. **Check Stoichiometric Ratios and Correct:**
Correct equilibrium calculations should reflect accurately (likely an error in re-checking the coefficient considerations):
- \( SO_3 \) = Initial concentration - \( x \)
- \[ [SO_3] = 0.2044 - x \]
- Using \( x = 0.140 \):
- \[ [SO_3] = 0.2044 - 0.140 = 0.0644 \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85650e2f-17ad-4aa4-9d56-acd675fbc675%2F6abccad7-7ab5-4cf2-9a29-6f29c55b1b57%2F4abnv1p_processed.png&w=3840&q=75)
Transcribed Image Text:### Chemistry Equilibrium Problem Explanation
**Problem Statement**
At a certain temperature, 0.920 moles of sulfur trioxide (\(SO_3\)) is placed in a 4.50 L container.
**Reaction:**
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
At equilibrium, 0.140 moles of oxygen (\(O_2\)) is present. Calculate the equilibrium constant \(K_c\).
**Steps to Solve:**
1. **Initial Concentration Calculation:**
- Calculate the initial concentration of \(SO_3\).
- \[ \text{Initial Concentration of } SO_3 = \frac{0.920 \, \text{mol}}{4.50 \, \text{L}} = 0.2044 \, \text{M} \]
2. **Change in Concentration:**
- Let the change in concentration of \(SO_3\) be \( -2x \) since 2 moles of \(SO_3\) produce 2 moles of \(SO_2\) and 1 mole of \(O_2\).
- At equilibrium, \(O_2\) concentration is given as 0.140 M, so \( x = 0.140 \, \text{M} \).
3. **Equilibrium Concentrations:**
- \( SO_3 \) = Initial concentration - \( 2x \)
- \[ [SO_3] = 0.2044 - 2(0.140) = 0.2044 - 0.280 \]
- \[ [SO_3] = -0.0756 \, \text{M} \rightarrow (\text{Not possible, needs re-evaluation}) \]
4. **Check Stoichiometric Ratios and Correct:**
Correct equilibrium calculations should reflect accurately (likely an error in re-checking the coefficient considerations):
- \( SO_3 \) = Initial concentration - \( x \)
- \[ [SO_3] = 0.2044 - x \]
- Using \( x = 0.140 \):
- \[ [SO_3] = 0.2044 - 0.140 = 0.0644 \,
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