0.017mol of butane is added to a 0.5L flask. What are the equilibrium concentrations of each species? (K.=2.5) Butane Isobutane Initial Change Equilibrium
0.017mol of butane is added to a 0.5L flask. What are the equilibrium concentrations of each species? (K.=2.5) Butane Isobutane Initial Change Equilibrium
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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![### Determining Equilibrium Concentrations of Butane and Isobutane
In this example, 0.017 moles of butane are added to a 0.5 L flask. We aim to find the equilibrium concentrations of butane and isobutane, given the equilibrium constant \( K_c = 2.5 \).
#### Table: ICE Method for Concentration Calculation
To solve this, we'll use the ICE (Initial, Change, Equilibrium) method. Here is the table structure which represents the initial amounts, changes, and equilibrium concentrations of the species involved:
| | Butane | ⇌ | Isobutane |
|-----------------|-----------------------|---|-------------------------|
| Initial | | | |
| Change | | | |
| Equilibrium | | | |
**Explanation of Table Columns:**
- **Initial:** Shows the initial concentration of butane and isobutane before the reaction reaches equilibrium.
- **Change:** Represents the change in concentration of butane and isobutane as the system progresses to equilibrium.
- **Equilibrium:** Contains the equilibrium concentrations of butane and isobutane.
**Calculation Steps:**
1. **Initial Concentration:**
- Butane: Before the reaction, butane's initial concentration is calculated by dividing moles by volume: \[ \text{Concentration of Butane} = \frac{0.017 \, \text{mol}}{0.5 \, \text{L}} = 0.034 \, \text{M} \]
- Isobutane: Initially, there is no isobutane, so its concentration is 0 M.
2. **Change:**
- Suppose the concentration of butane decreases by \( x \), then that amount converts to isobutane.
- Butane: \( -x \)
- Isobutane: \( +x \)
3. **Equilibrium:**
- Butane: \( 0.034 - x \)
- Isobutane: \( x \)
#### Utilizing the Equilibrium Constant:
\[ K_c = \frac{[\text{Isobutane}]}{[\text{Butane}]} = 2.5 \]
\[ 2.5 = \frac{x}{0.034 - x} \]
Solving for \( x \):
\[ 2.5 (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6373081f-17d9-417b-b535-bd8706276caa%2Fc4fbe2bf-7885-4f24-8206-025ebc688571%2Fzbn6eqc_processed.png&w=3840&q=75)
Transcribed Image Text:### Determining Equilibrium Concentrations of Butane and Isobutane
In this example, 0.017 moles of butane are added to a 0.5 L flask. We aim to find the equilibrium concentrations of butane and isobutane, given the equilibrium constant \( K_c = 2.5 \).
#### Table: ICE Method for Concentration Calculation
To solve this, we'll use the ICE (Initial, Change, Equilibrium) method. Here is the table structure which represents the initial amounts, changes, and equilibrium concentrations of the species involved:
| | Butane | ⇌ | Isobutane |
|-----------------|-----------------------|---|-------------------------|
| Initial | | | |
| Change | | | |
| Equilibrium | | | |
**Explanation of Table Columns:**
- **Initial:** Shows the initial concentration of butane and isobutane before the reaction reaches equilibrium.
- **Change:** Represents the change in concentration of butane and isobutane as the system progresses to equilibrium.
- **Equilibrium:** Contains the equilibrium concentrations of butane and isobutane.
**Calculation Steps:**
1. **Initial Concentration:**
- Butane: Before the reaction, butane's initial concentration is calculated by dividing moles by volume: \[ \text{Concentration of Butane} = \frac{0.017 \, \text{mol}}{0.5 \, \text{L}} = 0.034 \, \text{M} \]
- Isobutane: Initially, there is no isobutane, so its concentration is 0 M.
2. **Change:**
- Suppose the concentration of butane decreases by \( x \), then that amount converts to isobutane.
- Butane: \( -x \)
- Isobutane: \( +x \)
3. **Equilibrium:**
- Butane: \( 0.034 - x \)
- Isobutane: \( x \)
#### Utilizing the Equilibrium Constant:
\[ K_c = \frac{[\text{Isobutane}]}{[\text{Butane}]} = 2.5 \]
\[ 2.5 = \frac{x}{0.034 - x} \]
Solving for \( x \):
\[ 2.5 (
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