The equilibrium constant, K, for the following reaction is 123 at a certain temperature. H₂(g) + F₂(g) 2 HF(g) 0.727 mol of HF is placed in a 0.750-L flask. What is the concentration of HF in the flask at equilibrium?
The equilibrium constant, K, for the following reaction is 123 at a certain temperature. H₂(g) + F₂(g) 2 HF(g) 0.727 mol of HF is placed in a 0.750-L flask. What is the concentration of HF in the flask at equilibrium?
Chemistry
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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
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Ok i am trying to use the info from the last question but i am still struggling with the set up.
![The equilibrium constant, \( K_c \), for the following reaction is 123 at a certain temperature.
\[ \text{H}_2(g) + \text{F}_2(g) \rightleftharpoons 2 \text{HF}(g) \]
0.727 mol of HF is placed in a 0.750-L flask. What is the concentration of HF in the flask at equilibrium?
- A. 0.148 M
- B. 1.42 × 10\(^{-4}\) M
- C. 0.821 M
- D. 0.895 M
- E. 0.179 M](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb0ffec19-af41-4c50-af1b-66e883fc2bb8%2Fcc516aa3-f4ba-4f47-bf8c-c49c168ddcd8%2F0o61wed_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The equilibrium constant, \( K_c \), for the following reaction is 123 at a certain temperature.
\[ \text{H}_2(g) + \text{F}_2(g) \rightleftharpoons 2 \text{HF}(g) \]
0.727 mol of HF is placed in a 0.750-L flask. What is the concentration of HF in the flask at equilibrium?
- A. 0.148 M
- B. 1.42 × 10\(^{-4}\) M
- C. 0.821 M
- D. 0.895 M
- E. 0.179 M
![**Equilibrium Constant Calculation Example**
In this example, we determine the concentration of HF in a reaction at equilibrium. The given equilibrium constant (\( \text{K}_c \)) is 123 for the reaction:
\[
\text{H}_2(g) + \text{F}_2(g) \rightleftharpoons 2\text{HF}(g)
\]
**Initial Information Given:**
- Moles of HF: 0.727 moles
- Volume of the reaction vessel: 0.750 L
**Objective:**
Calculate the concentration of HF in the flask at equilibrium.
**Calculations:**
1. **Calculation of Initial Concentration:**
Moles of HF = 0.727 moles
Volume = 0.750 L
\[
\text{Concentration of HF} = \frac{0.727 \text{ moles}}{0.750 \text{ L}} = 0.969 \text{ M}
\]
2. **Using the Equilibrium Constant Expression:**
The equilibrium constant expression for this reaction is:
\[
\text{K}_c = \frac{[\text{HF}]^2}{[\text{H}_2][\text{F}_2]}
\]
3. **Solving for Concentrations:**
- Let the equilibrium concentrations of \(\text{H}_2\) and \(\text{F}_2\) be 0.100 M and each changes by \(-x\).
- Substitute the values in:
\[
123 = \frac{(0.969 - 2x)^2}{(0.100 - x)(0.100 - x)}
\]
**Solution Steps:**
- Substitute known values:
\[
123 = \frac{(0.969 - 2x)^2}{(0.100 - x)^2} = \frac{0.969^2 - 2(0.969)(2x) + (2x)^2}{0.100^2 - 0.200x + x^2}
\]
- The handwritten notes continue this process to derive the concentration values. Further simplification and calculation would reveal the exact equilibrium concentrations.
This example demonstrates the method for calculating concentrations at equilibrium using the equilibrium constant \( \text{K}_c](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb0ffec19-af41-4c50-af1b-66e883fc2bb8%2Fcc516aa3-f4ba-4f47-bf8c-c49c168ddcd8%2Fgirsxv9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Constant Calculation Example**
In this example, we determine the concentration of HF in a reaction at equilibrium. The given equilibrium constant (\( \text{K}_c \)) is 123 for the reaction:
\[
\text{H}_2(g) + \text{F}_2(g) \rightleftharpoons 2\text{HF}(g)
\]
**Initial Information Given:**
- Moles of HF: 0.727 moles
- Volume of the reaction vessel: 0.750 L
**Objective:**
Calculate the concentration of HF in the flask at equilibrium.
**Calculations:**
1. **Calculation of Initial Concentration:**
Moles of HF = 0.727 moles
Volume = 0.750 L
\[
\text{Concentration of HF} = \frac{0.727 \text{ moles}}{0.750 \text{ L}} = 0.969 \text{ M}
\]
2. **Using the Equilibrium Constant Expression:**
The equilibrium constant expression for this reaction is:
\[
\text{K}_c = \frac{[\text{HF}]^2}{[\text{H}_2][\text{F}_2]}
\]
3. **Solving for Concentrations:**
- Let the equilibrium concentrations of \(\text{H}_2\) and \(\text{F}_2\) be 0.100 M and each changes by \(-x\).
- Substitute the values in:
\[
123 = \frac{(0.969 - 2x)^2}{(0.100 - x)(0.100 - x)}
\]
**Solution Steps:**
- Substitute known values:
\[
123 = \frac{(0.969 - 2x)^2}{(0.100 - x)^2} = \frac{0.969^2 - 2(0.969)(2x) + (2x)^2}{0.100^2 - 0.200x + x^2}
\]
- The handwritten notes continue this process to derive the concentration values. Further simplification and calculation would reveal the exact equilibrium concentrations.
This example demonstrates the method for calculating concentrations at equilibrium using the equilibrium constant \( \text{K}_c
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