3. 0.250 moles of H2 and 0.300 moles of I2 are placed in a 4.25 liter container. At a temperature for which the equilibrium constant Ke is 54.0, determine the equilibrium concentration of HI(g).
3. 0.250 moles of H2 and 0.300 moles of I2 are placed in a 4.25 liter container. At a temperature for which the equilibrium constant Ke is 54.0, determine the equilibrium concentration of HI(g).
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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![**Problem Statement:**
0.250 moles of H₂ and 0.300 moles of I₂ are placed in a 4.25 liter container. At a temperature for which the equilibrium constant \( K_c \) is 54.0, determine the equilibrium concentration of HI(g).
**Explanation:**
This problem involves a chemical equilibrium scenario where hydrogen gas (H₂) and iodine gas (I₂) react to form hydrogen iodide (HI) gas. The equilibrium constant \( K_c \) is given, and we are asked to find the equilibrium concentration of HI(g).
### Steps to Solve:
1. **Write the Balanced Equation:**
\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \]
2. **Initial Concentrations:**
\[
[\text{H}_2]_0 = \frac{0.250 \text{ moles}}{4.25 \text{ L}} = 0.0588 \text{ M}
\]
\[
[\text{I}_2]_0 = \frac{0.300 \text{ moles}}{4.25 \text{ L}} = 0.0706 \text{ M}
\]
\[
[\text{HI}]_0 = 0 \text{ M (initially no HI)}
\]
3. **Change in Concentrations (Using ICE Table):**
\[
\begin{array}{c|c|c|c}
& \text{H}_2(g) & \text{I}_2(g) & \text{HI}(g) \\
\hline
\text{Initial (M)} & 0.0588 & 0.0706 & 0 \\
\text{Change (M)} & -x & -x & +2x \\
\text{Equilibrium (M)} & 0.0588-x & 0.0706-x & 2x \\
\end{array}
\]
4. **Equilibrium Expression:**
\[
K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = 54.0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16bfa673-9403-48ea-8f28-dfab5d100b97%2F9b195518-cb34-4e04-8b35-84989d667666%2Ficz7cd7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
0.250 moles of H₂ and 0.300 moles of I₂ are placed in a 4.25 liter container. At a temperature for which the equilibrium constant \( K_c \) is 54.0, determine the equilibrium concentration of HI(g).
**Explanation:**
This problem involves a chemical equilibrium scenario where hydrogen gas (H₂) and iodine gas (I₂) react to form hydrogen iodide (HI) gas. The equilibrium constant \( K_c \) is given, and we are asked to find the equilibrium concentration of HI(g).
### Steps to Solve:
1. **Write the Balanced Equation:**
\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \]
2. **Initial Concentrations:**
\[
[\text{H}_2]_0 = \frac{0.250 \text{ moles}}{4.25 \text{ L}} = 0.0588 \text{ M}
\]
\[
[\text{I}_2]_0 = \frac{0.300 \text{ moles}}{4.25 \text{ L}} = 0.0706 \text{ M}
\]
\[
[\text{HI}]_0 = 0 \text{ M (initially no HI)}
\]
3. **Change in Concentrations (Using ICE Table):**
\[
\begin{array}{c|c|c|c}
& \text{H}_2(g) & \text{I}_2(g) & \text{HI}(g) \\
\hline
\text{Initial (M)} & 0.0588 & 0.0706 & 0 \\
\text{Change (M)} & -x & -x & +2x \\
\text{Equilibrium (M)} & 0.0588-x & 0.0706-x & 2x \\
\end{array}
\]
4. **Equilibrium Expression:**
\[
K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = 54.0
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