Use the following equilibria 2CH4(g) C2Hg(g) + H2(g) K1 = 9.5 × 10-13 CH4(g) + H20(g) CH3OH(g) + H2(g) Kcz = 2.8 × 10-21 to calculate the value of Ke for the following reaction: 2CH3OH(g) + H2(g) - C2Hg(g) + 2H20(g)
Use the following equilibria 2CH4(g) C2Hg(g) + H2(g) K1 = 9.5 × 10-13 CH4(g) + H20(g) CH3OH(g) + H2(g) Kcz = 2.8 × 10-21 to calculate the value of Ke for the following reaction: 2CH3OH(g) + H2(g) - C2Hg(g) + 2H20(g)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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![**Equilibrium Calculations for Reaction Kc'**
In this exercise, we use two given equilibrium reactions to calculate the equilibrium constant, \( K_c' \), for a third reaction.
1. **Given Equilibria:**
- **Reaction 1:**
\[
2\text{CH}_4(g) \rightleftharpoons \text{C}_2\text{H}_6(g) + \text{H}_2(g) \quad K_{c1} = 9.5 \times 10^{-13}
\]
- **Reaction 2:**
\[
\text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CH}_3\text{OH}(g) + \text{H}_2(g) \quad K_{c2} = 2.8 \times 10^{-21}
\]
2. **Target Reaction for Calculation:**
- **Reaction:**
\[
2\text{CH}_3\text{OH}(g) + \text{H}_2(g) \rightleftharpoons \text{C}_2\text{H}_6(g) + 2\text{H}_2\text{O}(g)
\]
3. **Calculation Required:**
- Calculate \( K_c' \) for the above reaction using the provided equilibrium constants.
**Note:** Apply principles of chemical equilibrium and algebraic manipulation to find \( K_c' \) based on \( K_{c1} \) and \( K_{c2} \).
**Diagram Explanation:**
There are no diagrams or graphs accompanying this text.
**Equation Box:**
\[ K_c' = \boxed{} \]
Use the values and relationships of the given equilibria to solve for the unknown \( K_c' \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdad50ce8-459c-4858-95e9-e24f7f9ebbc0%2F81108f26-7f18-4abe-99ba-c030189dd92b%2Fhim9yc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Calculations for Reaction Kc'**
In this exercise, we use two given equilibrium reactions to calculate the equilibrium constant, \( K_c' \), for a third reaction.
1. **Given Equilibria:**
- **Reaction 1:**
\[
2\text{CH}_4(g) \rightleftharpoons \text{C}_2\text{H}_6(g) + \text{H}_2(g) \quad K_{c1} = 9.5 \times 10^{-13}
\]
- **Reaction 2:**
\[
\text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CH}_3\text{OH}(g) + \text{H}_2(g) \quad K_{c2} = 2.8 \times 10^{-21}
\]
2. **Target Reaction for Calculation:**
- **Reaction:**
\[
2\text{CH}_3\text{OH}(g) + \text{H}_2(g) \rightleftharpoons \text{C}_2\text{H}_6(g) + 2\text{H}_2\text{O}(g)
\]
3. **Calculation Required:**
- Calculate \( K_c' \) for the above reaction using the provided equilibrium constants.
**Note:** Apply principles of chemical equilibrium and algebraic manipulation to find \( K_c' \) based on \( K_{c1} \) and \( K_{c2} \).
**Diagram Explanation:**
There are no diagrams or graphs accompanying this text.
**Equation Box:**
\[ K_c' = \boxed{} \]
Use the values and relationships of the given equilibria to solve for the unknown \( K_c' \).
![At 25°C, \( K_c = 1 \times 10^{-85} \) for the reaction
\[ 7IO_3^-(aq) + 9H_2O + 7H^+(aq) \rightleftharpoons I_2(aq) + 5H_5IO_6(aq) \]
What is the value of \( K_c \) for the following reaction?
\[ I_2(aq) + 5H_5IO_6(aq) \rightleftharpoons 7IO_3^-(aq) + 9H_2O + 7H^+(aq) \]
\( K_c = \) _______](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdad50ce8-459c-4858-95e9-e24f7f9ebbc0%2F81108f26-7f18-4abe-99ba-c030189dd92b%2Fjop6duj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:At 25°C, \( K_c = 1 \times 10^{-85} \) for the reaction
\[ 7IO_3^-(aq) + 9H_2O + 7H^+(aq) \rightleftharpoons I_2(aq) + 5H_5IO_6(aq) \]
What is the value of \( K_c \) for the following reaction?
\[ I_2(aq) + 5H_5IO_6(aq) \rightleftharpoons 7IO_3^-(aq) + 9H_2O + 7H^+(aq) \]
\( K_c = \) _______
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