reaction AH 4CO, (g) + 4C,H,OH(1) → 2C,H,,0,(s) 2C,H,.0 H120;(s) O kJ 4C,H,,0,(s) 8CO, (g) + 8C, H,OH(1) O kJ 6 12 2' 2C0, (g) + 2C, H,OH() → C,H,,0,(s) 69. kJ 12 6
reaction AH 4CO, (g) + 4C,H,OH(1) → 2C,H,,0,(s) 2C,H,.0 H120;(s) O kJ 4C,H,,0,(s) 8CO, (g) + 8C, H,OH(1) O kJ 6 12 2' 2C0, (g) + 2C, H,OH() → C,H,,0,(s) 69. kJ 12 6
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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![**Thermochemistry: Using the General Properties of Reaction Enthalpy**
A chemist measures the enthalpy change (ΔH) during the following reaction:
\[ \text{C}_6\text{H}_{12}\text{O}_6(s) \rightarrow 2\text{CO}_2(g) + 2\text{C}_2\text{H}_5\text{OH}(l) \]
\[ \Delta H = -69. \text{kJ} \]
Use this information to complete the table below. Round each of your answers to the nearest kJ/mol.
| Reaction | ΔH |
|------------------------------------------------------------------------------------------------------|------------|
| \( 4\text{CO}_2(g) + 4\text{C}_2\text{H}_5\text{OH}(l) \rightarrow 2\text{C}_6\text{H}_{12}\text{O}_6(s) \) | □ kJ |
| \( 4\text{C}_6\text{H}_{12}\text{O}_6(s) \rightarrow 8\text{CO}_2(g) + 8\text{C}_2\text{H}_5\text{OH}(l) \) | □ kJ |
| \( 2\text{CO}_2(g) + 2\text{C}_2\text{H}_5\text{OH}(l) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(s) \) | 69 kJ |
Buttons: Explanation, Check
**Explanation:**
- The first reaction is the reverse of the given reaction, multiplied by 2. Therefore, ΔH should be \( +69 \text{ kJ} \times 2 = +138 \text{ kJ} \).
- The second reaction is the given reaction, multiplied by 4. Thus, ΔH should be \( -69 \text{ kJ} \times 4 = -276 \text{ kJ} \).
- The third reaction is already provided with its enthalpy change, confirming the value.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F69bf82c5-0864-4aa2-8974-c55ce610672b%2F887ea6bf-6a29-414b-a9b7-24ea304f202b%2Fhxhlgjr.jpeg&w=3840&q=75)
Transcribed Image Text:**Thermochemistry: Using the General Properties of Reaction Enthalpy**
A chemist measures the enthalpy change (ΔH) during the following reaction:
\[ \text{C}_6\text{H}_{12}\text{O}_6(s) \rightarrow 2\text{CO}_2(g) + 2\text{C}_2\text{H}_5\text{OH}(l) \]
\[ \Delta H = -69. \text{kJ} \]
Use this information to complete the table below. Round each of your answers to the nearest kJ/mol.
| Reaction | ΔH |
|------------------------------------------------------------------------------------------------------|------------|
| \( 4\text{CO}_2(g) + 4\text{C}_2\text{H}_5\text{OH}(l) \rightarrow 2\text{C}_6\text{H}_{12}\text{O}_6(s) \) | □ kJ |
| \( 4\text{C}_6\text{H}_{12}\text{O}_6(s) \rightarrow 8\text{CO}_2(g) + 8\text{C}_2\text{H}_5\text{OH}(l) \) | □ kJ |
| \( 2\text{CO}_2(g) + 2\text{C}_2\text{H}_5\text{OH}(l) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(s) \) | 69 kJ |
Buttons: Explanation, Check
**Explanation:**
- The first reaction is the reverse of the given reaction, multiplied by 2. Therefore, ΔH should be \( +69 \text{ kJ} \times 2 = +138 \text{ kJ} \).
- The second reaction is the given reaction, multiplied by 4. Thus, ΔH should be \( -69 \text{ kJ} \times 4 = -276 \text{ kJ} \).
- The third reaction is already provided with its enthalpy change, confirming the value.
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