C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(I) + 2800 kJ Occasionally, not all values are found in the table of thermodynamic data. For most substances it is impossible to go into a lab and directly synthesize a compound from its free elements. The heat of formation for the substance must be calculated by working backwards from its heat of combustion. Calculate the AHf of C6H1206(s) given the combustion reaction above along with the following information. Substance AH; ° (kJ/mol) CO2(g) -393.5 H2O(1) -285.8
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(I) + 2800 kJ Occasionally, not all values are found in the table of thermodynamic data. For most substances it is impossible to go into a lab and directly synthesize a compound from its free elements. The heat of formation for the substance must be calculated by working backwards from its heat of combustion. Calculate the AHf of C6H1206(s) given the combustion reaction above along with the following information. Substance AH; ° (kJ/mol) CO2(g) -393.5 H2O(1) -285.8
Chemistry
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Chapter1: Chemical Foundations
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![**Combustion Reaction of \( \text{C}_6\text{H}_{12}\text{O}_6 (\text{s})\):**
The reaction is as follows:
\[ \text{C}_6\text{H}_{12}\text{O}_6(\text{s}) + 6 \text{O}_2(\text{g}) \rightarrow 6 \text{CO}_2(\text{g}) + 6 \text{H}_2\text{O}(\text{l}) + 2800 \text{ kJ} \]
**Understanding Heat Formation and Combustion:**
Occasionally, not all values are found in the table of thermodynamic data. For most substances, it is impossible to go into a lab and directly synthesize a compound from its free elements. The heat of formation for the substance must be calculated by working backwards from its heat of combustion.
**Problem:**
Calculate the \( \Delta H_f \) of \( \text{C}_6\text{H}_{12}\text{O}_6(\text{s}) \) given the combustion reaction above along with the following information.
**Data Table:**
| Substance | \( \Delta H_f^\circ \) (kJ/mol) |
|-----------|--------------------|
| \( \text{CO}_2(\text{g}) \) | \(-393.5\) |
| \( \text{H}_2\text{O}(\text{l}) \) | \(-285.8\) |](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90b2ad7e-71be-4993-92e0-fe2030354ee3%2F8dda0471-3b3a-43e5-a5da-46889ac7f9bf%2Fq48xv1o_processed.png&w=3840&q=75)
Transcribed Image Text:**Combustion Reaction of \( \text{C}_6\text{H}_{12}\text{O}_6 (\text{s})\):**
The reaction is as follows:
\[ \text{C}_6\text{H}_{12}\text{O}_6(\text{s}) + 6 \text{O}_2(\text{g}) \rightarrow 6 \text{CO}_2(\text{g}) + 6 \text{H}_2\text{O}(\text{l}) + 2800 \text{ kJ} \]
**Understanding Heat Formation and Combustion:**
Occasionally, not all values are found in the table of thermodynamic data. For most substances, it is impossible to go into a lab and directly synthesize a compound from its free elements. The heat of formation for the substance must be calculated by working backwards from its heat of combustion.
**Problem:**
Calculate the \( \Delta H_f \) of \( \text{C}_6\text{H}_{12}\text{O}_6(\text{s}) \) given the combustion reaction above along with the following information.
**Data Table:**
| Substance | \( \Delta H_f^\circ \) (kJ/mol) |
|-----------|--------------------|
| \( \text{CO}_2(\text{g}) \) | \(-393.5\) |
| \( \text{H}_2\text{O}(\text{l}) \) | \(-285.8\) |
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