5. In question 1, you calculated the enthalpy change for the combustion of propane from combining 3 reactions with known enthalpy changes using Hess's Law. Alternatively, standard heats of formation are a general method to find the enthalpy change in ANY reaction from information that can be looked up in a table. CH₂(g) -103.85 -393.5 -285.8 -241.8 Use the tabulated heats of formation and the equation connecting standard enthalpy change of reaction to standard enthalpy change of formation of reactants and products to calculate AH for the combustion of propane in kJ/mol. Compare this with your answer to 1. b) Why did you not need a tabulated value of AH', for O₂ (g)? CO₂ (g) H₂O (1) H₂O(g) a)
5. In question 1, you calculated the enthalpy change for the combustion of propane from combining 3 reactions with known enthalpy changes using Hess's Law. Alternatively, standard heats of formation are a general method to find the enthalpy change in ANY reaction from information that can be looked up in a table. CH₂(g) -103.85 -393.5 -285.8 -241.8 Use the tabulated heats of formation and the equation connecting standard enthalpy change of reaction to standard enthalpy change of formation of reactants and products to calculate AH for the combustion of propane in kJ/mol. Compare this with your answer to 1. b) Why did you not need a tabulated value of AH', for O₂ (g)? CO₂ (g) H₂O (1) H₂O(g) a)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
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5. In question 1, you calculated the enthalpy change for the combustion of propane from combining 3 reactions with known enthalpy changes using Hess’s Law. Alternatively, standard heats of formation are a general method to find the enthalpy change in any reaction from information that can be looked up in a table.
| Compound | ΔHf° (kJ/mol) |
|----------|---------------|
| C3H8 (g) | -103.85 |
| CO2 (g) | -393.5 |
| H2O (l) | -285.8 |
| H2O (g) | -241.8 |
a) Use the tabulated heats of formation and the equation connecting standard enthalpy change of reaction to standard enthalpy change of formation of reactants and products to calculate ΔHrxn for the combustion of propane in kJ/mol. Compare this with your answer to 1.
b) Why did you not need a tabulated value of ΔHf° for O2 (g)?
---
**Explanation:**
- The table lists standard enthalpy of formation values (ΔHf°) for specific chemicals involved in the combustion of propane. These values are in kilojoules per mole (kJ/mol).
- These values are used to calculate the enthalpy change (ΔHrxn) of the reaction by applying the formula:
\[
\Delta H_{rxn} = \sum (\Delta H_{f}^{\circ} \text{ of products}) - \sum (\Delta H_{f}^{\circ} \text{ of reactants})
\]
- The question also asks why the standard enthalpy of formation for oxygen gas (O2) is not needed, which is because the standard enthalpy of formation for any element in its standard state is zero.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe361a5c2-e3a6-4cd8-bee8-5dcb720cb10b%2F930eb789-fe08-4e45-8523-59dc1d805452%2F4g1o7q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website**
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5. In question 1, you calculated the enthalpy change for the combustion of propane from combining 3 reactions with known enthalpy changes using Hess’s Law. Alternatively, standard heats of formation are a general method to find the enthalpy change in any reaction from information that can be looked up in a table.
| Compound | ΔHf° (kJ/mol) |
|----------|---------------|
| C3H8 (g) | -103.85 |
| CO2 (g) | -393.5 |
| H2O (l) | -285.8 |
| H2O (g) | -241.8 |
a) Use the tabulated heats of formation and the equation connecting standard enthalpy change of reaction to standard enthalpy change of formation of reactants and products to calculate ΔHrxn for the combustion of propane in kJ/mol. Compare this with your answer to 1.
b) Why did you not need a tabulated value of ΔHf° for O2 (g)?
---
**Explanation:**
- The table lists standard enthalpy of formation values (ΔHf°) for specific chemicals involved in the combustion of propane. These values are in kilojoules per mole (kJ/mol).
- These values are used to calculate the enthalpy change (ΔHrxn) of the reaction by applying the formula:
\[
\Delta H_{rxn} = \sum (\Delta H_{f}^{\circ} \text{ of products}) - \sum (\Delta H_{f}^{\circ} \text{ of reactants})
\]
- The question also asks why the standard enthalpy of formation for oxygen gas (O2) is not needed, which is because the standard enthalpy of formation for any element in its standard state is zero.
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