Discussion Overview
The discussion revolves around determining the standard enthalpy change for the reaction 4NH3 + 3O2 --> 2N2 + 6H2O. Participants explore the implications of using different standard heats of formation for water and the significance of specifying the physical states of the reactants and products.
Discussion Character
- Homework-related
- Debate/contested
- Technical explanation
Main Points Raised
- One participant presents a calculation for the standard enthalpy change using the given heats of formation, resulting in -1531 kJ/mol.
- Another participant challenges the value used for the standard heat of formation of water, suggesting that -285.83 kJ/mol refers to liquid water, while the gaseous value is -241.826 kJ/mol.
- Several participants note that the reaction does not specify the physical state of water, leading to ambiguity in the calculation.
- There is a discussion about whether the standard state of water is typically considered as liquid, with some participants suggesting that it is acceptable to assume gaseous water in this context.
- One participant raises the possibility of an error in the quiz grading, indicating that there may be discrepancies in how the problem was interpreted.
- There is uncertainty regarding whether the enthalpy change should be divided by four, with differing opinions on whether this is necessary based on the reaction stoichiometry.
Areas of Agreement / Disagreement
Participants express differing views on the appropriate value for the standard heat of formation of water and whether the physical state of water should be assumed. The discussion remains unresolved regarding the correct approach to calculating the enthalpy change and the conventions for reporting it.
Contextual Notes
There are limitations regarding the assumptions made about the physical states of the reactants and products, as well as the lack of clarity in the problem statement about whether the enthalpy change should be reported per mole of a specific reactant.
