Calc Enthalpy of Formation N2O5(g) from Data

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Discussion Overview

The discussion revolves around calculating the standard enthalpy of formation of N2O5(g) using provided thermodynamic data for various reactions involving nitrogen oxides. The focus is on the application of Hess's law and enthalpy changes associated with chemical reactions.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant presents a method to calculate the standard enthalpy of formation of N2O5(g) by manipulating the given reactions and applying Hess's law.
  • The proposed calculation involves scaling the first and third reactions and combining them to derive the enthalpy change for the formation of N2O5.
  • The calculated value for the standard enthalpy of formation of N2O5 is presented as 11.3 kJ/mol.
  • Other participants introduce unrelated topics, such as the heat of formation for lactic acid, which does not contribute to the main discussion.

Areas of Agreement / Disagreement

The discussion primarily features one participant's calculation method without explicit agreement or disagreement from others regarding the correctness of the approach or the final value. The introduction of unrelated topics suggests a lack of focus on the original question.

Contextual Notes

The calculation relies on specific enthalpy values and assumes that the reactions can be combined as proposed. The validity of the method and the final result may depend on the accuracy of the provided data and the assumptions made during the calculation.

Who May Find This Useful

This discussion may be useful for students or individuals interested in thermodynamics, particularly those looking to understand the application of Hess's law in calculating enthalpy changes for chemical reactions.

chemister
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How would you calculate the standard enthalpy of formation of N2O5(g) in kJ/mol from the following data (all species are gases):
2 NO + O2-----2 NO2 ∆H° = –114.1 kJ
4 NO2 + O2-----2 N2O5 ∆H° = –110.2 kJ
N2 + O2------2 NO ∆H° = +180.5 kJ

Any help is greatly appreciated!
 
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chemister said:
How would you calculate the standard enthalpy of formation of N2O5(g) in kJ/mol from the following data (all species are gases):
2 NO + O2-----2 NO2 ∆H° = –114.1 kJ
4 NO2 + O2-----2 N2O5 ∆H° = –110.2 kJ
N2 + O2------2 NO ∆H° = +180.5 kJ

Any help is greatly appreciated!
ok, here is your answer. this one is the easy one!
time first and third one with 2, then add three of them up, we have:
4 NO + 2O2 + 4NO2 + O2 + 2N2 + 2O2 -> 4NO2 +2N2O5 + 4NO

5O2 + 2N2 -> 2N2O5 ∆H° = 2*(-114.1) -110.2 + 2*180.5 = 22.6
-> standard enthalpy of formation of N2O5 = 22.6/2 = 11.3 kJ
 
heat of formation for lactic acid
 
what is the standard value of heat of formation for lactic acid
 

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