Enthalpy of formation

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  • #1
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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!
 

Answers and Replies

  • #2
DH
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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
 
  • #3
2
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heat of formation for lactic acid
 
  • #4
2
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what is the standard value of heat of formation for lactic acid
 

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