- #1

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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!

- Thread starter chemister
- Start date

- #1

- 21

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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!

- #2

- 21

- 0

ok, here is your answer. this one is the easy one!chemister said:

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!

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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