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Calculate the ∆H for the following reaction:
NO + O à NO2

You are given these three equations:
O2 à 2O ∆H = +495 kJ

2O3 à 3O2 ∆H = -427 kJ

NO + O3 à NO2 + O2 ∆H = -199 kJ

Respuesta :

jashae2003
As "O" is a react = FLIP FLIP to cancel "O2 & O3"

2O --> O2- TRIANGLE H = -495kJ
3O2 --> 2O3 TRIANGLE H = + 427 kJ
2NO + 2O3 --> 2NO2 + 2O2 TRIANGLE H = -398 kJ
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2 NO + 2 O --> 2NO2 TRIANGLE H= -466 kJ
Have to simplify TRIANGLE H= -233kJ

Hope you understand

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