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

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

2 NO + 2 O --> 2NO2 TRIANGLE H= -466 kJ
Have to simplify TRIANGLE H= -233kJ

Hope you understand

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