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Use the standard reaction enthalpies given below to determine DH°rxn for the following reaction: 2 NO(g) + O2?
You need to reverse N2(g) + O2(g) → 2 NO(g) DH°rxn = +183 kJ to get NO(g) on the left
Doing so you need to reverse the sign of DHº giving DH°rxn = -183 kJ
To this equation 1/2 N2(g) + O2(g) → NO2(g) DH°rxn = +33 kJ you need to double it to get 2NO2
Doing so you need to double the DHº giving DH°rxn = +66 kJ
If you add the two changed formulas together it will reduce to 2 NO(g) + O2(g) → 2 NO2(g)
so then DH°rxn = -183 kJ + 66kJ
DH°rxn = -117kJ
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