Chemistry
wilson12
22

Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) 1/2N2(g)+O2(g)→NO2(g), ΔH∘A=33.2 kJ 1/2N2(g)+12O2(g)→NO(g), ΔH∘B=90.2 kJ

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(1) Answers
Jesibobo1998

1/2 N2(g) + O2(g) --> NO2(g), deltaH = 33.2 kJ (should be multiplied by 2)  N2 + 2 O2 -> 2 NO2.........deltaH = +66,4 kJ  1/2 N2(g) + 1/2O2(g) --> NO(g), delta H = 90.2 kJ (should be multiplied by 2 and inverted)  2 NO -> N2 + O2..............deltaH = -180,4 kJ  Now you have:  N2 + 2 O2 -> 2 NO2.........deltaH = +66,4 kJ  2 NO -> N2 + O2..............deltaH = -180,4 kJ  --------------------------------------...  canceling both N2 and 2 O2 with O2, you will find the given reaction:  2 NO + O2 -> 2 NO2........deltaH = 66,4 - 180,4 = -114.0 kJ <--- Answer 

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