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Use the standard reaction enthalpies to determine delta h rxn...?
Use the standard reaction enthalpies given below to determine delta h rxn for the following reaction:
2NO + O2 --> 2NO2
Given:
N2 + O2 --> 2NO delta h rxn = +183 kJ
1/2N2 + O2 --> NO2 delta h rxn = +33 kJ
If someone could explain this to me and show steps, I would be very grateful! I don't understand this "heat" stuff.
3 Answers
- PhilippeLv 710 years agoFavorite Answer
Hi Cupcake,
I am french (Boulogne sur mer 62200 - FRANCE)
N2 + O2 --> 2NO . . . . . . .ΔH = +183 kJ
. . . then :
2 NO --> N2 + O2 . . . . . . ΔH = - 183 kJ
1/2N2 + O2 --> NO2 . . . . . . . . . . ΔH = + 33 kJ
. . . then :
. . . . . 2 x ( 1/2N2 + O2 --> NO2 ) . . . . . ΔH = (2 x 33) = + 66 kJ
. . or . . . . . . N2 + 2 O2 --> 2 NO2 . . . . .ΔH = (2 x 33) = + 66 kJ
. . . finally :
. . . . 2 NO --> N2 + O2 . . . . . . ΔH = - 183 kJ
N2 + 2 O2 --> 2 NO2 . . . . . . . .ΔH = (2 x 33) = + 66 kJ
- - - - - - - - - - - - - - - - - - - -
2NO + O2 --> 2NO2 . . . . . . . . ΔH = - 183 + 66 = - 117 kJ
I hope to have answered your question.
.
- Anonymous6 years ago
Question, so how do you get O2 by itself in the final reaction, that s getting me. Do you use the 2O2 from the second given and just ignore the 2 Moles?
- 6 years ago
You can cancel out one O2 from each equation so then your leftover will be 02 from the second equation. (You only remove one from each equation.
