Take equation 1 and add it to the reverse of equation 2 and add that to the reverse of 1/2 equation 3.
When reversing dH, change the sign.
When taking 1/2 equation, also take 1/2 dH.
Calculate the enthalpy change for the reactions
NO(g) + O(g) → NO2(g)
from the following data:
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -198.9 kJ
O3(g) → 1.5O2(g) ΔH = -142.3 kJ
O2(g) → 2O(g) ΔH = 495.0 kJ
Put the answer in kJ.
4 answers
530kj
-304.1kJ
NO+O3=NO2+O2 -198.9 kJ
O3=1.5 O2 -142.3 kJ
O2=2O 495.0 kJ
NO+O=NO2 ?
NO+O3=NO2+O2 -198.9 kJ
O3=1.5 O2 -142.3 kJ
2O=O2 -495.0 kJ
NO+O=NO2 ?
NO+O3=NO2+O2 -198.9 kJ
O3=1.5 O2 -142.3 kJ
1/2(2O=O2) 1/2(-495.0 kJ)
NO+O=NO2 ?
NO+ O̶3̶ = NO2+ ̶O̶2̶ -198.9 kJ
O̶3̶ =1̶.̶5̶O̶2̶ -142.3 kJ
O= ̶1̶/̶2̶ ̶O̶2̶ -247.5 kJ
NO+O=NO2 =-588.7 kJ
O3=1.5 O2 -142.3 kJ
O2=2O 495.0 kJ
NO+O=NO2 ?
NO+O3=NO2+O2 -198.9 kJ
O3=1.5 O2 -142.3 kJ
2O=O2 -495.0 kJ
NO+O=NO2 ?
NO+O3=NO2+O2 -198.9 kJ
O3=1.5 O2 -142.3 kJ
1/2(2O=O2) 1/2(-495.0 kJ)
NO+O=NO2 ?
NO+ O̶3̶ = NO2+ ̶O̶2̶ -198.9 kJ
O̶3̶ =1̶.̶5̶O̶2̶ -142.3 kJ
O= ̶1̶/̶2̶ ̶O̶2̶ -247.5 kJ
NO+O=NO2 =-588.7 kJ
1 answer