[caleb]

ok here is the problem.

calculate delta H for the reaction

           NO+O -----  NO2

GIVEN THE FOLLOWING INFO:

NO+ O3---NO2+O2    DELTA H = -198.9 KJ

O3----3/2 O2            DELTA H = -142.3 KJ

O2----2O                  DELTA H = 495.0 KJ

ANSWER - 304.1KJ

OK I JUST CAN FIGURE OUT WHY THEY ARE TURING O3--- 3/2 O2 AROUND. THAT IS TEH PROBLEM I AM HAVEING.YOU LEAVE THE FIRST ONE ALONE BECAUSE NO AND NO2 ARE ON TEH SIDES THEY NEED TO BE ON BUT WHY MOVE TEH SECONE ONE .

THANKS:CALEB

[savoy7]

Caleb with info given

NO + O₃ ---- NO₂ +  O₂             -198.9 KJ/mole

   3/2 O₂ ---- O₃                      +142.3 KJ/mole

        O  ---- 1/2 O₂                 -247.5 KJ/mole
_______________________________________________________________________

  NO + O ---- NO₂                     -304.1 KJ/mole

when a reactant and a product are the same, they (in a sense) can be eliminated from the equation.  Both ozone and oxygen gas were not part of the final equation, so they needed to be "canceled".  That is why you had to switch the 2nd equation and why you needed half of the final equation.