[halthedog]

hi:

Thanks for the help on the last question.  I am stumped on the following question-i think maybe there's a typo in the acs test book.

question:
What is standard enthalpy change for this reaction:
3H2(g) + O3(g) > 3H2O(l)

given:
1)     H2(g) + 1/2 O2(g) > H2O(l)        = -286kJ
2)     3O2(g) > 2 O3(g)                      = +271kJ
_____________________________________________________________
The acs book i am using calculates the answer to be -944 kJ.
I calculate -722kJ- which is one of the answer options - but not the correct answer according to the key.

I understand how the book came to -944kJ, but isn't the 2 equation
   " 3O2(g) > 2 O3(g)                = +271kJ"
suppose to be reversed - as O3(g) is on the products side of the equation that I'm trying to determing the standard enthalpy change = 3H2(g) + O3(g) > 3H2O(l) and there for the +271kJ should be switched to -271kJ.

Thanks for the help.  I appreciate it.

[mrdeadman]

H_total = sum of H_products - sum of H_reactants elements H's are zero

[plu]

To obtain the designed reaction equation, one must multiply equation (1) by a factor of 3 and equation (2) by a factor of -0.5 .  This will give you an overall enthalpy value of -994 kJ.