1. Chem II Chapter 5
Hess’s Law

2. N2 (g) + 2O2 (g) ------> 2NO2 (g) DH1 = 68kJ
The equation below….
N2 (g) + 2O2 (g) > 2NO2 (g) DH1 = 68kJ
….is derived from a combination of two other equations….
N2 (g) + O2 (g) > 2NO(g) DH2 = 180kJ
2NO(g) + O2 (g) > 2NO2 (g) DH3 =-112kJ

3. Notice that when the two equations are added together….
N2 (g) + O2 (g) > 2NO(g) DH2 = 180kJ +
2NO(g) + O2 (g) > 2NO2 (g) DH3 =-112kJ
….that the NO molecules cancel.

4. …after eliminating the NO’s on both sides…
N2 (g) + O2 (g) > DH2 = 180kJ +
O2 (g) > 2NO2 (g) DH3 =-112kJ
...now, add up what is leftover…..
N2 (g) + O2 (g) + O2 (g) > 2NO2 (g)
…we get the original equation
N2 (g) + 2O2 (g) > 2NO2 (g) DH1 = 68kJ

5. Let’s try another equation……
W(s) + C(graphite) > WC(s) DH=?? kJ
Using the following three equations….
2W(s) + 3O2 (g) > 2WO3(s) DH= kJ
C(graphite) + O2 (g) > CO2 (g) DH= kJ
2WC(s) + 5O2 (g) > 2WO3 (s) + 2CO2(g) DH= kJ
How can these equations be converted derive the original equation???

6. Dealing with the first equation…
2W(s) + 3O2 (g) > 2WO3(s) DH= kJ
…..need to keep W, but find a way to
eliminate the 3O2 (g)+ 2WO3(s) .
Take ½ of everything, including that of DH= kJ.
½ [2W(s) + 3O2 (g) > 2WO3(s) DH= kJ]
This leaves…
W(s) + 3/2O2 (g) > WO3(s) DH= kJ

7. Now the 2nd equation…
C(graphite) + O2 (g) > CO2 (g) DH= kJ
…need to keep C, but not O2 (g) and CO2 (g)
No need to alter the equation. The terms
O2 (g) and CO2 (g) will be eliminated later.

8. The 3rd equation requires a lot of work.
2WC(s) + 5O2 (g) > 2WO3 (s) + 2CO2(g) DH= kJ
1st, WC needs to become a product and everything else
eliminated. Additionally, take ½ of everything, including
DH= kJ.
Reverse the equation….
2WO3 (s) + 2CO2(g) > 2WC(s) + 5O2 (g) DH= kJ
…..then, take ½ of everything….
½ [2WO3 (s) + 2CO2(g) > 2WC(s) + 5O2 (g) DH= kJ]
This leaves…
WO3 (s) + CO2(g) > WC(s) + 5/2 O2 (g) DH= kJ

9. Now, put the three equations along with their
Enthalpy values. Notice all of the terms that
cancel out.
W(s) + 3/2O2 (g) > WO3(s) DH= kJ
C(graphite) + O2 (g) > CO2 (g) DH= kJ
WO3 (s) + CO2(g) > WC(s) + 5/2 O2 (g) DH= kJ
(2 ½ O2 molecules = 5/2 O2 molecules )
W(s) + C(graphite) > WC(s) DH= kJ