Hess’s Law
Start
Finish
A State Function: Path independent.
Both lines accomplished the same result,
they went from start to finish.
Net result = same.
1
Determine the heat of reaction for the reaction:
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g)  2NO(g)
H = 180.6 kJ
N2(g) + 3H2(g)  2NH3(g)
H = -91.8 kJ
2H2(g) + O2(g)  2H2O(g)
H = -483.7 kJ
Hint: The three reactions must be algebraically
manipulated to sum up to the desired reaction.
and.. the H values must be treated accordingly. 2
Goal:
4NH3(g) + 5O2(g) 
4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g)  2NO(g)
H = 180.6 kJ
N2(g) + 3H2(g)  2NH3(g)
H = -91.8 kJ
2H2(g) +
NH3:
O2 :
NO:
H2O:
O2(g)  2H2O(g)
H = -483.7 kJ
Reverse and x 2 4NH3  2N2 + 6H2 H = +183.6 kJ
Found in more than one place, SKIP IT (its hard).
x2
x3
2N2 + 2O2  4NO
6H2 + 3O2  6H2O
H = 361.2 kJ
H = -1451.1 kJ
3
Goal:
4NH3(g) + 5O2(g) 
4NO(g) + 6H2O(g)
NH3: Reverse and x2 4NH3  2N2 + 6H2 H = +183.6 kJ
O2 : Found in more than one place, SKIP IT.
NO: x2
2N2 + 2O2  4NO
H = 361.2 kJ
H2O: x3
6H2 + 3O2  6H2O
H = -1451.1 kJ
Cancel terms and take sum.
4NH3
+ 5O2

4NO
+ 6H2O
H = -906.3 kJ
Is the reaction endothermic or exothermic?
4
Determine the heat of reaction for the reaction:
C2H4(g) + H2(g)  C2H6(g)
Use the following reactions:
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
H = -1401 kJ
C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l)
H = -1550 kJ
H2(g) +
1/2O2(g)  H2O(l)
H = -286 kJ
Consult your neighbor if necessary.
5
Determine the heat of reaction for the reaction:
Goal: C2H4(g) + H2(g)  C2H6(g)
H = ?
Use the following reactions:
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
H = -1401 kJ
C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l)
H = -1550 kJ
H2(g) + 1/2O2(g)  H2O(l)
H = -286 kJ
C2H4(g) :use 1 as is C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ
H2(g) :# 3 as is
H2(g) + 1/2O2(g)  H2O(l)
H = -286 kJ
C2H6(g) : rev #2
2CO2(g) + 3H2O(l)  C2H6(g) + 7/2O2(g) H = +1550 kJ
C2H4(g) + H2(g)  C2H6(g)
H = -137 kJ
6
Summary:
enthalpy is a state function
and is path independent.
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Standard Enthalpies of formation:
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Thermodynamic Quantities of Selected Substances @ 298.15 K
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