Hess's law states that the total enthalpy change for a reaction is independent of the pathway between the initial and final states. To calculate the enthalpy change of a reaction, standard enthalpy change values can be used for the elementary steps that add up to the overall reaction through algebraic manipulation. For the reaction of 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), using standard enthalpy change values for the elementary steps gives an overall exothermic enthalpy change of -906.3 kJ.

1. 1
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.

2. 2
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.

3. 3
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
Goal:
NH3:
O2 :
NO:
H2O:
Reverse and x 2 4NH3  2N2 + 6H2 ∆H = +183.6 kJ
Found in more than one place, SKIP IT (its hard).
x2 2N2 + 2O2  4NO ∆H = 361.2 kJ
x3 6H2 + 3O2  6H2O ∆H = -1451.1 kJ

4. 4
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)Goal:
NH3:
O2 :
NO:
H2O:
Reverse and x2 4NH3  2N2 + 6H2 ∆H = +183.6 kJ
Found in more than one place, SKIP IT.
x2 2N2 + 2O2  4NO ∆H = 361.2 kJ
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?

5. 5
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.

6. 6
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

7. 7
Summary:
enthalpy is a state function
and is path independent.

8. 8

9. 9
Standard Enthalpies of formation:

10. 10
Thermodynamic Quantities of Selected Substances @ 298.15 K