1. Lecture 4: Hess’s Law
Reading: Zumdahl 9.5
Outline:
Definition of Hess’ Law
Using Hess’ Law (examples)

2. Q: What is Hess’s Law ?
• Recall (lecture 3) Enthalpy is a state function.
As such, H for going from some initial state to
some final state is pathway independent.
• Hess’s Law: H for a process involving the
transformation of reactants into products is not
dependent on pathway.
(This means we can calculate H for a reaction by a
single step, or by multiple steps)

4. Using Hess’s Law
N2 (g) + 2O2 (g) 2NO 2 (g)
When calculating H for
a chemical reaction as a
2NO 2 (g)
single step, we can use
combinations of
q reactions as “pathways”
N2 (g) + 2O2 (g)
to determine H for our
“single step” reaction.

5. The reaction of interest is:
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
This reaction can also be carried out in two steps:
N2 (g) + O2 (g) 2NO(g) H = +180 kJ
2NO (g) + O2(g) 2NO2(g) H = -112 kJ

6. • If we take the previous two reactions and add
them, we get the original reaction of interest:
N2 (g) + O2 (g) 2NO(g) H = +180 kJ
2NO (g) + O2 (g) 2NO2(g) H = -112 kJ
N2 (g) + 2O2 (g) 2NO2(g) H = + 68 kJ

7. Note: the important things about this example is that
the sum of H for the two reaction steps is equal
to the H for the reaction of interest.
Big idea: We can combine reactions of known
H to determine the H for the
overall
“combined” reaction.

8. Hess’s Law: An Important Detail
One can always reverse the direction of a
reaction when making a combined reaction.
When you do this, the sign of H changes.
N2(g) + 2O2(g) 2NO2(g) H = +68 kJ
2NO2(g) N2(g) + 2O2(g) H = - 68 kJ

9. One more detail:
• The magnitude of H is directly proportional to the
quantities involved. (This means H is an “extensive”
quantity).
• So, if the coefficients of a reaction are multiplied by a
number, the value of H is also multiplied by the
same number.
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
N2(g) + 4O2(g) 4NO2(g) H = 136 kJ

10. Using Hess’s Law: tips
• When trying to combine reactions to form a
reaction of interest, it is usually best to work
backwards from the reaction of interest.
• Example:
What is H for the following reaction?
3C (gr) + 4H2 (g) C3H8 (g)

11. 3C (gr) + 4H2 (g) C3H8 (g) H=?
You’re given the following reactions:
C (gr) + O2 (g) CO2 (g) H = -394 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220 kJ
H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ

12. Step 1. Only reaction 1 has C (gr). Therefore, we
will multiply by 3 to get the correct amount of C
(gr) with respect to our final equation.
(x3) C (gr) + O2 (g) CO2 (g) H = -394 kJ
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ

13. Step 2: To get C3H8 on the product side of the
reaction, we need to reverse reaction 2, and change
the sign of H.
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220 kJ
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220 kJ

14. Step 3: Add two “new” reactions together to see what
remains:
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220 kJ
2
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ

15. • Step 4: Compare previous reaction to final
reaction, and determine how to reach final
reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ
3C (gr) + 4H2 (g) C3H8 (g)
Need to multiply second reaction by 4

16. Example (cont.)
• Step 4: Compare previous reaction to final
reaction, and determine how to reach final
reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ
3C (gr) + 4H2 (g) C3H8 (g)

17. 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ
3C (gr) + 4H2 (g) C3H8 (g) H = -106 kJ
Which is the one step reaction of interest

18. Another Example:
• Calculate H for the following reaction:
H2(g) + Cl2(g) 2HCl(g)
Given the following:
NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ
N2 (g) + 3H2 (g) 2NH3 (g) H = - 92 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = - 629 kJ

19. Step 1: Only the first reaction contains the product
of interest (HCl), but as a reactant.
Therefore, reverse this reaction and multiply by 2
to get stoichiometry correct.
NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ
2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = +352 kJ

20. Step 2: Need Cl2 as a reactant, therefore, add
reaction 3 to result from step 1 and see what is
left.
2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = 352 kJ
N2(g) + 4H2(g) + Cl2(g) 2NH4Cl(s) H = -629 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)
H = -277 kJ

21. Step 3: Use remaining known reaction in
combination with the result from Step 2 to get
final reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ
? ( N2 (g) + 3H2(g) 2NH3(g) H = -92 kJ)
H2(g) + Cl2(g) 2HCl(g) H=?
Key: need to reverse the middle reaction

22. • Step 3. Use remaining known reaction in
combination with the result from Step 2 to
get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ
1
2NH3(g) 3H2 (g) + N2 (g) H = +92 kJ
H2(g) + Cl2(g) 2HCl(g) H = -185 kJ
This is the desired reaction and resultant H!