1. Enthalpy and Hess’s Law
By Shawn P. Shields, Ph.D.
This work is licensed by Shawn P. Shields-Maxwell under a
Creative Commons Attribution-NonCommercial-ShareAlike

2. Hess’s Law and the Conservation
of Energy
Since enthalpy (H) is a state function,
the change in the enthalpy of reaction
(∆Hrxn) does not depend on how
reactants are converted to products (i.e.,
in one step, two steps, or more).

3. Hess’s Law and the Conservation
of Energy
∆Hrxn is only dependent on two factors:
The mass of the substances that are
reacting and their physical states (solid,
liquid, gas, aqueous solution, etc.)

4. Hess’s Law and ∆Hrxn
Consider:
N2O4(g) → 2NO2(g) ∆Hrxn = qp = +57.2 kJ/mol
½N2O4(g) → NO2(g) ∆Hrxn = qp = + = + 28.6 kJ/mol
2NO2(g) → N2O4(g) ∆Hrxn = qp = − 57.2 kJ/mol
What do you notice?


5. Hess’s Law and ∆Hrxn
Consider:
N2O4(g) → 2NO2(g) ∆Hrxn = qp = +57.2 kJ/mol
Let’s use the first as the original reaction so we can
compare it to the others.
½N2O4(g) → NO2(g) ∆Hrxn = qp = + = + 28.6 kJ/mol
If we cut the amount of N2O4 reacting in half,
only half the heat is absorbed!


6. Hess’s Law and ∆Hrxn
Consider:
N2O4(g) → 2NO2(g) ∆Hrxn = qp = +57.2 kJ/mol
Let’s use this as the original reaction.
2NO2(g) → N2O4(g) ∆Hrxn = qp = − 57.2 kJ/mol
If we reverse the reaction so that 1 mol N2O4 is
produced rather than consumed, the same amount
of heat is released in the reaction as was absorbed
in the original.

7. Hess’s Law
Hess’s Law states that when two or more reactions
are added to give a new chemical reaction, addition of
the corresponding enthalpies will provide the enthalpy
of the new reaction.
In other words, whether the reaction is done in one
step, two steps, or more, the overall enthalpy change
of the reaction will be the same.

8. “Rules” for Hess’s Law Calculations
Rule 1: If we add two chemical
equations together, then we need to
add the corresponding ∆Hrxn values.
A + B → C ∆Hrxn = +20 kJ/mol
C + D → F ∆Hrxn = +10 kJ/mol
A + B + D → F ∆Hrxn = 20 + 10 = +30
kJ/mol

9. “Rules” for Hess’s Law Calculations
Rule 2: If we reverse the direction
of a chemical reaction, then the sign
of ∆Hrxn should be reversed.
A + B → C ∆Hrxn = +20 kJ/mol
C → A + B ∆Hrxn = − 20 kJ/mol

10. “Rules” for Hess’s Law Calculations
Rule 3: If we multiply a chemical reaction
by a factor, then we need to multiply ∆Hrxn
by the same factor.
Original reaction C → A + B ∆Hrxn = − 20 kJ/mol
2(C → A + B) multiply each coefficient by 2
2C → 2A + 2B ∆Hrxn = 2(− 20) = − 40 kJ/mol

11. Hess’s Law Example Calculation
Suppose we need ∆Hrxn for the following “target”
reaction
N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol
We have information for the following two reactions
N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol
2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol
Calculate ∆Hrxn for the target reaction using
this information.

12. Hess’s Law Example
We need to add the two reactions so that they match
our target reaction.
N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol
Look at the first reaction to see if reactants and/or products
are on the correct side of the equation (matching the target):
N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol
2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol

13. Hess’s Law Example
We need to add the two reactions so that they
match our target reaction.
N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol
N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol
The reactants are on the correct side of the equation,
so move on to the second equation:
2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol

14. Hess’s Law Example
We need to add the two reactions so that they match
our target reaction.
N2(g) + O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol
N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol
2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol
The products are on the correct side of the equation, and
the NO2 should cancel out.
Add the two equations together.

15. Hess’s Law Example
We need to add the two reactions so that they match our
target reaction.
N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol
N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol
2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol
N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = 66.4 + (− 57.2 )
∆Hrxn = 9.20 kJ/mol

16. Why Does Hess’s Law Work?
∆H is a state function. Only the initial and final
states matter, not how we get there!
 The initial state is the moles of each reactant.
 The final state is the moles of each product.
 We end up with the same reaction whether we
run it directly or do it in two steps (i.e., add two
reactions together.)

17. Why Does Hess’s Law Work?
∆H is a state
function.
Only the initial and
final states matter,
not how we get there!
∆Hrxn
H
(kJ/mol)
N2(g) +
2O2(g)
2NO2(g)
N2O4(g)
Target reaction
Reaction 2
Reaction
1

18. What You Should Be Able to Do (so
far)
Describe how Hess’s Law can be used to calculate
the enthalpy of a reaction.
Use Hess’s Law to calculate the enthalpy of
reaction for a target reaction. (∆Hrxn)
Be able to use the rules for manipulating chemical
equations in Hess’s Law calculations.
Determine whether the target reaction is exo- or
endothermic from the sign of ∆H.