Hess’ Law
Reading: p 519-522
Outline
Definition of Hess’ Law
Using Hess’ Law (examples)
First Law of Thermodynamics
First Law: Energy of the Universe is Constant
E=q+w
q = heat. Transferred between two bodies
w = work. Force acting over a distance (F x d)
Changes in Enthalpy
Consider the following expression for a chemical
process:
DH = Hproducts - Hreactants
If DH >0, then qp >0. The reaction is endothermic
If DH <0, then qp <0. The reaction is exothermic
Hess’ Law: An Example
Using Hess’ Law
N2 (g) + 2O2 (g)
2NO2 (g)
2NO2 (g)
q
N2 (g) + 2O2 (g)
When calculating DH
for a chemical reaction
as a single step, we
can use combinations
of reactions as
“pathways” to
determine DH for our
“single step” reaction.
Example (cont.)
Our reaction of interest is:
N2(g) + 2O2(g)
2NO2(g)
DH = 68 kJ
• This reaction can also be carried out in two
steps:
N2 (g) + O2 (g)
2NO (g) + O2 (g)
2NO(g) DH = 180 kJ
2NO2(g) DH = -112 kJ
Example (cont.)
If we take the previous two reactions and add
them, we get the original reaction of interest:
N2 (g) + O2 (g)
2NO (g) + O2 (g)
N2 (g) + 2O2 (g)
2NO(g)
2NO2(g)
DH = 180 kJ
DH = -112 kJ
2NO2(g) DH =
68 kJ
Changes in Enthalpy
Consider the following expression for a chemical
process:
DH = Hproducts - Hreactants
If DH >0, then qp >0. The reaction is endothermic
If DH <0, then qp <0. The reaction is exothermic
Example (cont.)
Note the important things about this example,
the sum of DH for the two reaction steps is
equal to the DH for the reaction of interest.
Big point: We can combine reactions of known
DH to determine the DH for the “combined”
reaction.
Hess’ Law: Details
Once can always reverse the direction of a
reaction when making a combined
reaction. When you do this, the sign of
DH changes.
N2(g) + 2O2(g)
2NO2(g)
2NO2(g)
DH = 68 kJ
N2(g) + 2O2(g) DH = -68 kJ
Details (cont.)
The magnitude of DH is directly proportional
to the quantities involved (it is an “extensive”
quantity).
As such, if the coefficients of a reaction are
multiplied by a constant, the value of DH is
also multiplied by the same integer.
N2(g) + 2O2(g)
2NO2(g)
DH = 68 kJ
2N2(g) + 4O2(g)
4NO2(g)
DH = 136 kJ
Using Hess’ Law
When trying to combine reactions to form
a reaction of interest, one usually works
backwards from the reaction of interest.
Example:
What is DH for the following reaction?
3C (g) + 4H2 (g)
C3H8 (g)
Example (cont.)
C3H8 (g) DH = ?
3C (g) + 4H2 (g)
•
You’re given the following reactions:
C (g) + O2 (g)
C3H8 (g) + 5O2 (g)
H2 (g) + 1/2O2 (g)
CO2 (g)
DH = -394 kJ
3CO2 (g) + 4H2O (l)
H2O (l)
DH = -2220 kJ
DH = -286 kJ
Example (cont.)
Step 1. Only reaction 1 has C (g).
Therefore, we will multiply by 3 to get the
correct amount of C (g) with respect to
our final equation.
Initial:
C (g) + O2 (g)
CO2 (g)
DH = -394 kJ
Final:
3C (g) + 3O2 (g)
3CO2 (g)
DH = -1182 kJ
Example (cont.)
Step 2. To get C3H8 on the product side of
the reaction, we need to reverse reaction
2.
Initial:
C3H8 (g) + 5O2 (g)
3CO2 (g) + 4H2O (l)
DH = -2220 kJ
Final:
3CO2 (g) + 4H2O (l)
C3H8 (g) + 5O2 (g) DH = +2220 kJ
Example (cont.)
Step 3: Add two “new” reactions together
to see what is left:
3C (gr) + 3O2 (g)
3CO2 (g)
DH = -1182 kJ
3CO2 (g) + 4H2O (l)
C3H8 (g) + 5O2 (g) DH = +2220 kJ
3C (gr) + 4H2O (l)
C3H8 (g) + 2O2 DH = +1038 kJ
2
Example (cont.)
Step 4: Compare previous reaction to final
reaction, and determine how to reach final
reaction:
3C (g) + 4H2O (l)
C3H8 (g) + 2O2 DH = +1038
kJ
H2 (g) + 1/2O2 (g)
H2O (l)
DH = -286 kJ
3C (g) + 4H2 (g)
C3H8 (g)
Need to multiply second reaction by 4
Example (cont.)
Step 4: Compare previous reaction to final
reaction, and determine how to reach final
reaction:
3C (g) + 4H2O (l)
C3H8 (g) + 2O2 DH = +1038kJ
4H2 (g) + 2O2 (g) 4H2O (l)
DH = -1144 kJ
3C (g) + 4H2 (g)
C3H8 (g)
Example (cont.)
Step 4 (cont.):
3C (g) + 4H2O (l)
4H2 (g) + 2O2 (g)
3C (g) + 4H2 (g)
C3H8 (g) + 2O2 DH = +1038 kJ
4H2O (l)
DH = -1144 kJ
C3H8 (g)
DH = -106 kJ
Changes in Enthalpy
Consider the following expression for a chemical
process:
DH = Hproducts - Hreactants
If DH >0, then qp >0. The reaction is endothermic
If DH <0, then qp <0. The reaction is exothermic
Another Example
Calculate DH for the following reaction:
H2(g) + Cl2(g)
2HCl(g)
Given the following:
NH3 (g) + HCl (g)
NH4Cl(s) DH = -176 kJ
N2 (g) + 3H2 (g)
2NH3 (g) DH = -92 kJ
N2 (g) + 4H2 (g) + Cl2 (g)
2NH4Cl(s) DH = -629 kJ
Another Example (cont.)
Step 1: Only the first reaction contains
the product of interest (HCl). Therefore,
reverse the reaction and multiply by 2 to
get stoichiometry correct.
NH3 (g) + HCl (g)
2NH4Cl(s)
NH4Cl(s) DH = -176 kJ
2NH3 (g) + 2HCl (g) DH = 352 kJ
Another Example (cont.)
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) DH = 352 kJ
N2 (g) + 4H2 (g) + Cl2 (g)
N2 (g) + 4H2 (g) + Cl2 (g)
2NH4Cl(s) DH = -629 kJ
2NH3(g) + 2HCl(g)
DH = -277 kJ
Another Example (cont.)
Step 3. Use remaining known reaction
in combination with the result from Step
2 to get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g)
( N2 (g) + 3H2(g)
2NH3(g) + 2HCl(g)
2NH3(g)
H2(g) + Cl2(g)
2HCl(g)
DH = -277 kJ
DH = -92 kJ)
DH = ?
Need to take middle reaction and reverse it
Another Example (cont.)
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)
1
2NH3(g)
3H2 (g) + N2 (g)
H2(g) + Cl2(g)
2HCl(g)
DH = -277 kJ
DH = +92 kJ
DH = -185 kJ
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