Hess’ Law
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Reading: p 519-522
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Outline
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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
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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)
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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.)
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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.)
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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.)
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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
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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.)
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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
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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.)
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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.)
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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.)
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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
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3C (g) + 4H2 (g)
C3H8 (g)
Example (cont.)
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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
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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.)
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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.)
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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.)
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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.)
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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