Lecture 3: Hess’ Law
• Reading: Zumdahl 9.5
• 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)
Definition of Enthalpy
• Thermodynamic Definition of Enthalpy (H):
H = E + PV
E = energy of the system
P = pressure of the system
V = volume of the system
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
Heat Capacity, energy and enthalpy
Ideal Monatomic Gas
• Cv = 3/2R
• Cp = Cv + R = 5/2 R
Polyatomic Gas
• Cv > 3/2R
• Cp > 5/2 R
•
•
•
•
DE = q + w
w = -PextDV (for now)
DE = nCvDT = qV
DH = nCpDT = qP
Thermodynamic State Functions
• Thermodynamic State Functions:
Thermodynamic properties that are dependent on
the state of the system only. (Example: DE and
DH)
• Other variables will be dependent on pathway
(Example: q and w). These are NOT state
functions. The pathway from one state to the other
must be defined.
Hess’ Law Defined
• From lecture 3: Enthalpy is a state function. As
such, DH for going from some initial state to some
final state is pathway independent.
• Hess’ Law: DH for a process involving the
transformation of reactants into products is not
dependent on pathway. Therefore, we can pick
any pathway to calculate DH for a reaction.
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) DH = 180 kJ
2NO2(g) 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.
• 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 (gr) + 4H2 (g)
C3H8 (g)
Example (cont.)
3C (gr) + 4H2 (g)
•
C3H8 (g) DH = ?
You’re given the following reactions:
C (gr) + 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 (gr).
Therefore, we will multiply by 3 to get the
correct amount of C (gr) with respect to our
final equation.
Initial:
C (gr) + O2 (g)
CO2 (g)
DH = -394 kJ
Final:
3C (gr) + 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 (gr) + 4H2O (l)
H2 (g) + 1/2O2 (g)
3C (gr) + 4H2 (g)
C3H8 (g) + 2O2 DH = +1038 kJ
H2O (l)
DH = -286 kJ
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 (gr) + 4H2O (l)
4H2 (g) + 2O2 (g)
3C (gr) + 4H2 (g)
C3H8 (g) + 2O2 DH = +1038 kJ
4H2O (l)
DH = -1144 kJ
C3H8 (g)
Example (cont.)
• Step 4 (cont.):
3C (gr) + 4H2O (l)
4H2 (g) + 2O2 (g)
3C (gr) + 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
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