# Lecture 4: Hess� Law by 0K3gdl

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```									       Lecture 4: Hess’ Law

• Outline
– Definition of Hess’ Law
– Using Hess’ Law (many examples)
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 a
single-step pathway to calculate DH for a reaction.
Hess’ Law: An Example
Using Hess’ Law
• When calculating DH for
N2 (g) + 2O2 (g)     2NO 2 (g)     a chemical reaction as a
single step, we can use
2NO 2 (g)
combinations of
reactions as “pathways”
q      to determine DH for our
N2 (g) + 2O2 (g)                 “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) DH = 180 kJ
2NO (g) + O2 (g)     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) DH = 180 kJ
2NO (g) + O2 (g)      2NO2(g) DH = -112 kJ

N2 (g) + 2O2 (g)      2NO2(g) DH =       68 kJ
Example (cont.)
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
• One 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)    DH = 68 kJ

2NO2(g)         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)      CO2 (g)    DH = -394 kJ

C3H8 (g) + 5O2 (g)       3CO2 (g) + 4H2O (l)   DH = -2220 kJ

H2 (g) + 1/2O2 (g)       H2O (l)    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
2
3C (gr) + 4H2O (l)        C3H8 (g) + 2O2 DH = +1038 kJ
Example (cont.)
• Step 4: Compare previous reaction to final
reaction, and determine how to reach final
reaction:
3C (gr) + 4H2O (l)    C3H8 (g) + 2O2 DH = +1038 kJ
H2 (g) + 1/2O2 (g)    H2O (l)       DH = -286 kJ

3C (gr) + 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 (gr) + 4H2O (l)    C3H8 (g) + 2O2 DH = +1038 kJ
4H2 (g) + 2O2 (g)     4H2O (l)       DH = -1144 kJ

3C (gr) + 4H2 (g)       C3H8 (g)
Example (cont.)
• Step 4 (cont.):

3C (gr) + 4H2O (l)   C3H8 (g) + 2O2 DH = +1038 kJ
4H2 (g) + 2O2 (g)    4H2O (l)       DH = -1144 kJ

3C (gr) + 4H2 (g)     C3H8 (g)    DH = -106 kJ
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)   NH4Cl(s) DH = -176 kJ

2NH4Cl(s)     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)    2NH4Cl(s) DH = -629 kJ

N2 (g) + 4H2 (g) + Cl2 (g)    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)   2NH3(g) + 2HCl(g)       DH = -277 kJ

( N2 (g) + 3H2(g)      2NH3(g)              DH = -92 kJ)

H2(g) + Cl2(g)      2HCl(g)         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)   DH = -277 kJ
1
2NH3(g)         3H2 (g) + N2 (g)         DH = +92 kJ

H2(g) + Cl2(g)      2HCl(g)                DH = -185 kJ

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