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7/31/2019 Hess Law Yes
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Lecture : Hess Law
Outline
Definition of Hess Law
Using Hess Law (examples)
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First Law of Thermodynamics
First Law: Energy of theUniverse is Constant
E = q + w
q = heat. Transferredbetween two bodies
w = work. Force acting overa distance (F x d)
wq
EEE initialfinalsys
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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
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Changes in Enthalpy Consider the following expression for a chemical
process:
DH = Hproducts - Hreactants
IfDH >0, then qp >0. The reaction is endothermic
IfDH
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Thermodynamic State Functions
Thermodynamic State Functions:
Thermodynamic properties that are dependent on
the state of the system only. (Example: DE andDH)
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.
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Hess Law Defined
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.
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Hess Law: An Example
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Using Hess Law
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.
2NO2 (g)
N2 (g) + 2O2 (g)
q
2NO2 (g)N2 (g) + 2O2 (g)
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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
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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
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Changes in Enthalpy Consider the following expression for a chemical
process:
DH = Hproducts - Hreactants
IfDH >0, then qp >0. The reaction is endothermic
IfDH
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Example (cont.)
Note the important things about this example, the
sum ofDH for the two reaction steps is equal to
theD
H for the reaction of interest.
We can combine reactions of known DH to
determine the DH for the combined reaction.
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Hess Law: Details
Once can always reverse the direction of a
reaction when making a combined reaction.
When you do this, the sign ofDH changes.
N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ
2NO2(g) N2(g) + 2O2(g) DH = -68 kJ
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Details (cont.)
The magnitude ofDH is directly proportional tothe quantities involved (it is an extensive
quantity).
As such, if the coefficients of a reaction are
multiplied by a constant, the value ofDH 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
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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)
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Example (cont.)
3C (gr) + 4H2 (g) C3H8 (g) DH = ?
Youre 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
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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 ourfinal equation.
C (gr) + O2 (g) CO2 (g) DH = -394 kJ
3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ
Initial:
Final:
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Example (cont.)
Step 2. To get C3H8 on the product side of
the reaction, we need to reverse reaction 2.
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ
Initial:
Final:
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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 kJ2
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ
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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
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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)
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Example (cont.)
Step 4 (cont.):
3C (gr) + 4H2O (l) C
3H
8(g) + 2O
2DH = +1038 kJ
4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ
3C (gr) + 4H2 (g) C3H8 (g)D
H = -106 kJ
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Changes in Enthalpy Consider the following expression for a chemical
process:
DH = Hproducts - Hreactants
IfDH >0, then qp >0. The reaction is endothermic
IfDH
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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 kJN2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ
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Another Example (cont.)
Step 1: Only the first reaction contains the
product of interest (HCl). Therefore,
reverse the reaction and multiply by 2 to getstoichiometry correct.
NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ
2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ
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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
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Another Example (cont.)
Step 3. Use remaining known reaction incombination 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
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Another Example (cont.)
Step 3. Use remaining known reaction incombination with the result from Step 2 to
get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ
2NH3(g) 3H2 (g) + N2 (g) DH = +92 kJ
H2(g) + Cl2(g) 2HCl(g) DH = -185 kJ
1
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In the above enthalpy diagram note that
H1 = H2 + H3
Hesss Law
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Enthalpies of Formation If 1 mol of compound is formed from its
constituent elements, then the enthalpy change for
the reaction is called the enthalpy of formation,
DHof.
Standard conditions (standard state): 1 atm and 25oC
Standard enthalpy, DHo, is the enthalpy measured
when everything is in its standard state.
Standard enthalpy of formation: 1 mol of
compound is formed from substances in their
standard states.
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Enthalpies of Formation Standard enthalpy of formation of the most stable
form of an element is zero.
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Hrxn =
H1 + H2 + H3
Enthalpies of FormationUsing Enthalpies of Formation to Calculate
Enthalpies of Reaction
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Enthalpies of FormationUsing Enthalpies of Formation to Calculate
Enthalpies of ReactionFor a reaction:
HOrxn = n HO
f(products) - n HO
f(reactants)
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Changes in Enthalpy Consider the following expression for a chemical
process:
DH = Hproducts - Hreactants
IfDH >0, then qp >0. The reaction is endothermic
IfDH