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Chapter 19Chemical
Thermodynamics
John D. BookstaverSt. Charles Community CollegeSt. Peters, MO2006, Prentice Hall, Inc.Modified by S.A. Green, 2006Modified by D. Amuso 2011
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Thermodynamics
The study of the relationships between heat, work, and the energy of a system.
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First Law of Thermodynamics
• You will recall from Chapter 5 that energy cannot be created nor destroyed.
• Therefore, the total energy of the universe is a constant.
• Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.
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Second Law of Thermodynamics
The total Entropy of the universe increases in any spontaneous, irreversible process.
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Entropy• Entropy can be thought of as a
measure of the randomness or disorder of a system.
• It is related to the various modes of motion in molecules (microstates).
Vibrational
Rotational
Translational
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Entropy
Molecules have more entropy (disorder) when:
1)Phase Changes from: S L G
Example: Sublimation
CO2(s) CO2(g)
2)Dissolving occurs (solution forms):
Example:
NaCl(s) Na+(aq) + Cl-(aq)
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Entropy3) Temperature increases
Example:
Fe(s) at 0 oC Fe(s) at 25 oC
4) For Gases ONLY, when
Volume increases or Pressure decreases
Examples:
2 Liters He(g) 4 Liters He(g)
3 atm He(g) 1 atm He(g)
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Entropy5) Rx results in more molecules/moles of gas
Examples:
2NH3(g) N2(g) + 3H2(g)
CaCO3(s) CaO(s) + CO2(g)
N2O4(g) 2 NO2(g)
This one is difficult to predict:
N2(g) + O2(g) 2 NO(g)
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Entropy6) When there are more moles
Example:
1 mole H2O(g) 2 moles H2O(g)
7) When there are more atoms per molecule
Examples:
1 mole Ar(g) 1 mole HCl(g)
1 mole NO2(g) 1 mole N2O4(g)
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Entropy8) When an atom has a bigger atomic
number
1 mole He(g) 1 mole Ne (g)
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Spontaneous Processes
• Spontaneous processes are those that can proceed without any outside intervention.
• The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously move to just one vessel.
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Spontaneous Processes
• Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.
• Above 0C it is spontaneous for ice to melt.• Below 0C the reverse process is spontaneous.
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Spontaneous Processes
Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.
Spontaneous processes are irreversible.
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Second Law of Thermodynamics
A reversible process results in no overall change in Entropy while an irreversible, spontaneous process results in an overall increase in Entropy.
Reversible:
Irreversible (Spontaneous):
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Third Law of Thermodynamics
The Entropy of a pure crystalline substance at absolute zero is zero. why?
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Third Law of Thermodynamics
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Standard Entropies
• Standard Conditions:
298 K, 1 atm, 1 Molar
• The values for Standard Entropies (So) are expressed in J/mol-K.
• Note: Increase with increasing molar mass.
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Standard Entropies
Larger and more complex molecules have greater entropies.
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Entropy Changes
So = n Soproducts - mSo
reactants
Be careful: S°units are in J/mol-K
Note for pure elements:
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Gibbs Free Energy
Use G to decide if a process is spontaneous
G = negative value = spontaneous G = zero = at equilibrium
G = positive value = not spontaneous
Note: equation can be used w/o the o too.
Go = Ho – TSo
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Gibbs Free Energy
1. If G is negative
G = maximum amt of energy ‘free’ to do
work by the reaction
2. If G is positive
G = minimum amt of work needed
to make the reaction happen Go = Ho – TSo
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Gibbs Free Energy
In our tables, units are:
Go = kJ/mol
Ho = kJ/mol
So = J/mol-K
Go = Ho – TSo
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Free Energy and Temperature
• There are two parts to the free energy equation: H— the enthalpy term TS — the entropy term
• The temperature dependence of free energy comes from the entropy term.
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What causes a reaction to be spontaneous?
• Think Humpty Dumpty• System tend to seek:
Minimum Enthalpy
Exothermic Rx, H = negative
Maximum Entropy
More disorder, S = positive• Because: Go = Ho – TSo
- = (-) - (+)
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Free Energy and Temperature
By knowing the sign (+ or -) of S and H, we can get the sign of G and determine if a reaction is spontaneous.
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At Equilibrium
Go = zero Therefore: Ho = TSo
Or: SoHo
T
Use this equation when asked to calculate enthalpy of vaporization or enthalpy of fusion.
Go = Ho – TSo
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Go = n Gof products - mGo
f reactants
Standard Free Energy Changes
Be careful: Values for Gf are in kJ/mol
G can be looked up in tables or calculated from S° and H.
Go = Ho – TSo
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Free Energy and Equilibrium
Remember from above:If G is 0, the system is at equilibrium.
So G must be related to the equilibrium constant, K (chapter 15). The standard free energy, G°, is directly linked to Keq by:
Go = RT ln K Where R = 8.314 J/mol-K
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Go = RT ln K
If the free energy change is a negative value, the reaction is spontaneous, ln K must be a
positive value, and K will be a large number meaning the equilibrium mixture is mainly products.
If the free energy change is zero, ln K = zero and K = one.
Relationships
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Free Energy and Equilibrium
Under non-standard conditions, we need to use G instead of G°.
Q is the reaction quotient from chapter 15.