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Chemical Thermodynamics
Spontaneous Processes
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First Law of Thermodynamics
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Energy is Conserved – ΔE = q + w
Need value other than ΔE to determine if a
process if favored (spontaneous)
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Spontaneous processes have a direction
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Spontaneity can depend on temperature
Reversible Processes
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Reversible Process
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When a change in a
system is made in
such a way that the
system can be
restored to its original
state by exactly
reversing the change.
Irreversible Processes
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Irreversible Process
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A process that cannot
simply be reversed to
restore the system
and surroundings.
Must take alternative
pathway
Only system is
returned to its original
state
Entropy
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Processes where the disorder of the system
increases tend to occur spontaneously
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Ice melting
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Salts dissolving
Change in disorder and change in energy
determine spontaneity
Entropy (S) is a state function (ΔS) that
measures of disorder
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Units – J/K
Determining Entropy Change
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Predict whether ΔS is positive or negative for
each of the following processes.
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H2O(l) → H2O(g)
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Ag+ (aq) + Cl-(aq) → AgCl(s)
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4Fe(s) + 3O2(g) → 2Fe2O3(s)
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CO2(s) → CO2(g)
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CaO(s) + CO2(g) → CaCO3(s)
Calculating Entropy
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In a reversible process there is only one heat
for both processes (qr e v)
When a process occurs at constant temperature
entropy is related to both heat and absolute
temperature – ΔS = qr e v / T
When 1mol of water is converted to 1mol of
steam at 1 atm, ΔHv a p = 40.67kJ/mol, what is
the change in entropy?
Calculating Entropy
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The element mercury, Hg, is a silvery liquid at
room temperature. The normal freezing point of
mercury is -38.9°C, and its molar enthalpy of
fusion is ΔHf u s = 2.29 kJ/mol. What is the
entropy change of the system when 50.0 g of
Hg(l) freezes at the normal freezing point?
Answer: -2.44 J/K
Calculating Entropy
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The normal boiling point of ethanol is 78.3°C
and its molar enthalpy of vaporization is 38.56
kJ/mol. What is the change in entropy when
68.3g of ethanol at 1 atm condenses to liquid at
the normal boiling point?
Answer: -163 J/K
Second Law of Thermodynamics
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Second law = Entropy of the universe increases
in any spontaneous process.
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ΔSuniv = ΔSsys+ ΔSsurr = 0 → reversible process
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ΔSuniv = ΔSsys+ ΔSsurr > 0 → irreversible process
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Unlike energy, entropy is not conserved
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Examples
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Straightening up your room
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4Fe(s) + 3O2(g) → 2Fe2O3(s)
Exceptions – Isolated Systems
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ΔSsys = 0 → reversible process
Calculating Entropy
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Consider the reversible melting of 1 mol of ice
in a large isothermal water bath at 0°C and 1
atm pressure. The enthalpy of fusion of ice is
6.01 kJ/mol. Calculate the entropy change in
the system and in the surroundings, and the
overall change in entropy of the universe for this
process.
Answer = 22.0 J/K, 0 J/K
Molecular Interpretation of Entropy
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Why does the entropy
increase when a gas
expands?
Why does entropy
decrease in the
following reaction 2NO(g) + O2(g) →
2NO2(g)?
Degrees of Freedom
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Translational Motion
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Vibrational Motion
Molecular Interpretation of Entropy
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Lowering temperature
decreases energy
which lowers the
degrees of freedom.
Third law of
thermodynamics =
entropy of a pure
crystalline substance
at absolute zero is
zero.
Entropy generally
increases with
Entropy Changes in Reactions
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Entropy increases for processes in which:
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Liquids or solutions are formed from solids.
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Gases are formed from either solids of liquids.
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The number of molecules of gas increases during a
chemical reaction.
Choose the sample of matter that has greater
entropy and explain your choice.
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1 mol of NaCl(s) or 1 mol HCl(g) at 25°C
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1 mol of HCl(g) or 1 mol Ar (g) at 25°C
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1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C
Predicting Entropy Changes
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Predict whether the entropy change of the
system in each of the following isothermal
reactions if positive or negative.
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CaCO3(s) → CaO(s) + CO2(g)
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N2(g) + 3H2(g) → 2NH3(g)
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N2(g) + O2(g) → 2NO(g)
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HCl(g) + NH3(g) → NH4Cl(s)
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2SO2(g) + O2(g) → 2SO3(g)
Entropy Changes in Chemical
Reactions
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Using variation of heat capacity
with temperature absolute
entropy are measured.
Molar entropy at standard
states (S°)
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Molar entropies of elements are
not zero
Molar entropies of gases are
greater than liquids and solids
Molar entropies generally
increase with increasing molar
mass.
Entropy Changes in Chemical
Reactions
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Entropy Change in a reaction can be calculated
using a table of values
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ΔS°= ΣnS°(prod)-ΣmS°(react)
Calculate ΔS° for the synthesis of ammonia
from nitrogen and hydrogen at 298K.
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N2(g) + 3H2(g) → 2NH3(g)
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Answer: -198.3 J/K
Entropy Changes in the
Surroundings
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Surroundings are essentially a large constant
temperature heat source.
The change in entropy will then depend on how
much heat is absorbed or given off by the
system.
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ΔSs u r r = -qs y s /T
If the reaction happens at constant P, what is q?
Calculate the ΔSu n i v given the heat of formation
of ammonia (-46.19kJ/mol)
Gibbs Free Energy
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Spontaneity depends on enthalpy and entropy.
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Gibbs Free Energy – G = H – TS
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In a chemical reaction at constant T
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ΔG = ΔH – TΔS
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Algebra fun
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When both T and P are constant
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If ΔG is negative, the reaction is spontaneous in the
forward direction
If ΔG is zero, the reaction is at equilibrium
If ΔG is positive, the forward reaction is
nonspontaneous, work must be done to make it
Standard Free Energy Changes
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Gibbs free energy is a state function.
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For Standard free energies
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Free energies for standard states are set to zero.
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Gases should be at 1 atm
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Solutions should be 1M in concetration
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Solids and Liquids should be in their pure forms
ΔG°= ΣnG°(prod)-ΣmG°(react)
Calculating Standard Free Energies
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Using the data from Appendix C, calculate the
standard free-energy change for the following
reaction at 298K:
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P4(g) + 6Cl2(g) → 4PCl3(g)
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What about the reverse reaction?
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Answer: -1054.0kJ
Calculating Standard Free Energies
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Without using data from appendix C, predict
whether ΔG° for this reaction is more or less
negative than ΔH°
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C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH°=2220kJ
Using the data from appendix C, calculate the
standard free energy change for this reaction.
What your prediction correct?
Answer: -2108kJ
Calculating Standard Free Energies
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Consider the combustion of propane to form
CO2(g) and H2O(g) at 298K. Would you expect
ΔG° to be more negative or less negative than
ΔH°?
Free Energy and Temperature
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ΔG = ΔH – TΔS
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Ice melting
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ΔG° = ΔH° – TΔS°
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Use to estimate at
other temperatures
Homework
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Part 1 - 2, 5, 7, 17, 20, 23, 25, 29, 31, 34, 37
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Part 2 - 41, 43, 47, 49, 53, 55, 56