AP Notes Chapter 19

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AP Chemistry Chapter 19 Notes
CHEMICAL THERMODYNAMICS
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REVIEW 1st Law of Thermodynamics Energy in the
universe is conserved
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REVIEW Path Function value depends on how
process takes place (i.e. q, w)
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heat q mCp?T
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REVIEW State Function value independent of path -
is a defined reference or zero point (i.e. H)
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Enthalpy - ?H

• zero point
• ?Hf0 of elements in natural form at 250C

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Enthalpy - ?H types ?Hrxn ?n?Hf0(prod) -
?n?Hf0(react)
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Enthalpy - ?H types ?Hfus , ?Hvap
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ENTROPY S

• a measure of randomness or disorder of a system

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Entropy is NOT conserved. The universe seeks
maximum disorder.
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2nd Law of Thermodynamics for a
spontaneous process, entropy increases
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Entropy - state function

• zero reference S 0
• for a perfect crystal at absolute zero

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Ssolid lt Sliquid ltlt Sgas
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S0 standard entropy of elements cmpds at P
1 atm T 250C units J/K mol Appendix
L, text
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Calculate ?S using Hess Law ?S(rxn)
?nS0(prod) - ?nS0(react)
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Example Ca(s) C(gr) 3/2 O2 ? CaCO3(s)
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?Suniv ?Ssys ?Ssurr if ?Suniv gt 0 process
spontaneous
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?Suniv ?Ssys ?Ssurr if ?Suniv lt 0 process ?
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?Suniv ?Ssys ?Ssurr if ?Suniv 0 process ?
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to relate ?S and ?H consider
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H2O(s) ? H2O(l) where the water is the system
everything else is the surroundings
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Temperature Dependence ?S J/mol (1/T) ?Ssurr
-?H/T
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?Ssys ?Ssurr ?Suniv spon
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?Ssys ?Ssurr ?Suniv spon
Y - -
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?Ssys ?Ssurr ?Suniv spon
Y - - -
N -
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?Ssys ?Ssurr ?Suniv spon
Y - - -
N - ?
D -
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?Ssys ?Ssurr ?Suniv spon
Y - - -
N - ?
D - ? D
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For a phase change
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GIBBS FREE ENERGY ?G
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most abstract of thermodynamic state functions
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w1 ?w reversible
work w2
P
V
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Definition G w PV w reversible work PV
pressure/volume work isothermal, reversible path
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?G ?w P?V V?P at constant P ?P 0 so
V?P 0 ?G ?w P?V
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?G ?w P?V V?P at constant P ?G ?w
P?V at constant V ?V 0 so P?V 0 ?G ?w
useful work
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G cannot be measured

• must measure ?G over a process

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ZERO REFERENCE ?G 0 for elements in stable form
under Standard Thermodynamic Conditions T
25oC P 1 atm
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?Gf0 standard Free Energy of Formation from the
elements Appendix L, text
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?G follows Hess Law ?G0 (rxn) ?n?Gf0(p) -
?n?Gf0(r)
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Summary of Laws of Thermodynamics Zeroth
Law Heat Gain Heat Loss
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Summary of Laws of Thermodynamics First Law Law
of Conservation of Energy
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Summary of Laws of Thermodynamics Second
Law Defines Entropy
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Summary of Laws of Thermodynamics Third
Law Defines Absolute Zero
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GIBBS HELMHOLTZ EQUATION
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combine
?G
?H
T
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?G -aT ?H ?G, ?H are state functions, thus
a must be a state function ?G -?S(T) ?H
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Gibbs Helmholtz Equation ?G ?H - T?S
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Units on the State Functions
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thus, a process is spontaneous if and only if ?G
is negative
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Spontaneity controlled by enthalpy (minimum
energy)
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Sponaneity controlled by enthalpy entropy (maximu
m disorder)
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Sponaneity controlled by enthalpy entropy both
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Predict Spontaneity IF ?H(-) and ?S() ?G -?H
- T(?S) ?G lt 0, gt spontaneous
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Predict Spontaneity IF ?H() and ?S(-) ?G ?H
- T(-?S) ?G gt 0, gt NOT spontan
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Summary of Spontaneity ?H ?S ?G
Spont. - - yes
- no or -
? - - or - ?
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Uses of the Gibbs Helmholtz Equation
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1. Find the molar entropy of formation
for ammonia.
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2. Elemental boron, in thin fibers, can be
made from a boron halide BCl3(g) 3/2
H2(g) -gtB(s) 3HCl(g)
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Calculate ?H0, ?S0 and ?G0.

• Spontaneous?
• Driving force?

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3. Using thermodynamic information, determine
the boiling point of bromine.
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Thermodynamic Definition of Equilibrium ?Geq 0
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by definition G H - TS H E PV
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thus, G E PV - TS take derivative of both
sides
dG dE PdV VdP - TdS - SdT
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for a reversible process TdS ?q
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derivative used for state function while partial
derivative used for path function
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if the only work is PV work of expansion PdV ?w
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First Law of Thermodynamics E q - w dE ?q -
?w 0
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thus ?q ?w or TdS PdV
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by substitution dG 0 PdV VdP
- PdV - SdT or dG VdP - SdT
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lets assume we have a gaseous system
at equilibrium, therefore, examine Kp at
that constant temperature
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at constant T SdT 0 thus dG VdP
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but
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if condition 2 is at standard thermodynamic
conditions, then G2 G0 and P2 1 atm
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thus
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determine ?G for aA bB ? cC dD where all
are gases
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?G Gprod - Greact cGC dGD - aGA - bGB
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but
and likewise for the others
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and
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but
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Thus, DG DG0 (RT) ln Q
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But aA bB ?? cC dD ?G 0
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thus
or in general
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• THERMODYNAMICS
• EQUILIBRIUM

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3. Using thermodynamic information, determine
the boiling point of bromine.
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Thermodynamics and Keq

• FACT Product-favored systems have Keq gt 1.

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Thermodynamics and Keq

• Therefore, both ?Grxn and Keq are related to
  reaction favorability.

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Thermodynamics and Keq

• Keq is related to reaction favorability and thus
  to ?Gorxn.
• The larger the value of K the more negative the
  value of ?Gorxn

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Thermodynamics and Keq

• ?Gorxn - RT lnK
• where R 8.314 J/Kmol

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?Gorxn - RT lnK

• Calculate K for the reaction
• N2O4 ?2 NO2
• ?Gorxn 4.8 kJ
• K 0.14
• When ?G0rxn gt 0, then K lt 1

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?G, ?G, and Keq

• ?G is change in free energy at non-standard
  conditions.
• ?G is related to ?G
• ?G ?G RT ln Q where Q reaction quotient

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?G, ?G, and Keq

• When Q lt K or Q gt K, reaction is spontaneous.
• When Q K reaction is at equilibrium
• When ?G 0 reaction is at equilibrium
• Therefore, ?G - RT ln K

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?G, ?G, and Keq

• Product favored reaction
• ?Go and K gt 1
• In this case
• ?Grxn is lt ?Gorxn , so state with both reactants
  and products present is MORE STABLE than complete
  conversion.

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• Product-favored reaction.
• 2 NO2 ? N2O4
• ?Gorxn 4.8 kJ
• Here ?Grxn is less than ?Gorxn , so the state
  with both reactants and products present is more
  stable than complete conversion.

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Thermodynamics and KeqOverview

• ?Gorxn is the change in free energy when
  reactants convert COMPLETELY to products.

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• Keq is related to reaction favorability.
• When ?Gorxn lt 0, reaction moves energetically
  downhill

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4. For the following reaction, calculate the
temperature at which the reactants are favored.
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• THERMODYNAMICS
• OF
• CHEMICAL
• REACTIONS

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5. How much useful work can be obtained from
an engine fueled with 75.0 L of hydrogen at 10 C
at 25 atm?
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6. The reaction to split water into hydrogen and
oxygen can be promoted by first reacting silver
with water.
2 Ag(s) H2O(g) ?Ag2O(s) H2(g) Ag2O(s) ?
2 Ag(s) 1/2 O2(g)
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Calculate ?H0, ?S0 and ?G0 for each reaction.
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Combine the reactions and calculate ?H0 and ?G0
for the combination. Is the combination
spontaneous?
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At what temperature does the second reaction
become spontaneous?
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7. The conversion of coal into hydrogen for
fuel is C(s) H2O(g) ? CO(g) H2(g)
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Calculate ?G0 and Kp at 250C. Is the reaction
spontaneous?
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At what temperature does the reaction become
spontaneous?
Calculate the temperature at which K 1.0 x 10-4.
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8. The generation of nitric acid in the upper
atmosphere might destroy the ozone layer
by NO(g) O3(g) ? NO2(g) O2(g)
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Calculate ?G0 (reaction) and K at 250C.