Download (AP Chem Unit 11

Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 11 (Chp 5,8,19):
Thermodynamics
(∆H, ∆S, ∆G, K)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
Energy (E)
Enthalpy (H)
(kJ)
Entropy (S)
(J/K)
Free Energy (G)
(kJ)
Energy (E)
What is it?
• ability to do work OR transfer heat
(w) moving
(q) due to DT
• Energy is neither created nor destroyed.
(universe)
(no transfer matter/energy) (conserved)
• total energy of an isolated system is constant
DE = q + w
• System:
molecules to be studied
(reactants & products)
on/by
• Surroundings:
in/out +/–
everything else
+/–
(container, thermometer)
Enthalpy of Reaction
enthalpy is…
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
DH = q
…the heat transfer
in/out of a system
(at constant P)
exergonic
endergonic
exothermic
endothermic
DH = Hproducts − Hreactants
DHf = 0 for all elements in standard state
o
Endothermic &
Exothermic
• Endothermic: DH > 0 (+)
DH(+) =
Hfinal
− Hinitial
products
reactants
• Exothermic: DH < 0 (–)
DH(–) = Hfinal −
products
is thermodynamically
Hinitial
reactants
Potential Energy of Bonds
High PE
Low PE
(energy released
when bonds form)
+
High PE
(energy absorbed
when bonds break)
+
Enthalpies of Reaction (∆H)
BE: ∆H for the breaking of a bond (all +)
To determine DH for a reaction:
• compare the BE of bonds broken (reactants)
to the BE of bonds formed (products).
DHrxn = (BEreactants)  (BEproducts)
(bonds broken)
(stronger)
DH(+) = BEreac − BEprod
DH(–) =
BEreac −
BEprod
(stronger)
(bonds formed)
(released)
(NOT on
equation sheet)
Calorimetry
We can’t know the exact enthalpy
of reactants and products, so we
calculate DH by calorimetry,
the measurement of heat flow.
Calorimeter
nearly
isolated
By reacting (in solution) in a calorimeter,
we indirectly determine DH of system
by measuring ∆T & calculating q of the
surroundings (calorimeter).
heat (J)
(on equation
q = mcDT
sheet)
oC)
T
f
–
T
i
(
mass (g)
[of sol’n] [of surroundings]
(thermometer)
Specific Heat Capacity (c)
• specific heat capacity,(c):
(or specific heat)
energy required to
raise temp of 1 g by 1C.
(for water)
c = 4.18 J/goC
Metals have much lower
c’s b/c they transfer heat
and change temp easily.
+ 4.18 J
of heat
Calorimetry
(in
J)
of
calorimeter
q = mcDT
or surroundings
– q = DHrxn (in kJ/mol) of system
When 4.50 g NaOH(s) is dissolved 200. g of
water in a calorimeter, the temp. changes
from 22.4oC to 28.3oC. Calculate the molar
heat of solution, ∆Hsoln (in kJ/mol NaOH).
q = (4.50 + 200)(4.18)(28.3–22.4)
qsurr = 5040 J
DH = –5.04 kJ
DHsys = –5.04 kJ
0.1125 mol
4.50 g NaOH x 1 mol = 0.1125 mol NaOH = –44.8 kJ
mol
40.00 g
Chp. 5,8: Calculate ∆H (4 Ways)
1) Bond Energies
(NOT DH
rxn = (BEreactants)  (BEproducts)
given)
(+ broken)
(– formed)
2) Hess’s Law
(NOT DH
overall = DHrxn1 + DHrxn2 + DHrxn3 …
given)
3) Standard Heats of Formation (Hf )
(given)

DH  = nHf(products) – nHf(reactants)
4) Calorimetry (lab)
(given)
q = mc∆T (surroundings or thermometer)
–q = ∆H ∆H/mol = kJ/mol (molar enthalpy)
Big Idea #5: Thermodynamics
Bonds break and form
to lower free energy (∆G).
Chemical and physical processes
are driven by:
• a decrease in enthalpy (–∆H), or
• an increase in entropy (+∆S).
• Thermodynamically Favorable processes
proceed without any outside intervention.
(spontaneous)
1st Law of Thermodynamics
• total energy of the universe is constant.
DHsystem = –DHsurroundings
DHuniv = DHsystem + DHsurroundings = 0
(DHuniv = 0)
2nd Law of Thermodynamics
• All favorable processes increase
the entropy of the universe.
DSuniv = DSsystem + DSsurroundings > 0
(DSuniv > 0)
Entropy (J/K)
S : dispersal of matter & energy at T
DS = + therm fav
DS = – not therm fav
• entropy increases as number of microstates:
↑Temperature (motion as KEavg)
↑Volume (motion in space)
↑Number of particles (motion as KEtotal)
↑Size of particles (motion of bond vibrations)
↑Types of particles (slaqg)
3rd Law of Thermodynamics
The entropy of a pure crystalline substance
at absolute zero is 0. (not possible)
S = k lnW
S = k ln(1)
S=0
increase
only 1
microstate
0K
S=0
Temp.
>0K
S>0
Thermodynamically Favorable
Chemical and physical processes are driven by:
•
decrease in enthalpy (–∆Hsys) causes (+∆Ssurr)
•
increase in entropy (+∆Ssys)
(+) + DS (+)
DSuniv = DSsystem
surroundings > 0
•
Thermodynamically Favorable: (defined as)
increasing entropy of the universe (∆Suniv > 0)
∆Suniv > 0
(+Entropy Change of the Universe)
(∆Suniv) & (∆Gsys)
–TDSuniv = DHsys – TDSsys
DGsys = DHsys – TDSsys
(Gibbs free energy equation)
• Gibbs defined TDSuniv as the change in
free energy of a system (DGsys) or DG.
• Free Energy (DG) is more useful than
DSuniv b/c all terms focus on the system.
• If –DGsys , then +DSuniverse . Therefore…
–DG is thermodynamically favorable.
“Bonds break & form to lower free energy (∆G).”
o
(∆G )
Standard Free Energy
and Temperature (T)
(on equation
sheet)
(consists DG = DH – TDS
of 2 terms)
free
enthalpy entropy units
energy
term
term
convert
(kJ/mol) (kJ/mol) (J/mol∙K) to kJ!!!
max energy
energy
energy
used for
transferred dispersed
work
as heat
as disorder
The temperature dependence of free energy
comes from the entropy term (–TDS).
o
(∆G )
Standard Free Energy
and Temperature (T)
DG
Thermodynamic
o
∆G
Favorability
(fav. at high T) (high T) –
(unfav. at low T) (low T) +
(unfav. at ALL T)
+
(fav. at ALL T)
–
(unfav. at high T) (high T) +
(fav. at low T) (low T) –
= DH  TDS
= (∆Ho) – T(∆So)
( + ) –T( + )
= ( + ) – T( + )
= ( + ) – T( – )
= ( – ) – T( + )
– ) –T( – )
(
= ( – ) – T( – )
Calculating ∆Go (4 ways)
1) Standard free energies of formation, Gf :
DG = nG
f (products) – mG
f (reactants)
(given equation)
2) Gibbs Free Energy equation:
DG = DH – TDS
(given equation)
3) From K value DG = –RT ln K
(given equation)
4) From voltage, Eo (next Unit)
Free Energy (∆G) & Equilibrium (K)
DG = –RT ln K
(on equation sheet)
–1∙K–1
R
=
8.314
J∙mol
If DG in kJ,
then R in kJ……… = 0.008314 kJ∙mol–1∙K–1
–∆Go = ln K
RT
Solved
for K :
–∆Go
RT
K = e^
(NOT on
equation sheet)
Free Energy (∆G) & Equilibrium (K)
DG = –RT ln K
∆Go = –RT(ln K)
– = –RT ( + )
K
@ Equilibrium
> 1 product favored
(favorable forward)
+ = –RT ( – ) < 1
reactant favored
(unfavorable forward)
∆Go & Rxn Coupling
Rxn Coupling:
Unfav. rxns (+∆Go) combine with
Fav. rxns (–∆Go) to make a
Fav. overall (–∆Gooverall ).
(zinc ore)  (zinc metal)
ZnS(s)  Zn(s) + S(s)
S(s) + O2(g)  SO2(g)
goes up
if
coupled
(NOT therm.fav.)
∆Go = +198 kJ/mol
∆Go = –300 kJ/mol
ZnS(s) + O2(g)  Zn(s) + SO2(g) ∆Go = –102 kJ/mol
(therm.fav.)
Thermodynamic vs Kinetic Control
Kinetic Control: (path 2: A  C )
A very high Ea causes a thermodynamically
favored process (–ΔGo) to have no product.
∆Go = +10
Ea = +20
(kinetic product)
Free Energy (G) 
A  B
(initially pure reactant A)
path 1
A
B
+10 kJ
–50 kJ
(low Ea , Temp , time)
∆Go = –50
path 2
Ea = +50
C (thermodynamic product)
A  C
(–∆Go, Temp, Q<<K, time)
Energy (E)
Enthalpy (H)
(kJ)
+
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
ΔE = q + w
PΔV = –w
(at constant P)
ΔH = q
(heat)
Entropy (S)
(J/K)
=
Free Energy (G)
(kJ)
(disorder)
microstates
–T∆Suniv as: ΔHsys
& ΔSsys at a T
dispersal of
matter &
energy at T
max work done
by favorable rxn
∆Suniv = +
K > 1 means
–∆Gsys & +∆Suniv
ΔS = ΔH
ΔG = ΔH – TΔS
sys
T –T∆Suniv sys