1. Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K)
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.

2. Energy (E)
Enthalpy (H)
(kJ)
Entropy (S)
(J/K)
Free Energy (G) (kJ)

3. Energy (E) What is it? ability to do work OR transfer heat
(w) moving
(q) due to DT
Energy is neither created nor destroyed.
total energy of an isolated system is constant
(universe)
(no transfer matter/energy)
(conserved)
System:
molecules to be studied (reactants & products)
Surroundings:
everything else
(container, thermometer)
E = q + w
on/by
+/–
in/out
+/–

4. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Enthalpy of Reaction
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
enthalpy is…
…the heat transfer in/out of a system
(at constant P)
H = q
exergonic
exothermic
endergonic
endothermic
H = Hproducts − Hreactants o
Hf = 0 for all elements in standard state

5. Endothermic & Exothermic
Endothermic: H > 0 (+)
H(+) = Hfinal − Hinitial
products
reactants
Exothermic: H < 0 (–)
H(–) = Hfinal − Hinitial
reactants
products
H(–) is thermodynamically favorable

6. Potential Energy of Bonds
High PE
chemical bond
Low PE
(energy released when bonds form)
High PE
(energy absorbed when bonds break) + +

7. Enthalpies of Reaction (∆H)
BE: ∆H for the breaking of a bond (all +)
To determine H for a reaction:
compare the BE of bonds broken (reactants) to the BE of bonds formed (products).
Hrxn = (BEreactants)  (BEproducts)
(bonds broken)
(bonds formed)
(released)
(stronger)
H(+) = BEreac − BEprod
(NOT on equation sheet)
H(–) = BEreac − BEprod
(stronger)

8. Calorimeter nearly isolated
Calorimetry
We can’t know the exact enthalpy of reactants and products, so we calculate H by calorimetry,
the measurement of heat flow.
By reacting (in solution) in a calorimeter, we indirectly determine H of system by measuring ∆T & calculating q of the surroundings (calorimeter).
Calorimeter nearly isolated
(on equation sheet)
heat (J)
q = mcT
Tf – Ti (oC)
[of surroundings]
(thermometer)
mass (g)
[of sol’n]

9. 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
J of heat
Metals have much lower c’s b/c they transfer heat and change temp easily.

10. (in J) of calorimeter or surroundings
Calorimetry
(in J) of calorimeter or surroundings
q = mcT
– q = Hrxn
(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 = ( )(4.18)(28.3–22.4)
qsurr = 5040 J
H = –5.04 kJ
mol
Hsys = –5.04 kJ
4.50 g NaOH x 1 mol = mol NaOH
40.00 g
= –44.8 kJ
mol

11. Chp. 5,8: Calculate ∆H (4 Ways)
1) Bond Energies
Hrxn = (BEreactants)  (BEproducts)
2) Hess’s Law
Hoverall = Hrxn1 + Hrxn2 + Hrxn3 …
3) Standard Heats of Formation (Hf )
H = nHf(products) – nHf(reactants)
4) Calorimetry (lab)
q = mc∆T (surroundings or thermometer)
–q = ∆H ∆H/mol = kJ/mol (molar enthalpy)
(NOT given)
(+ broken)
(– formed)
(NOT given)    
(given)
(given)

12. 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)

13. 1st Law of Thermodynamics
total energy of the universe is constant.
Hsystem = –Hsurroundings
Huniv = Hsystem + Hsurroundings = 0
(Huniv = 0)
2nd Law of Thermodynamics
All favorable processes increase the entropy of the universe.
Suniv = Ssystem + Ssurroundings > 0
(Suniv > 0)

14. S : dispersal of matter & energy at T
Entropy (J/K)
S : dispersal of matter & energy at T
S = + therm fav
S = – 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)

15. 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
Temp.
only 1 microstate
0 K
S = 0
> 0 K
S > 0

16. Thermodynamically Favorable
Chemical and physical processes are driven by:
decrease in enthalpy (–∆Hsys)
increase in entropy (+∆Ssys)
causes (+∆Ssurr)
(+)
(+)
Suniv = Ssystem + Ssurroundings > 0
Thermodynamically Favorable: (defined as)
increasing entropy of the universe (∆Suniv > 0)
∆Suniv > 0
(+Entropy Change of the Universe)

17. (∆Suniv) & (∆Gsys) –TSuniv = Hsys – TSsys Gsys = Hsys – TSsys
(Gibbs free energy equation)
Gibbs defined TDSuniv as the change in
free energy of a system (Gsys) or G.
Free Energy (G) is more useful than
Suniv b/c all terms focus on the system.
If –Gsys , then +Suniverse . Therefore…
–G is thermodynamically favorable.
“Bonds break & form to lower free energy (∆G).”

18. Standard Free Energy (∆Go) and Temperature (T)
(on equation sheet)
(consists of 2 terms)
DG = DH – TS
free energy (kJ/mol)
enthalpy term (kJ/mol)
entropy term (J/mol∙K)
units convert to kJ!!!
max energy used for work
energy transferred as heat
energy dispersed as disorder
The temperature dependence of free energy comes from the entropy term (–TS).

19. Standard Free Energy (∆Go) and Temperature (T)
DG = DH  TS
Thermodynamic Favorability
∆Go =
(∆Ho)
∆So
– T( )
( ) –T( )
(high T) –
(low T) +
(fav. at high T)
(unfav. at low T) + + =
( ) – T ( ) + + –
(unfav. at ALL T) + =
( ) – T( ) +
(fav. at ALL T) – =
( ) – T( ) – +
(high T) +
(low T) –
( ) –T( )
(unfav. at high T)
(fav. at low T) – – =
( ) – T ( ) – –

20. Calculating ∆Go (4 ways)
Standard free energies of formation, Gf :
Gibbs Free Energy equation:
From K value
From voltage, Eo (next Unit)
DG = SnG(products) – SmG(reactants) f
(given equation)
DG = DH – TS
(given equation)
G = –RT ln K
(given equation)

21. Free Energy (∆G) & Equilibrium (K)
G = –RT ln K
(on equation sheet)
If G in kJ,
then R in kJ………
R = J∙mol–1∙K–1
= kJ∙mol–1∙K–1
–∆Go RT
= ln K
–∆Go RT
Solved for K :
(NOT on equation sheet)
K = e^

22. Free Energy (∆G) & Equilibrium (K)
G = –RT ln K
∆Go = –RT(ln K) K
@ Equilibrium – + =
–RT ( )
> 1
product favored
(favorable forward) + – =
–RT ( )
< 1
reactant favored
(unfavorable forward)

23. ∆Go & Rxn Coupling Rxn Coupling:
Unfav. rxns (+∆Go) combine with Fav. rxns (–∆Go) to make a Fav. overall (–∆Gooverall ).
goes up if coupled
(zinc ore)  (zinc metal)
(NOT therm.fav.)
ZnS(s)  Zn(s) + S(s)
∆Go = +198 kJ/mol
S(s) + O2(g)  SO2(g)
∆Go = –300 kJ/mol
ZnS(s) + O2(g)  Zn(s) + SO2(g)
∆Go = –102 kJ/mol
(therm.fav.)

24. Thermodynamic vs Kinetic Control
Kinetic Control: (path 2: A  C )
A very high Ea causes a thermodynamically favored process (–ΔGo) to have no product.
A  B ∆Go = +10
Ea = +20
(kinetic product)
(initially pure reactant A)
path 1
(low Ea , Temp , time) B A
A  C ∆Go = –50
Ea = +50
(thermodynamic product)
Free Energy (G) 
+10 kJ
path 2
–50 kJ C
(–∆Go, Temp, Q<<K, time)

25. + = Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K)
Free Energy (G) (kJ)
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
(disorder)
microstates
–T∆Suniv as: ΔHsys & ΔSsys at a T
dispersal of matter & energy at T
max work done by favorable rxn
ΔE = q + w
PΔV = –w
(at constant P)
K > 1 means –∆Gsys & +∆Suniv
∆Suniv = +
ΔH = q
(heat)
ΔS = ΔH T
ΔG = ΔH – TΔS
sys
sys
–T∆Suniv