1. Enthalpy (H)
The heat transferred sys ↔ surr during a chemical constant P
Can’t measure H, only ΔH
At constant P, ΔH = q = mCΔT, etc.
Literally, ΔH = Hproducts - Hreactants
ΔH = + (endothermic)
Heat goes from surr into sys
ΔH = - (exothermic)
Heat leaves sys and goes into surr

2. Reactant + Energy Product
In this example, the energy of the system (reactants and products) ↑, while the energy of the surroundings ↓
Notice that the total energy does not change
Reactant + Energy Product
Endothermic Reaction
Surroundings
Surroundings
System
Notice that E must be added, and thus is like a reactant
Energy
System
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41

3. Reactant Product + Energy
In this example, the energy of the system (reactants and products) ↓, while the energy of the surroundings ↑
Notice again that the total energy does not change
Reactant Product + Energy
Exothermic Reaction
Surroundings
Surroundings
System
Notice that E is released and thus is like a product
System
Energy
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41

4. Energy released to the surrounding as heat
Burning of a Match
System
Surroundings
D(PE)
(Reactants)
Potential energy
Energy released to the surrounding as heat
(Products)
Exothermic Reaction
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293

5. Reaction Coordinate Diagrams: Endothermic Reaction
Activation Energy Ea
D(PE)
Products
ΔHrxn= + PE
Reactants
Progress of the Reaction

6. Reaction Coordinate Diagrams: Exothermic Reaction
Activation Energy Ea
ΔHrxn= -
Reactants
D(PE) PE
Products
Progress of the Reaction

7. Reaction Coordinate Diagrams
Draw the reaction coordinate diagram for the following rxn:
C(s) + O2(g)  CO kJ
Activation Energy
EXOTHERMIC Ea
ΔHrxn= kJ
C + O2
D(PE) PE
CO2
Progress of the Reaction

8. Enthalpies of Reaction
All reactions have some ΔH associated with it
H2(g) + ½ O2(g) → H2O(l) ΔH = kJ
How can we interpret this ΔH?
Amount of energy released or absorbed per specific reaction species
Use balanced equation to find several definitions kJ
1 mol H2 kJ
½ mol O2 kJ
1 mol H2O or or
Able to use like conversion factors in stoichiometry

9. Enthalpies of Reaction
Formation of water
H2(g) + ½ O2(g) → H2O(l) ΔH = kJ
ΔH is proportional to amount used and will change as amount changes
2H2(g) + O2(g) → 2H2O(l)
For reverse reactions, sign of ΔH changes
2H2O(l) → 2H2(g) + O2(g)
Treat ΔH like reactant or product
ΔH = kJ
ΔH = kJ
H2(g) + ½ O2(g) → H2O(l) kJ (exo)

10. Enthalpies of Reaction Practice
Consider the following rxn:
C(s) + 1/2O2(g)  CO kJ
Is the ΔH for this reaction positive or negative?
NEGATIVE (E released as a product)
What is the ΔH for 2.00 moles of carbon, if all the carbon is used?
2.00 mol C kJ
= kJ
1 mol C
What is the ΔH if 50.0g of oxygen is used?
50.0 g O2
1 mol O2 kJ
= kJ
32.0 g O2
0.5 mol O2
What is the ΔH if 50.0 g of carbon monoxide decompose, in the reverse reaction?
50.0 g CO
1 mol CO
458.1 kJ
= kJ
28.0 g CO
1 mol CO

11. Hess’s law
Hess’s Law states that the enthalpy of a whole reaction is equivalent to the sum of it’s steps.
For example: C + O2  CO2
This can occur as 2 steps
C + ½O2  CO H = – kJ
CO + ½O2  CO2 H = – kJ
C + CO + O2  CO + CO2 H = – kJ
I.e. C + O2  CO2 H = – kJ
Hess’s law allows us to add equations.
We add all reactants, products, & H values.

12. Why? Because enthalpy is a state function
Hess’s Law
Reactants  Products
The change in enthalpy is the same whether the reaction takes place in one step or a series of steps
Why? Because enthalpy is a state function
Victor Hess
To review:
1. If a reaction is reversed, ΔH is also reversed
2 CH4 + O2  2 CH3OH ΔHrxn = -328 kJ
2 CH3OH  2 CH4 + O2 ΔHrxn = +328 kJ
2. If the coefficients of a reaction are multiplied by an integer, ΔH is multiplied by that same integer
CH4 + 2 O2  CO2 + 2 H2O ΔHrxn = kJ
2(CH4 + 2 O2  CO2 + 2 H2O) ΔHrxn = kJ

13. Example: Methanol-Powered Cars
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g) ΔHrxn = ?
2 CH4(g) + O2(g)  2 CH3OH(l) ΔHrxn = -328 kJ
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) ΔHrxn = kJ
2 CH3OH(l)  2 CH4(g) + O2(g) ΔHrxn = +328 kJ
2(CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)) ΔHrxn = kJ
2 CH3OH(l) + 2 CH4(g) + 4 O2(g)  2 CH4(g) + O2(g) + 2 CO2(g) + 4 H2O(g) 3
2 CH3OH + 3 O2  2 CO2 + 4 H2O ΔHrxn = kJ

14. Tips for applying Hess’s Law…
Look at the final equation that you are
trying to create first…
Find a molecule from that eq. that is only in one of the given equations
Make whatever alterations are necessary to those
Once you alter a given equation, you will not alter it again
Continue to do this until there are no other options
Next, alter remaining equations to get things to cancel that do not appear in the final equation

15. 1. Given the following data:
S(s) + 3/2O2(g) → SO3(g) ΔH = kJ
2SO2(g) + O2(g) → 2SO3(g) ΔH = kJ .
Calculate ΔH for the following reaction:
S(s) + O2(g) → SO2(g)
S(s) + 3/2O2(g) → SO3(g) ΔH = kJ
2SO2(g) + O2(g) → 2SO3(g) ΔH = kJ
2SO3(g) → O2(g) + 2SO2(g) ΔH = kJ
SO3(g) → ½ O2(g) + SO2(g) ΔH = kJ
S(s) + O2(g) → SO2(g)
ΔH = kJ

16. 2. Given the following data:
C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = kJ
C(s) + O2(g) → CO2(g) ΔH = -394 kJ
H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ
Calculate ΔH for the following reaction:
2C(s) + H2(g) → C2H2(g)
C(s) + O2(g) → CO2(g) ΔH = -394 kJ
2C(s) + 2O2(g) → 2CO2(g) ΔH = -788 kJ
C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = kJ
2CO2(g) + H2O(l) → C2H2(g) + 5/2O2(g) ΔH = kJ
H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ
2C(s) + H2(g) → C2H2(g) ΔH = +226kJ

17. 3. Given the following data:
  O3(g) → 3O2(g) ΔH = kJ
O2(g) → 2O(g) ΔH = kJ
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = kJ
Calculate ΔH for the following reaction:
NO(g) + O(g) → NO2(g)
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = kJ
O2(g) → 2O(g) ΔH = kJ
O(g) → ½ O2(g) ΔH = kJ
2O3(g) → 3O2(g) ΔH = kJ
3/2 O2(g) → O3(g) ΔH = kJ
NO(g) + O(g) → NO2(g) ΔH = +233kJ

18. Heats of Formation, ΔH°f
The enthalpy change when one mole of a compound is formed from the elements in their standard states
° = standard conditions
Gases at 1 atm pressure
All solutes at 1 M concentration
Pure solids and pure liquids
f = a formation reaction
1 mole of product formed
From the elements in their standard states (1 atm, 25°C)
For all elements in their standard states, ΔH°f = 0
What’s the formation reaction for adrenaline, C9H12NO3(s)? 9
Cgr + 6
H2(g) +
1/2
N2(g) +
3/2
O2(g)
 C9H12NO3(s)

19. Thermite Reaction Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l) ΔHrxn = ?
Welding railroad tracks

20. ΔHrxn = nΔH°f(products) - nΔH°f(reactants)
Thermite Reaction
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
Reactants
Elements
(standard states)
Products
Fe2O3(s)
2 Fe(s)
2 Fe(l)
2 Al(s)
3/2 O2(g)
Al2O3(s)
2 Al(s)
ΔHrxn = 2ΔH°f(Fe(l)) + ΔH°f(Al2O3(s)) - ΔH°f(Fe2O3(s)) - 2ΔH°f(Al(s))
ΔHrxn = [2(15 kJ) + (-1676 kJ)] – [(-822 kJ) – 2(0)]
ΔHrxn = -824 kJ
ΔHrxn = nΔH°f(products) - nΔH°f(reactants)

21. ∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
ΔH°f Example Problems
∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
1. CH4(g) + 2 Cl2(g)  CCl4(g) + 2 H2(g)
ΔHrxn = ?
(-74.8) 2
(0)
(-106.7) 2
(0)
∆H = [(-106.7) + 0] – [(-74.8)+0] =
= kJ/mol
2. 2 KCl(s) + 3 O2(g)  2KClO3(s)
ΔHrxn = ? 2
(-435.9) 3
(0) 2
(-391.2)
∆H = [(2)(-391.2)] – [(2)(-435.9) + (3)(0)] =
= kJ/mol

22. 3. AgNO3(s) + NaCl (aq)  AgCl(s) + NaNO3(aq)
ΔH°f Example Problems
∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
3. AgNO3(s) + NaCl (aq)  AgCl(s) + NaNO3(aq)
ΔHrxn = ?
(-124.4)
(-127.0)
(-446.2)
(-407.1)
∆H = [(-127.0) + (-446.2)] – [(-124.4) + (-407.1)] =
= kJ/mol
4. C2H5OH(l) + 7/2 O2(g)  2CO2(g) + 3H2O(g)
ΔHrxn = ?
(-277.7)
(7/2)
(0)
(2)
(-393.5)
(3)
(-241.8)
∆H = [(2)(-393.5) + (3)(-241.8)] – [(-277.7) + (7/2)(0)] =
= kJ/mol

23. Bond Energies Chemical reaction ⇔ Bond breakage & bond formation
Bond energy = energy required to break a bond
Bond breaking is endothermic (raises potential energy)
Bond formation is exothermic (lowers PE)
Average energy for one type of bond in different molecules
Table 8.4 in Zumdahl
C-H : 413 kJ/mol ∙ C=O : 799 kJ/mol
O=O : 495 kJ/mol ∙ O-H : 467 kJ/mol
ΔHrxn = (bonds broken) – (bonds formed)
Energy required
Energy released

24. Bond Energies CH4 + 2 O2  CO2 + 2 H2O
ΔHrxn = (bonds broken) – (bonds formed)
CH4 + 2 O2  CO2 + 2 H2O
C-H : 413 kJ/mol C=O : 799 kJ/mol
O=O : 495 kJ/mol O-H : 467 kJ/mol
ΔHrxn = [4(C-H) + 2(O=O)] – [2(C=O) + 4(O-H)]
ΔHrxn = [4(413 kJ) + 2(495 kJ)] – [2(799 kJ) + 4(467 kJ)]
ΔHrxn = -824 kJ
Compare to ΔHrxn = kJ

26. Entropy (S) = a measure of randomness or disorder
MATTER IS ENERGY.
ENERGY IS INFORMATION.
EVERYTHING IS INFORMATION.
PHYSICS SAYS THAT STRUCTURES... BUILDINGS, SOCIETIES, IDEOLOGIES... WILL SEEK THEIR POINT OF LEAST ENERGY.
THIS MEANS THAT THINGS FALL.
THEY FALL FROM HEIGHTS OF ENERGY AND STRUCTURED INFORMATION INTO MEANINGLESS, POWERLESS DISORDER.
THIS IS CALLED ENTROPY.
Entropy (S)
= a measure of randomness or disorder

27. Entropy: Tendency toward disorder

28. Entropy: Tendency toward disorder

29. Second Law of Thermodynamics
occurs without outside intervention
In any spontaneous process, the entropy of the universe increases or is +
ΔSuniverse > 0
ΔSuniverse = ΔSsystem + ΔSsurroundings

30. Entropy of the Universe
ΔSuniverse = ΔSsystem + ΔSsurroundings
Positional disorder
Energetic disorder
ΔSuniverse > spontaneous process
Both ΔSsys and ΔSsurr positive
Both ΔSsys and ΔSsurr negative
ΔSsys negative, ΔSsurr positive
ΔSsys positive, ΔSsurr negative
spontaneous process
nonspontaneous process
depends
depends

31. ΔSsys: Positional Disorder and Probability
Probability of 1 particle in left bulb = ½
" 2 particles both in left bulb = (½)(½) = ¼
" 3 particles all in left bulb = (½)(½)(½) = 1/8
" 4 " all " = (½) 4 = 1/16
" 10 " all " = (½)10 = 1/1024
" 20 " all " = (½)20 = 1/
" a mole of " all " = (½)6.02 x1023
The arrangement with the greatest entropy is the one with the highest probability (most “spread out”).

32. Entropy of the System: Positional Disorder
Ludwig Boltzmann
Ordered
state
Low probability
(few ways)
Ludwig Boltzmann
Low S
Disordered
state
High probability
(many ways)
High S
Ssystem is proportional to positional disorder
S increases with increasing # of possible positions
Ssolid < Sliquid <<<< Sgas

33. Entropy of the Surroundings (Energetic Disorder)
System
Entropy
Heat
ΔSsurr > 0
ΔHsys < 0
Surroundings
System
Surroundings
Heat
Entropy
ΔSsurr < 0
ΔHsys > 0
Low T → large entropy change (surroundings)
High T → small entropy change (surroundings)

34. The Third Law of Thermodynamics
The entropy of a perfect crystal at 0 K is zero
Everything locked into place
No molecular motion whatsoever
Crystallization of Water into Ice

35. Entropy Curve ΔHvap (l ↔ g) Δfus (s ↔ l)
Solid
Liquid
Gas S
(J/K)
ΔHvap (l ↔ g)
Δfus (s ↔ l)
Temperature (K)
S° (absolute entropy) can be calculated for any substance

36. Entropy Increases with...
Melting (fusion) Sliquid > Ssolid
ΔHfus/Tfus = ΔSfus
Vaporization Sgas > Sliquid
ΔHvap/Tvap = ΔSvap
Increasing ngas in a reaction
Heating ST2 > ST1 if T2 > T1
Dissolving (usually) Ssolution > (Ssolvent + Ssolute)
Molecular complexity more bonds, more entropy
Atomic complexity more e-, protons, neutrons

37. Entropy Practice C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)
For the above reaction, predict the sign of ΔSsys
We can predict S values based on phases…
Remember that Ssolid< Sliquid <<<<<< Sgas and that
ΔSsys = Sproducts – Sreactants
ΔSsys. = S (12 mol gas) – S (6 mol gas + 1 mol solid)
Solid is negligible, more gas products, ↑ disorder
Therefore…ΔSsys = +

38. Entropy Practice 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
For the above reaction, predict the sign of ΔSsys
We can predict S values based on phases…
Remember that Ssolid< Sliquid <<<<<<< Sgas
ΔSsys = Sproducts – Sreactants
ΔSsys = S (2 mol solid) – S (4 mol solid + 3 mol gas)
Solid is negligible, gas R → solid P , ↓ disorder
Therefore…ΔSsys = -

39. Predicting ΔSsys sign summary
Use relative S values for phases
Ssolid< Saqueous< Sliquid < Sgas
Gases always have greater entropy
Consider number of moles, especially with gases
You cannot predict for some reactions
H2(g) + Cl2(g) → 2HCl(g)
2 mol gas → 2 mol gas
Unable to predict sign ΔSsys for this reaction

40. C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
Calculating Entropy Quantitatively
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
Calculate the value of ΔS°sys:
Compound
C6H12O6(s)
O2(g)
CO2(g)
H2O(g)
S° (J/mol K)
212
205
214
189
ΔS°sys = ΣnS°(products) - ΣnS°(reactants)
= [6 S°(CO2(g)) + 6 S°(H2O(g))] – [S°(C6H12O6(s)) + 6 S°(O2(g))]
= [6(214) + 6(189)] – [(212) + 6(205)] J/K
ΔS°sys = 976 J/K

41. C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
Is this reaction spontaneous at 298K?
YES!
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
ΔSuniverse = ΔSsys + ΔSsurr
and
ΔSsurr = - ΔH/T
Compound
C6H12O6(s)
O2(g)
CO2(g)
H2O(g)
ΔH°f (kJ/mol)
-1275
-393.5
-242
S° (J/mol K)
212
205
214
189
ΔH°rxn = ΣnΔH°f (products) - ΣnΔH°f(reactants)
= [6 ΔH°f(CO2(g)) + 6 ΔH°f(H2O(g))] – [ΔH°f(C6H12O6(s)) + 6 ΔH°f(O2(g))]
= [6(-393.5) + 6(-242)] – [(-1275) + 6(0)] kJ
ΔH°rxn = kJ
ΔSsys = 976 J/K from last problem
ΔSuniverse = kJ/K + -(-2538 kJ)/298K =
9.49 kJ/K

42. Entropy (S) Review ΔSuniverse > 0 for spontaneous processes
ΔSuniverse = ΔSsystem + ΔSsurroundings
positional
energetic
We can find the absolute entropy value for a substance
S° values for elements & compounds in their standard states are tabulated (Thermodynamic Appendix)
For any chemical reaction, we can calculate ΔS°rxn:
ΔS°rxn = ΣnS°(products) - ΣnS°(reactants)

43. Recap: Characteristics of Entropy
S is a state function (we can use final – initial)
S is extensive (more stuff, more entropy)
At 0 K, S = 0 (we can determine absolute entropy)
S > 0 for elements and compounds in their standard states
Raise T  increase S
Increase ngas  increase S
More complex systems  larger S

44. A process (at constant T, P) is spontaneous if free energy decreases
Gibbs Free Energy (G)
G = H – TS
At constant temperature,
ΔG = ΔH – TΔS
(system’s point of view)
ΔG = ΔH – TΔS
Divide both sides by –T
-ΔG/T = -ΔH/T + ΔS
ΔSuniverse = ΔS – ΔH/T
–ΔG means +ΔSuniv
A process (at constant T, P) is spontaneous if free energy decreases
Josiah Gibbs

45. ΔG and Chemical Reactions
ΔG = ΔH – TΔS
If ΔG < 0, the reaction is spontaneous
If ΔG > 0, the reaction is not spontaneous
(The reverse reaction is spontaneous)
If ΔG = 0, the reaction is at equilibrium
Neither the forward nor the reverse reaction is favored
Both reactions are occurring simultaneously and at equal rates
ΔG is an extensive, state function
Depends on how much stuff
Depends on final and initial states only

46. Ba(OH)2(s) + 2NH4Cl(s)  BaCl2(s) + 2NH3(g) + 2 H2O(l)
Is the following reaction spontaneous at 298 K?
Ba(OH)2(s) + 2NH4Cl(s)  BaCl2(s) + 2NH3(g) + 2 H2O(l)
ΔH°rxn = 50.0 kJ (per mole Ba(OH)2)
ΔS°rxn = 328 J/K (per mole Ba(OH)2)
ΔG = ΔH - TΔS
ΔG° = 50.0 kJ – 298 K(0.328 kJ/K)
ΔG° = – 47.7 kJ
Spontaneous
At what T does the reaction stop being spontaneous?
The T where ΔG = 0
ΔG = 0 = 50.0 kJ – T(0.328 kJ/K)
50.0 kJ = T(0.328 kJ/K)
T = 152 K
not spontaneous below 152 K

47. Effect of ΔH and ΔS on Spontaneity
ΔG = ΔH – TΔS
ΔG (-) → spontaneous reaction ΔH – + ΔS + –
Spontaneous?
Spontaneous at all temps
Spontaneous at high temps
Reverse reaction spontaneous at low temps
Spontaneous at low temps
Reverse reaction spontaneous at high temps
Not spontaneous at any temp

48. Ways to Calculate ΔG°rxn
1. ΔG° = nΔG°f(products) - nΔG°f(reactants)
ΔG°f = free energy change when forming 1 mole of compound from elements in their standard states (see Thermodynamics Appendix for values)
2. ΔG° = ΔH° - TΔS°
3. ΔG° can be calculated by combining ΔG° values for several reactions
Using Hess’s Law! Your favorite!

49. Calculate ΔG° for the following reaction: 2 H2(g) + O2(g)  2 H2O(g)
1. ΔG° = ΔG°f(products) - ΔG°f(reactants)
ΔG°f(O2(g)) = 0
ΔG°f(H2(g)) = 0
ΔG°f(H2O(g)) = -229 kJ/mol
ΔG° = (2(-229 kJ) – 2(0) – 0) kJ = -458 kJ
2. ΔG° = ΔH° - TΔS°
ΔH° = -484 kJ
ΔS° = -89 J/K
ΔG° = -484 kJ – 298 K( kJ/K) = -457 kJ

50. 2H2(g) + O2(g)  2 H2O(g) Method 1: Method 2: Method 3:
3. ΔG° = combination of ΔG° from other reactions (using Hess’s Law)
2H2O(l)  2H2(g) + O2(g) ΔG°1 = 475 kJ
H2O(l)  H2O(g) ΔG°2 = 8 kJ
ΔG° = - ΔG°1 + 2(ΔG°2)
ΔG° = -475 kJ + 16 kJ = -459 kJ
Method 1:
Method 2:
Method 3:
Method 1: kJ
Method 2: kJ
Method 3: kJ

51. What is Free Energy, Really?
NOT just “another form of energy”
Free Energy is the energy available to do useful work
If ΔG is negative, the system can do work (wmax = ΔG)
If ΔG is positive, then ΔG is the work required to make the process happen
Example: Photosynthesis
6 CO2 + 6 H2O  C6H12O6 + 6 O2
ΔG = 2870 kJ/mol of glucose at 25°C
Thus, 2870 kJ of work is required to photosynthesize 1 mole of glucose