1. Chem. 1B – 10/18 Lecture

2. Announcements I Exam 2: SacCT/Website Lab:
Next Week on Thursday (10/27)
Will Cover Titrations, Solubility, Complex Ions (from Ch. 16) + Chapter 17 (Thermodynamics)
SacCT/Website
Updated bonus points + quiz keys (1 to 5 posted)
Lab:
Lab Midterm and Exp 3 report on Wed./Thurs.
Note: Syllabus says Lab midterm is on Exp 1, 2, 5, and 7, but will also have questions related to pre-lab on Experiment 3

3. Announcements II Today’s Lecture Thermodynamics Gibb’s Free Energy
Reversible Reactions
Relating Gibb’s Free Energy Changes to Equilibrium Constants

4. Chem 1B – Thermodynamics Chapter 17 – Gibbs Free Energy
Meaning of Gibbs Free Energy
DG < 0 means a spontaneous process
DG > 0 means a non-spontaneous process
Does this match our previous knowledge?
DH < 0 is generally spontaneous (and gives rise to DG < 0 as DG = DH – TDS)
DS > 0 is generally spontaneous (and gives rise to DG < 0 as DG = DH – TDS) 4

5. Chem 1B – Thermodynamics Chapter 17 – Gibbs Free Energy
How Temperature Affects Reaction Direction
We have 4 combinations of signs of DH and DS:
DH < 0 and DS > 0 (DG always < 0)
DH < 0 and DS < 0 (DG < 0 at low T only)
DH > 0 and DS > 0 (DG < 0 at high T only)
DH > 0 and DS < 0 (DG always > 0)
When DH and DS have the same sign, process depends on T 5

6. Chem 1B – Thermodynamics Chapter 17 – Gibbs Free Energy
Some Example Problems
Under what temperature regimes will these reactions be spontaneous?
N2(g) + O2(g) ↔ 2NO(g) DH° = 91.3 kJ/mol
N2(g) + 3H2(g) ↔ 2NH3(g) DH° = kJ/mol
NH4NO3(s) ↔ NH4+(aq) + NO3- (aq) DH° = 25.7 kJ/mol
N2(g) + 2H2(g) ↔ N2H4(g) DH ° = 95.4 kJ/mol
2C2H6(g) + 7O2(g) ↔ 4CO2(g) + 6H2O(g)
DH ° = -2,855 kJ/mol 6

7. Chem 1B – Thermodynamics Chapter 17 – Gibbs Free Energy
Some Example Problems
From Standard Tables, Calculate DG° for the following reaction:
N2(g) + 3H2(g) ↔ 2NH3(g)
At what temperature is DG° = 0 for the above reaction?
Note: can also calculate DG from standard values
Species
N2(g)
H2(g)
NH3(g)
S° (J/mol K)
191.6
130.7
192.8
DH° (kJ/mol)
-45.9 7

8. Chem 1B – Thermodynamics Chapter 17 – Reversibility
Reactions can be reversible or irreversible
For a process to use 100% of available energy (Gibbs free energy) for work, process must be reversible
Example of reversible process:
H2O(s) ↔ H2O(l) at 0°C
at T > 0°C, this process is spontaneous and non-reversible
When chemical reactions are used to generate energy (e.g. electrochemical reactions in a battery), some energy becomes heat 8

9. Chem 1B – Thermodynamics Chapter 17 – Reversibility
Battery Example – cont.
The greater the power draw on the battery (current), the less efficient (or reversible) the energy use will be 9

10. Chem 1B – Thermodynamics Chapter 17 – Gibbs Free Energy
Free Energy Under Nonstandard Conditions
DG = DG° + RTlnQ
where Q = reaction quotient
Example AgCl(s) ↔ Ag+(aq) + Cl- (aq)
Standard conditions DG° = 72.9 kJ/mol
would require [Ag+] = [Cl-] = 1 M
Q = [Ag+][Cl-] 10

11. Chem 1B – Thermodynamics Chapter 17 – Gibbs Free Energy and Equilibrium
Under Nonstandard Conditions:
DGrxn = DGºrxn + RTlnQ
Example Reaction:
CaSO4(s) ↔ Ca2+(aq) + SO42-(aq)
DGºrxn = 23.7 kJ/mol and Q = [Ca2+][SO42-]
If we put CaSO4(s) in water Q = 0 (before reaction)
lnQ is undefined but negative, making DGrxn negative even though DGºrxn is positive
As CaSO4(s) dissolves, [Ca2+] and [SO42-] increase, and DGrxn increases (e.g. when [Ca2+] = M, DGrxn = kJ/mol) 11

12. Chem 1B – Thermodynamics Chapter 17 – Gibbs Free Energy and Equilibrium
Example Reaction – cont.
CaSO4(s) ↔ Ca2+(aq) + SO42-(aq)
We also could start by mixing high concentrations of Ca2+(aq) + SO42-(aq) (e.g. 2 M each)
In this case DGrxn > DGºrxn and reaction goes even stronger to reactants Q > 1
At Equilibrium DGrxn = 0 = DGºrxn + RTlnK
or DGºrxn = -RTlnK which allows us to determine K from DGº or visa versa 12

13. Chem 1B – Thermodynamics Chapter 17 – Example Problems
If for the reaction: 2H2(g) + N2(g) ↔ N2H4(g) has DHº > 0 and DSº < 0, how can N2H4(g) ever be formed?
Not from 2H2(g) + N2(g) (in more than small quantities), but can’t rule out other sources
For example: N2O(g) + 3H2(g) ↔ N2H4(g) + H2O(g) calculate DGº to show
What caused DGº < 0? Favorability in forming H2O(g) (Note: this reaction is energetically favorable, but may not occur) 13

14. Chem 1B – Thermodynamics Chapter 17 – Example Problems
Catalysts can help energetically favorable reactions occur, but can not allow products to form if DGº of products is higher than reactants. Which of the following hydrocarbons can be produced by syngas (CO + H2) – assume H2O forms if needed?
CH4 C2H6 CH3OH C2H2
How do we solve? Make balanced reactions and calculate DGº (or DHº and DSº)
What are the best conditions for these reactions? 14

15. Chem 1B – Thermodynamics Chapter 17 – Example Problems
HI has a DGfº = 1.7 kJ/mol at 298 K
Can it be formed from H2(g) + I2(s)?
What is K for the reaction: H2(g) + I2(g) ↔ 2HI(g) if DGfº(I2(g)) = 19.3 kJ/mol
Does increasing T favor reactants or products?
Water is sprayed into a reaction flask at equilibrium and absorbs 99% of the HI but little of the other gases. Explain what this will do to DG. 15

16. Chem 1B – Thermodynamics Chapter 17 – Equilibrium and Temperature
We know DGº changes with temperature according to:
DGº = DHº – TDSº (note: DHº and DSº may change with T – but generally not a lot)
We also know that DGº = -RTlnK
-RTlnK = DHº – TDSº or lnK = -DHº/RT + DSº/R
A Plot of lnK vs. 1/T would give m (slope) = DHº/R and b (y-intercept) = +DSº/R
What would a positive slope in the above plot mean?
What would a positive y-intercept mean? 16

17. Chem 1B – Thermodynamics Chapter 17 – Equilibrium and Temperature
A chemist has designed a catalyst allowing ethanol to be made from CO + H2. The catalyst will only work at T > 150°C. At that temperature will the product still be favored? Determine the K at that temperature. 17