Chem. 1B – 10/18 Lecture

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

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

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

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

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° = -91.8 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

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

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

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

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

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+] = 0.001 M, DGrxn = -10.5 kJ/mol) 11

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

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

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

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

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

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