CHAPTERS 16-18,21, AND 22 AP CHEMISTRY

FREE ENERGY CHANGE G = G°+ RT In(P) ∆G = ∆G°+ RT In( P p ) (P R )(P R ) (P R )(P R ) ∆G= ∆G° + RT In(Q) If G ∆ = 0 and Q = k, then ∆G ° = -RTIn(k) If G ∆ = 0 and Q = k, then ∆G ° = -RTIn(k) Page 775 Table 16.6 Page 775 Table 16.6 ∆G° = ∆H° - T∆S° Remember S° is in J not in kJ Remember S° is in J not in kJ

CALCULATION OF ∆G FROM ∆G° ∆G = ∆G° + RT In Q Q = reaction quotient (section 15.5) Q = reaction quotient (section 15.5) R= J/Kmol R= J/Kmol T = absolute temperature T = absolute temperature-38kJ/mol Example on page 776 Example on page 776

RELATION BETWEEN ∆G° AND k ∆G° = -RTInk k = equilibrium constant If k> 1, ∆G°< 0, spontaneous at standard conditions If k<1, ∆G°>0, non-spontaneous at standard conditions If k =1, ∆G°= 0, equilibrium at standard conditions

REDOX When oxidation number increases it is oxidized When oxidation number decreases it is reduced When oxidation number decreases it is reduced Substances that goes through oxidation are called the reducing agent Substances that goes through oxidation are called the reducing agent Substances that that goes through reduction are called the oxidizing agent Substances that that goes through reduction are called the oxidizing agent Balancing redox Balancing redox

VOLTAIC CELLS ► A spontaneous reaction used to produce electrical energy ► Salt bridge cells ► Zn(s) + Cu > Zn 2+ (aq) + Cu(s) ► Must design cells to make electron transfer occur indirectly

Continue Anode: Zn(s) --> Zn 2+ (aq) + 2e - Cathode: Cu 2+ (aq) + 2e - --> Cu(s) The salt bridge allows a current to flow, but prevents any contact between the zinc metal and copper (II) ions. This would short circuit the cell The salt bridge allows a current to flow, but prevents any contact between the zinc metal and copper (II) ions. This would short circuit the cell Why? Why?

STANDARD VOLTAGE E°E°E°E° Cell voltage when all the species are at standard concentration (1atm for gases, 1M for solutions in water) Cell voltage when all the species are at standard concentration (1atm for gases, 1M for solutions in water) E° cell = E° oxidation + E° reduction E° cell = E° oxidation + E° reduction Zn(s) + 2H + (aq, 1M) -->Zn 2+ (aq, 1M) + H 2 (g, 1 atm) Zn(s) + 2H + (aq, 1M) -->Zn 2+ (aq, 1M) + H 2 (g, 1 atm) E° = V = E° ox Zn + E° red H + E° = V = E° ox Zn + E° red H + Cu(s) --> Cu 2+ (aq) + 2e - Cu(s) --> Cu 2+ (aq) + 2e - E° ox Cu = -E° red Cu 2+ = V E° ox Cu = -E° red Cu 2+ = V Relative strengths of oxidizing and reducing agents Relative strengths of oxidizing and reducing agents The larger (more +) the value E° red, the stronger the oxidizing agent The larger (more +) the value E° red, the stronger the oxidizing agent The smaller the E° red (more -) the stronger the reducing agent The smaller the E° red (more -) the stronger the reducing agent

Continue Table 17.1 Calculation of E° Calculation of E° E° = E° ox + E° red E° = E° ox + E° red

RELATION BETWEEN E, G, AND k ∆G° = -n  E lnk = n E°/ lnk = n E°/  = 96.5 kJ/V mole of e -  = 96.5 kJ/V mole of e - If E° is positive, then ∆G° is negative If E° is positive, then ∆G° is negative If lnk is positive, then k > 1 If lnk is positive, then k > 1 Nernst equation Nernst equation E = E° - (RT/n  ) lnQ = E° - (0.0257/n)(lnQ) E = E° - (RT/n  ) lnQ = E° - (0.0257/n)(lnQ) When using Q remember that gases are entered as partial pressure in atm and solutes are concentrations (moles per liter) When using Q remember that gases are entered as partial pressure in atm and solutes are concentrations (moles per liter)

ELECTROLYTIC CELLS Electrical energy supplied to bring abut a nonspontaneous redox reaction Amount of products formed in electrolysis Amount of products formed in electrolysis Ag + (aq) + e - --> Ag (s) Ag + (aq) + e - --> Ag (s) 1 mole of electrons = C = 1 mol Ag 1 mole of electrons = C = 1 mol Ag # coulombs = # amperes X # of seconds # coulombs = # amperes X # of seconds # Joules = # coulombs X # of volts # Joules = # coulombs X # of volts

Continue Flow chart Current and time --> quantity of charge in coulombs --> moles of electrons --> moles of elements --> grams of element Current and time --> quantity of charge in coulombs --> moles of electrons --> moles of elements --> grams of element Commercial cells Commercial cells Electrolysis of aqueous NaCl Electrolysis of aqueous NaCl 2H 2 O(l) + 2Cl - (aq) --> H 2 (g) +Cl 2 (g) + 2OH - (aq) 2H 2 O(l) + 2Cl - (aq) --> H 2 (g) +Cl 2 (g) + 2OH - (aq) How much voltage is required How much voltage is required

LEAD STORAGE BATTERIES Anode Pb(s) + SO42-(aq) --> PbSO4(s) + 2e- Cathode PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- --> PbSO4(s) + 2H2O(l) Overall reaction Pb(s) + 2SO42-(aq) +PbO2(s) +4H+(aq)-->2PbSO4(s) + 2H2O(l) As the cell discharges, concentration of the sulfuric acid and the density of the battery will decrease

RADIOACTIVE DECAY Beta particles or β or e - # of protons increase by one, mass stays constant # of protons increase by one, mass stays constant 14 6 C --> 14 7 N e C --> 14 7 N e - Zone of stability Zone of stability Up to 83 Up to 83 All nuclides with 84 or more protons are unstable All nuclides with 84 or more protons are unstable Small atoms that have a 1 to 1 ratio are stable. As they become larger more neutrons are needed to keep the protons in the nucleus (glue). After 84 no matter how many neutrons are used the nucleus will break apart Small atoms that have a 1 to 1 ratio are stable. As they become larger more neutrons are needed to keep the protons in the nucleus (glue). After 84 no matter how many neutrons are used the nucleus will break apart

Continue Alpha particles or α or 4 2 He 2+ Nuclides lose 2 protons and 2 neutrons. These particles are slower Nuclides lose 2 protons and 2 neutrons. These particles are slower Th --> 4 2 He Ra Th --> 4 2 He Ra Gamma ray Gamma ray High radiation with no loss of mass High radiation with no loss of mass U --> 4 2 He Th + γ U --> 4 2 He Th + γ Positron Positron Nuclides below the zone of stability (ratio is too small). Nuclides below the zone of stability (ratio is too small). Same mass as an electron but opposite charge Same mass as an electron but opposite charge Proton # decreases by one Proton # decreases by one Na --> 0 1 e Ne Na --> 0 1 e Ne

ELECTRON CAPTURE Inner-orbital electron captured by the nucleus Hg e- --> Au + γ Hg e- --> Au + γ

RATE OF DECAY ln(N/N o ) = -kt N = # of nuclides ln(N/N o ) = -kt N = # of nuclides Half life t 1/2 = 0.693/k Half life t 1/2 = 0.693/k 1Ci = X atoms/s 1Ci = X atoms/s

Continue Page 849 table 18.3 Age of organic material; measure of C--14 content Age of organic material; measure of C--14 content 14 7 N n ---> 14 6 C H 14 7 N n ---> 14 6 C H 14 6 C ---> 14 7 N β; t 1/2 = 5720 yr 14 6 C ---> 14 7 N β; t 1/2 = 5720 yr TABLE 18.5 radiotracers

Continue Mass defect = mass(n + p) - mass of nucleus C-12: Fission occurs with very heavy nuclei Fusion occurs with light nuclei Fission 23592U + 10n --> 9037Rb Cs + 210n Many different isotopes are formed

Continue More neutrons are produced than consumed, leading to a chain reaction. In nuclear reactors, excess neutrons are absorbed by cadmium rods Nuclei produced have too many neutrons and hence are intensely radioactive. Nuclei produced have too many neutrons and hence are intensely radioactive Rb --> 0 -1 e Sr Rb --> 0 -1 e Sr This is the principal danger with nuclear reactors This is the principal danger with nuclear reactors

CHAPTER 21 Colors of the ions Pages 946, Pages 946, Oxidation states Oxidation states Vanadium has 4 states (common one 5 + ), manganese has 4 states, and nitrogen has 5 states Coordination #’s pages Coordination numbers and ligands Optical isomerism page Chiral 964 Go over page 968 Read and take notes 965 to end of chapter

CHAPTER 22 Go over table 22.1 Exercise 22.3 Exercise 22.4 Exercise 22.5 Read and take notes 1008 to end of chapter