CHAPTERS 16-18,21, CHAPTERS 16-18,21, AND 22 AND 22

AP CHEMISTRYAP CHEMISTRY

FREE ENERGY CHANGEFREE ENERGY CHANGE

G = GG = G°°+ RT In(P)+ RT In(P)

∆∆G = G = ∆∆GG°°+ RT In+ RT In( P( Ppp ) )

(P(PRR)(P)(PRR))

∆∆G= G= ∆∆GG°° + RT In(Q) + RT In(Q) If G If G ∆∆ = 0 and Q = k, then = 0 and Q = k, then ∆∆G G °° = -RTIn(k) = -RTIn(k) Page 775 Table 16.6Page 775 Table 16.6∆∆GG°° = = ∆∆HH°° - T - T∆∆SS°° Remember SRemember S°° is in J not in kJ is in J not in kJ

CALCULATION OF CALCULATION OF ∆∆G FROM G FROM ∆∆GG°°

∆∆G = G = ∆∆GG°° + RT In Q + RT In Q

Q = reaction quotient (section 15.5)Q = reaction quotient (section 15.5)

R= 8.314 J/KmolR= 8.314 J/Kmol

T = absolute temperatureT = absolute temperature

-38kJ/mol-38kJ/mol

Example on page 776Example on page 776

RELATION BETWEEN RELATION BETWEEN ∆∆GG°° AND k AND k

∆∆GG°° = -RTInk k = equilibrium constant = -RTInk k = equilibrium constant

If k> 1, If k> 1, ∆∆GG°°< 0, spontaneous at standard < 0, spontaneous at standard conditionsconditions

If k<1, If k<1, ∆∆GG°°>0, non-spontaneous at >0, non-spontaneous at standard conditionsstandard conditions

If k =1, If k =1, ∆∆GG°°= 0, equilibrium at standard = 0, equilibrium at standard conditionsconditions

REDOXREDOX

When oxidation number increases it is oxidizedWhen oxidation number increases it is oxidized

When oxidation number decreases it is reducedWhen oxidation number decreases it is reduced

Substances that goes through oxidation are called the Substances that goes through oxidation are called the reducing agentreducing agent

Substances that that goes through reduction are called Substances that that goes through reduction are called the oxidizing agentthe oxidizing agent

Balancing redoxBalancing redox

VOLTAIC CELLSVOLTAIC CELLS

► A spontaneous reaction used to produce A spontaneous reaction used to produce electrical energyelectrical energy

► Salt bridge cells Salt bridge cells ► Zn(s) + CuZn(s) + Cu2+2+---> Zn---> Zn2+2+(aq) + Cu(s)(aq) + Cu(s)► Must design cells to make electron transfer Must design cells to make electron transfer

occur indirectlyoccur indirectly

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Anode: Zn(s) --> ZnAnode: Zn(s) --> Zn2+2+(aq) + 2e(aq) + 2e--

Cathode: CuCathode: Cu2+2+(aq) + 2e(aq) + 2e-- --> Cu(s) --> Cu(s)

The salt bridge allows a current to flow, The salt bridge allows a current to flow, but prevents any contact between the zinc but prevents any contact between the zinc metal and copper (II) ions. This would metal and copper (II) ions. This would short circuit the cellshort circuit the cell

Why?Why?

STANDARD VOLTAGESTANDARD VOLTAGE

EE°° Cell voltage when all the species are at standard concentration Cell voltage when all the species are at standard concentration

(1atm for gases, 1M for solutions in water)(1atm for gases, 1M for solutions in water) EE°°cellcell = E = E°°oxidationoxidation + E + E°°reductionreduction Zn(s) + 2HZn(s) + 2H++(aq, 1M) -->Zn(aq, 1M) -->Zn2+2+(aq, 1M) + H(aq, 1M) + H22(g, 1 atm)(g, 1 atm) EE°° = +.0762V = E = +.0762V = E°°oxoxZn + EZn + E°°redredHH++

Cu(s) --> CuCu(s) --> Cu2+2+(aq) + 2e(aq) + 2e--

EE°°oxoxCu = -ECu = -E°°redredCuCu2+2+ = -0.34 V = -0.34 V Relative strengths of oxidizing and reducing agentsRelative strengths of oxidizing and reducing agents The larger (more +) the value EThe larger (more +) the value E°°redred, the stronger the oxidizing agent, the stronger the oxidizing agent The smaller the EThe smaller the E°°redred (more -) the stronger the reducing agent (more -) the stronger the reducing agent

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Table 17.1Table 17.1

Calculation of ECalculation of E°°

EE°° = E = E°°oxox + E + E°°redred

RELATION BETWEEN E, G, AND RELATION BETWEEN E, G, AND kk

∆∆GG°° = -n = -n E E lnk = n Elnk = n E°°/0.0257/0.0257 = 96.5 kJ/V mole of e= 96.5 kJ/V mole of e--

If EIf E°° is positive, then is positive, then ∆∆GG°° is negative is negative If lnk is positive, then k > 1If lnk is positive, then k > 1 Nernst equationNernst equation E = EE = E°° - (RT/n - (RT/n ) lnQ = E) lnQ = E°° - (0.0257/n)(lnQ) - (0.0257/n)(lnQ) When using Q remember that gases are When using Q remember that gases are entered as partial pressure in atm and solutes entered as partial pressure in atm and solutes are concentrations (moles per liter)are concentrations (moles per liter)

ELECTROLYTIC CELLSELECTROLYTIC CELLS

Electrical energy supplied to bring abut a Electrical energy supplied to bring abut a nonspontaneous redox reaction nonspontaneous redox reaction Amount of products formed in electrolysisAmount of products formed in electrolysis AgAg+ + (aq) + e(aq) + e-- --> Ag (s) --> Ag (s) 1 mole of electrons = 96485 C = 1 mol Ag1 mole of electrons = 96485 C = 1 mol Ag # coulombs = # amperes X # of seconds# coulombs = # amperes X # of seconds # Joules = # coulombs X # of volts# Joules = # coulombs X # of volts

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Flow chartFlow chart

Current and time --> quantity of charge in Current and time --> quantity of charge in coulombs --> moles of electrons --> moles of coulombs --> moles of electrons --> moles of elements --> grams of elementelements --> grams of element

Commercial cellsCommercial cells Electrolysis of aqueous NaClElectrolysis of aqueous NaCl

2H2H22O(l) + 2ClO(l) + 2Cl--(aq) --> H(aq) --> H22(g) +Cl(g) +Cl22(g) + 2OH(g) + 2OH--(aq)(aq)

How much voltage is requiredHow much voltage is required

LEAD STORAGE BATTERIESLEAD STORAGE BATTERIES

AnodeAnode

Pb(s) + SOPb(s) + SO442-2-(aq) --> PbSO(aq) --> PbSO44(s) + 2e(s) + 2e--

CathodeCathode

PbOPbO22(s) + 4H(s) + 4H++(aq) + SO(aq) + SO442-2-(aq) + 2e(aq) + 2e-- --> PbSO --> PbSO44(s) + (s) +

2H2H22O(l)O(l)

Overall reactionOverall reaction

Pb(s) + 2SOPb(s) + 2SO442-2-(aq) +PbO(aq) +PbO22(s) +4H(s) +4H++(aq)-->2PbSO(aq)-->2PbSO44(s) + (s) +

2H2H22O(l)O(l)

As the cell discharges, concentration of the sulfuric acid As the cell discharges, concentration of the sulfuric acid and the density of the battery will decreaseand the density of the battery will decrease

RADIOACTIVE DECAYRADIOACTIVE DECAY

Beta particles or Beta particles or ββ or e or e--

# of protons increase by one, mass stays constant# of protons increase by one, mass stays constant

141466C --> C --> 1414

77N + N + 00-1-1ee--

Zone of stabilityZone of stability Up to 83Up to 83 All nuclides with 84 or more protons are unstableAll nuclides with 84 or more protons are unstable Small atoms that have a 1 to 1 ratio are stable. As they Small atoms that have a 1 to 1 ratio are stable. As they become larger more neutrons are needed to keep the become larger more neutrons are needed to keep the protons in the nucleus (glue). After 84 no matter how protons in the nucleus (glue). After 84 no matter how many neutrons are used the nucleus will break apart many neutrons are used the nucleus will break apart

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Alpha particles or Alpha particles or αα or or 4422HeHe2+2+

Nuclides lose 2 protons and 2 neutrons. These particles Nuclides lose 2 protons and 2 neutrons. These particles are slowerare slower 230230

9090Th --> Th --> 4422HeHe2+2+ + + 226226

8888RaRa Gamma rayGamma ray High radiation with no loss of mass High radiation with no loss of mass 238238

9292U --> U --> 4422HeHe2+2+ + + 230230

9090Th + Th + γγ PositronPositron 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 chargeSame mass as an electron but opposite charge Proton # decreases by oneProton # decreases by one 2222

1111Na --> Na --> 0011e + e + 2222

1010Ne Ne

ELECTRON CAPTUREELECTRON CAPTURE

Inner-orbital electron captured by the Inner-orbital electron captured by the nucleusnucleus

2012018080Hg + Hg + 00

-1-1e- --> e- --> 2012017979Au + Au + γγ

RATE OF DECAYRATE OF DECAY

ln(N/Nln(N/Noo) = -kt N = # of nuclides) = -kt N = # of nuclides Half life tHalf life t1/21/2 = 0.693/k = 0.693/k 1Ci = 3.700 X 101Ci = 3.700 X 101010 atoms/s atoms/s

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Page 849 table 18.3Page 849 table 18.3

Age of organic material; measure of C--14 Age of organic material; measure of C--14 contentcontent

141477N + N + 11

00n ---> n ---> 141466C + C + 11

11HH

141466C ---> C ---> 1414

77N + N + 00-1-1ββ; t; t1/21/2 = 5720 yr = 5720 yr

TABLE 18.5 radiotracers TABLE 18.5 radiotracers

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Mass defect = mass(n + p) - mass of Mass defect = mass(n + p) - mass of nucleusnucleus C-12:C-12:Fission occurs with very heavy nucleiFission occurs with very heavy nuclei Fusion occurs with light nucleiFusion occurs with light nuclei FissionFission

2352359292U + U + 11

00n --> n --> 90903737Rb + Rb + 144144

5555Cs + 2Cs + 21100nn

Many different isotopes are formedMany different isotopes are formed

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More neutrons are produced than More neutrons are produced than consumed, leading to a chain reaction. In consumed, leading to a chain reaction. In nuclear reactors, excess neutrons are nuclear reactors, excess neutrons are absorbed by cadmium rodsabsorbed by cadmium rods Nuclei produced have too many neutrons Nuclei produced have too many neutrons and hence are intensely radioactive.and hence are intensely radioactive.

90903737Rb --> Rb --> 00

-1-1e- + e- + 90903838SrSr

This is the principal danger with nuclear This is the principal danger with nuclear reactors reactors

CHAPTER 21CHAPTER 21

Colors of the ions Colors of the ions Pages 946, 951-955Pages 946, 951-955 Oxidation states 952-954Oxidation states 952-954Vanadium has 4 states (common one 5Vanadium has 4 states (common one 5++), manganese ), manganese has 4 states, and nitrogen has 5 stateshas 4 states, and nitrogen has 5 statesCoordination #’s pages 955-959Coordination #’s pages 955-959Coordination numbers and ligands 956-960Coordination numbers and ligands 956-960Optical isomerism page 961-965Optical isomerism page 961-965Chiral 964 Chiral 964 Go over page 968 Go over page 968 Read and take notes 965 to end of chapterRead and take notes 965 to end of chapter

CHAPTER 22CHAPTER 22

Go over table 22.1Go over table 22.1

Exercise 22.3Exercise 22.3

Exercise 22.4Exercise 22.4

Exercise 22.5Exercise 22.5

Read and take notes 1008 to end of Read and take notes 1008 to end of chapterchapter