Chapters 6,16 Enthalpy, Entropy, Free Energy 1 st Law of Thermodynamics: Energy of the universe is constant. 2 nd Law of Thermodynamics: In any spontaneous process, there is always an increase in entropy in the universe. ( Suniv = Ssystem + Ssurroundings) ( Ssurroundings = - H/T) 3 rd Law of Thermodynamics: Entropy of a perfect crystal at zero Kelvin is zero. G Free Energy (energy available to do work) - exergonic spontaneous release free energy + endergonic nonspontaneous absorb free energy H Enthalpy (heat) - exothermic release heat + endothermic absorb heat S Entropy (randomness) G = H T S + increasing randomness + + - decreasing randomness or + + or + + + + + + + G = G + R T lnq change in free energy change in free energy 8.31 J Kelvin Reaction at a nonstandard at standard molk Temp Quotient condition condition @ equilibrium G = 0, so G = R T lnk equal reactants and products G = 0 if K = 1 favor products/spon G < 0 if K > 1 favor reactants/nonspon G > 0 if K < 1 AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 1
H = CHANGE in HEAT energy (enthalpy) in rxn. Rxn absorbs energy H > 0 ENDOthermic Rxn releases energy H < 0 EXOthermic Ex. N2(g) + 2O2(g) 2NO2(g) H = 68 kj [ENDOthermic] 2H2O2(l) 2H2O(l) + O2(g) H = -196.1 kj [EXOthermic] E E Reactants Products ENDOthermic Also: N2(g) + 2O2(g) 2NO2(g) 2N2(g) + 4O2(g) 4NO2(g) 2NO2(g) N2(g) + 2O2(g) Stoichiometry Reactants Products EXOthermic H = 68 kj H = 2(68) kj H = -68 kj N2(g) + 2O2(g) + 68 kj 2NO2(g) 3.0 mol mol kj mol mol mol 34 kj mol g 48.0 g kj g AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 2
Finding H of a rxn. Hf = Heat of formation of ONE MOLE substance from its elements in their STANDARD STATES. (see Appendix A19 ) Hf of ELEMENTS in standard state = 0 kj [ie- C(s), O2(g), F2(g), Br2(l), etc.] Ex. Hf of CH4(g) = -75 kj or C(s) + 2H2(g) 1 CH4(g) Hf = -75 kj Ex. Hf of CO2(g) = -393.5 kj or C(s) + O2(g) 1 CO2(g) Hf = -393.5 kj Ex. Hf of H2O(g) = -242 kj or H2(g) + 1/2O2(g) 1 H2O(g) Hf = -242 kj H = [sum Hf (products)] - [sum Hf (reactants)] CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Hf : (-75) 2(0) (-393.5) 2(-242) (-75) (-877.5) H = -877.5 (-75) kj = -802.5 kj Bond Energy (Reactants Products) or (Brokendo Formexo) (see p. 351) Stoichiometry... CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802.5 kj (exothermic) Ex. 2 mol kj Ex. 16.0 g kj AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 3
FREE ENERGY energy of a system available to do work. exergonic - releases Free Energy SPONTANEOUS endergonic - absorbs Free Energy NONSPONTANEOUS G = H T S Appendix A19 Appendix A19 Appendix A19 Find G (@ 25 C) of C3H8(g) + 5 O2(g) 3CO2(g) + 4H2O(g) 1) G = Gf (prod) Gf (react) = [3(-394) +4(-229)] kj/mol [(-24) + 5(0.0)]kJ/mol = -2074 kj/mol *matching 2) H = Hf (prod) Hf (react) = [3(-393.5) + 4(-242)kJ/mol] [(-104) +5(0.0)]kJ/mol = -2044.5 kj/mol S = Sf (prod) Sf (react) = [3(214) + 4(189)] J/mol K [(270) + 5(205)]J/mol K = 103 J/mol K G = H T S = -2044.5 kj/mol 298K[0.103 kj/mol K] = -2075 kj/mol *matching Entropy Sign Predictions Predict the sign for S for: CaCO3(s) CaO(s) + CO2(g) 2H2(g) + O2(g) 2H2O(l) AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 4
Chapter 17 Electrochemistry Reduction charge = gain e - (can lose oxygen) Oxidation charge = lose e - (can gain oxygen) Standard Reduction Potentials (SRP) See p. 796 or P.T. Ref packet E > 0 spontaneous E < 0 nonspontaneous SPONTANEOUS (E > 0) chemical energy electrical energy Galvanic Cell Voltaic Cell Electrochemical Cell Chemical Cell Half Reactions: Voltage (E ) Cathode red Cu +2 (aq) + 2e - Cu(s) 0.34 V Anode ox Zn(s) Zn +2 (aq) + 2e - +0.76 V Overall Cu +2 (aq) + Zn(s) Cu(s) + Zn +2 (aq) 1.10 V Line Notation Zn(s) Zn +2 (aq) Cu +2 (aq) Cu(s) Anode Cathode AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 5
Half Reactions: Voltage (E ) Cathode red 2Ag + (aq) +2 e - 2Ag(s) 0.80 V* Anode ox H2(g) 2H + (aq) + 2e - 0.00 V Overall 2Ag + (aq) + H2(g) 2Ag(s) + 2H + (aq) 0.80 V *Notes: Do not multiply voltage only switch sign if reversed If there is no solid form of an element in the reaction (iehydrogen), then a platinum electrode is used instead. Line Notation Pt(s) H2(g), H + (aq) Ag + (aq) Ag(s) G = -nfe where F (Faraday) = 96485 Coulombs so G = -n F E = -R T lnk for standard conditions and G = -n F E = -R T lnq for nonstandard conditions AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 6
You Try It: Fill in the missing information for an aluminum and magnesium Galvanic cell. Half Reactions: Voltage (E ) Cathode red V Anode ox V Overall V Line Notation G = K = AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 7
The voltage you found is based on 1 M concentrations of the aqueous ions find the voltage if [Al +3 ] = 0.50 M and [Mg +2 ] = 1.50 M using the Nernst Equation E = voltage with new concentrations E = standard voltage with 1 M concentrations n = number of electrons transferred in balanced reaction Q = reaction quotient using balanced equation R = 8.31 J/mol K T = Temperature in Kelvin F = Faraday = 96485 Coulombs E = E (RT/nF) ln Q E = E (0.0592/n) log Q (@ 25 C) AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 8
NONSPONTANEOUS (E < 0) electrical energy chemical energy Electroplating Electrolysis Electrolytic Cu +2 (aq) + 2e - Cu(s) q = I t charge (C) = Current (A) time (s) and conversions Electroplating: How much copper can be plated onto a key from a copper solution when a current of 10.0 A (Ampere = Coulomb/sec) is passed for 30.0 minutes? http://www.youtube.com/watch?v=fnj0v7b7nko q = I t 18000 C = 10.0 A 30.0 min 60 s min 18000 C 1 mol e- 1 mol Cu = 0.0933 mol Cu or 5.93 g Cu 96500 C 2 mol e- You Try It: For how long must a current of 15.0 A be applied to a solution of Ag + to produce 25.0 g of silver metal onto a metal spoon? AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 9
Electrolytic Cells Electrolysis of H2O http://www.youtube.com/watch?v=znojwohg6ww 2H2(g) + O2(g) 2H2O(l) is spontaneous (with activation energy), so the opposite is nonspontaneous. (In order to pass electricity you need to add an electrolyte usually H2SO4) Half Reactions Voltage (E ) Anode (ox) 2H2O(l) O2(g) + 4H + + 4e - -1.23 V Cathode (red) 4H2O(l) + 4e - 2H2(g) + 4OH - (aq) -0.83 V Overall 6H2O(l) 2H2(g) + O2(g) + 4H + (aq) + 4OH - (aq) 6H2O(l) 2H2(g) + O2(g) + 4 H2O(l) 2H2O(l) 2H2(g) + O2(g) -2.06 V Electrolysis of Molten NaCl(l) mixed with CaCl2 to lower m.p. Voltage (E ) Half Reactions Anode (ox) 2Cl - (l) Cl2(g) + 2e - -1.36 V Cathode (red) 2Na + (l) + 2e - 2Na + -2.71 V Overall 2Na + (l) + 2Cl - (l) 2Na(s) + Cl2(g) -4.07 V Note: The Na(s) could actually be a liquid b/c of T and is reactive with O2(g) or H2O(l) Electrolysis of Aqueous NaCl(aq) Half Reactions Figure 1 Molten NaCl Voltage (E ) Anode (ox) 2Cl - (aq) Cl2(g) + 2e - -1.36 V Cathode (red) H2O (l) + 2e - H2(g) + 2OH - (aq) -0.83 V Overall 2Cl - (aq) + H2O (l) H2(g) + Cl2(g) + 2OH - (aq) -2.19 V Note: H2O(l) is reduced instead of Na + in the previous example b/c E H2O = -0.83 V, which is more spontaneous than E Na + = -2.71 V remember Na(s) is too reactive with water to be formed in aqueous solution. Final solution has Na + (aq), Cl - (aq), OH - (aq). AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 10
Chapter 22 Organic Chemistry Hydrocarbons compounds composed only of hydrogen and carbon # carbons Alkane Single bonds Alkene 1 Double Bond Alkyne 1 Triple Bond CnH2n+2 CnH2n CnH2n-2 Name Formula Name Formula Name Formula 1 methane CH4 ------- ------- ------- ------- 2 ethane C2H6 ethene C2H4 ethyne C2H2 3 propane C3H8 propene C3H6 propyne C3H4 4 butane C4H10 butene C4H8 butyne C4H6 5 pentane pentene pentyne 6 hexane hexene hexyne 7 heptane heptene heptyne 8 octane octene octyne 9 nonane nonene nonyne 10 decane decene decyne saturated all single bonds unsaturated multiple bonds exist Isomers 2 compounds with the same formula but different structures. Ex. C4H10 n-butane (normal butane) Isobutane (2-methyl propane) AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 11
Ex. C5H10 pentene 2-pentene 2-methyl butene 3-methyl butene 2-methyl 3-hexene 2-methyl 2-butene 2-methyl 2-butene 2-pentene cyclopentane Nomenclature (naming organic compounds) 1. Look for LONGEST carbon chain (even going around the corner!) 2. Count the carbons so that the attached groups are on the LOWEST # carbon possible. AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 12
Organic Chemistry Functional Groups (see p. 1011 and 1020 for polymers) Class Halohydrocarbons Functional Group X (F, Cl, Br, I) General Formula R X Examples C4H9Cl butyl chloride 1-chloro butane 2-chloro butane Alcohols OH, this is NOT a base! OH R OH C3H7OH Propanol (primary) propyl alcohol 2-propanol (secondary) 2-propyl alcohol Ethers There is something on ether side of an O. O R O R CH3OC2H5 ethyl methyl ether CH3OCH3 dimethyl ether Aldehydes Al hydes when he CHO s down. CHO R CHO HCHO methanal C4H9CHO pentanal AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 13
Organic Chemistry Functional Groups (continued) (see p. 1011 and 1020 for polymers) Class Ketones You need the keys to the CORR. Functional Group General Formula Examples CO R CO R CH3COCH3 propanone dimethyl ketone acetone C2H5COC3H7 3-hexanone ethyl propyl ketone Carboxylic Acids I knoic COOH do 2. COOH R COOH CH3COOH HC2H3O2 ethanoic acid acetic acid HCOOH methanoic acid formic acid Esters Ester wears glasses and her last name is -anoate. COO R COO R C4H9COOC2H5 ethyl pentanoate CH3COOC3H7 propyl ethanoate propyl acetate Amines NH2 R NH2 C5H11NH2 pentyl amine CH3NH2 methyl amine AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 14
Reactions of Organic Molecules Substitution for saturated carbon compounds and benzene C2H6 + Br2 C2H5Br + HBr C6H6 + Br2 C6H5Br + HBr Addition for unsaturated carbon compounds w/(cl2, HOH, HCl ) C2H4 + Br2 C2H4Br2 Esterification (ester formation) Instead of drinking acid and alcohol, ester drinks water. Ester s last name is anoate. acid alcohol ester water C3H7COOH + CH3OH C3H7COOCH3 + H2O butanoic acid methanol methyl butanoate AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 15
ws - Organic Naming Practice Directions: Name each of the following organic molecules and write their formulas. AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 16
ws Organic Structure Practice Directions: Draw the structures of the molecules below. 1. 2-butene 2. 2,3-dimethyl hexane 3. methane 4. 3-ethyl-2-methyl decane 5. 2-pentyne 6. 2,3-hexadiene 7. n-pentane 8. 3,3-diethyl heptane 9. 4-methyl-3-octene 10. 2-methyl -4-propyl nonane AP CHEMISTRY Chapter 6.16.17.18.22 Scotch Plains-Fanwood High School Page 17
ws Organic Structure Practice 2 Directions: Draw the structures of the molecules below and write their formulas. 1. 3-octanone 2. 6,9-dichloro-2-nonene 3. butanal 4. hexanoic acid 5. ethyl propyl ether 6. pentyl butanoate 7. cyclopentane 8. octyl amine 9. methyl pentyl ketone 10. 3-decyl alcohol AP CHEMISTRY Chapter 6.16.17.18.22 Scotch Plains-Fanwood High School Page 18
Chapter 18 Nuclear Chemistry Mass # AXZ Atomic # Alpha particle (α) Beta particle (electron) (β) [ 1 n0 1 p + 1 + 0 e - -1] Gamma radiation (γ) Positron emission [ 1 p + 1 1 n0 + 0 e1] 4 He2 0 e - -1 0 γ0 0 e1 proton 1 p + 1 neutron 1 n0 Equation Type Notes 238 U 234 Th + Stopped by paper 14 C 14 N + Stopped by wood [ 1 n0 1 p + 1 + 0 e - -1] 8 B 8 Be + [ 1 p + 1 1 n0 + 0 e1] 238 U 234 Th + 4 He + 2 Stopped by lead 201 Hg + 201 Au + 0 γ electron capture From inner-orbital 1 n + 235 U 141 Ba + 92 Kr + 3 Fission Neutron bombardment 1 H + 1 H 2 H + 0 e Fusion Too hot for earth occurs on the sun AP CHEMISTRY Chapter 6.16.17.18.22 Scotch Plains-Fanwood High School Page 19
Half Life (t1/2) Time needed for half of a given amount of radioactive substance to decay. Amount 100. g 50. g 25 g 12.5 g 6.125 g time 0 1 half life 2 half lives 3 half lives 4 half lives All nuclear decay is 1 st order Recall: ln[a] = -kt + ln[a]o and t1/2 = 0.693/k Binding Energy 16 O should weigh 8(mass of p + ) + 8(mass of n) = Total mass of 16 O 8(1.67262(10) -24 ) g + 8(1.67493(10) -24 ) g = 2.67804(10) -23 g but it actually weighs 2.65535(10) -23 g This difference is called the mass defect and 0.02269(10) -23 g is the mass released when forming 16 O from 8p + and 8n E = m c 2 2.04(10) -11 J/nucleus = 2.269(10) -28 kg/nucleus (3(10) 8 m/s) 2 This is called the binding energy per nucleus (Note: kg (m 2 /s 2 ) = kg (m/s 2 ) m = N m = J (kg m/s 2 = Newton) AP CHEMISTRY Chapter 6.16.17.18.22 Scotch Plains-Fanwood High School Page 20
Concentration Cell This is a cell with the same solution in both anode and cathode, but with different concentrations, which can generate an electrical current. Example: Find the voltage of a cell made of 0.10 M Cu +2 and 0.01 M Cu +2 Half Reactions: Voltage (E ) Cathode red V Anode ox V Overall V Line Notation AP CHEMISTRY Chapter 6.16.17.18.22 Scotch Plains-Fanwood High School Page 21