Chapters 6,16 Enthalpy, Entropy, Free Energy

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1 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 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

2 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 = 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

3 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) = kj or C(s) + O2(g) 1 CO2(g) Hf = 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 = (-75) kj = kj Bond Energy (Reactants Products) or (Brokendo Formexo) (see p. 351) Stoichiometry... CH4(g) + 2O2(g) CO2(g) + 2H2O(g) kj (exothermic) Ex. 2 mol kj Ex g kj AP CHEMISTRY Chapters 6, 16-18, 22 Scotch Plains-Fanwood High School Page 3

4 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 = kj/mol *matching 2) H = Hf (prod) Hf (react) = [3(-393.5) + 4(-242)kJ/mol] [(-104) +5(0.0)]kJ/mol = 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 = kj/mol 298K[0.103 kj/mol K] = 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

5 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 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

6 Half Reactions: Voltage (E ) Cathode red 2Ag + (aq) +2 e - 2Ag(s) 0.80 V* Anode ox H2(g) 2H + (aq) + 2e 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) = 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

7 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

8 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 = 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

9 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? q = I t C = 10.0 A 30.0 min 60 s min C 1 mol e- 1 mol Cu = mol Cu or 5.93 g Cu 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

10 Electrolytic Cells Electrolysis of H2O 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 V Cathode (red) 4H2O(l) + 4e - 2H2(g) + 4OH - (aq) 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) V Electrolysis of Molten NaCl(l) mixed with CaCl2 to lower m.p. Voltage (E ) Half Reactions Anode (ox) 2Cl - (l) Cl2(g) + 2e V Cathode (red) 2Na + (l) + 2e - 2Na V Overall 2Na + (l) + 2Cl - (l) 2Na(s) + Cl2(g) 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 V Cathode (red) H2O (l) + 2e - H2(g) + 2OH - (aq) V Overall 2Cl - (aq) + H2O (l) H2(g) + Cl2(g) + 2OH - (aq) V Note: H2O(l) is reduced instead of Na + in the previous example b/c E H2O = V, which is more spontaneous than E Na + = 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

11 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 CH 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

12 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

13 Organic Chemistry Functional Groups (see p 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

14 Organic Chemistry Functional Groups (continued) (see p 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

15 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

16 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

17 ws Organic Structure Practice Directions: Draw the structures of the molecules below 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 methyl -4-propyl nonane AP CHEMISTRY Chapter Scotch Plains-Fanwood High School Page 17

18 ws Organic Structure Practice 2 Directions: Draw the structures of the molecules below and write their formulas 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 decyl alcohol AP CHEMISTRY Chapter Scotch Plains-Fanwood High School Page 18

19 Chapter 18 Nuclear Chemistry Mass # AXZ Atomic # Alpha particle (α) Beta particle (electron) (β) [ 1 n0 1 p e - -1] Gamma radiation (γ) Positron emission [ 1 p n0 + 0 e1] 4 He2 0 e 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 e - -1] 8 B 8 Be + [ 1 p n0 + 0 e1] 238 U 234 Th + 4 He + 2 Stopped by lead 201 Hg Au + 0 γ electron capture From inner-orbital 1 n 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 Scotch Plains-Fanwood High School Page 19

20 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 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( (10) -24 ) g + 8( (10) -24 ) g = (10) -23 g but it actually weighs (10) -23 g This difference is called the mass defect and (10) -23 g is the mass released when forming 16 O from 8p + and 8n E = m c (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 Scotch Plains-Fanwood High School Page 20

21 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 Scotch Plains-Fanwood High School Page 21