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Enthalpy, Entropy and Free Energy Advancedplacement Chemistry Enthalpy, Entropy and Free Energy Advanced Placement 2014 2 Ar Kr 10 18 36 54 86 Xe He Ne Rn (222) 83.80 4.0026 20.179 39.948 131.29 ) 3 I 9 F Cl At Br 17 35 53 85 Lr 71 Lu 10 (210 19.00 79.90 (260) 35.453 126.91 174.97 ) 8 S O Te 16 34 52 84 Se Po 70 Yb No 102 (209 16.00 32.06 78.96 (259) 127.60 173.04 7 P N Bi 15 33 51 83 As Sb 69 Md Tm 101 74.92 (258) 14.007 30.974 121.75 208.98 168.93 1 2 0 6 C Si 8.71 14 32 50 82 Sn Pb Er Ge 68 Fm 10 28.09 72.59 207. (257) 12.01 11 167.26 11 §Not yet named 5 B Tl In Al 4.82 13 31 49 81 67 99 Es Ga Ho (252) 26.98 69.72 10.8 11 204.38 164.93 2 § 30 48 80 Zn Cd Hg Cf 66 98 Dy 11 12.41 (277) 65.39 (251) 1 200.59 162.50 § 29 47 79 Ag Au Cu 65 97 Tb Bk 111 (272) 63.55 (247) 107.87 196.97 158.93 ) 0 § Ni Pt 28 46 78 Pd 64 96 Gd 11 Cm (269 (247) 58.69 106.42 195.08 157.25 2 Ir 27 45 77 Mt Co Rh 63 95 Eu 109 Am ble of the Elements (266) 58.93 192. (243) 102.91 151.97 Ta 1 2 4 26 44 76 Fe Hs Ru Os 62 94 Pu 108 Sm (265) 55.85 101. 190. (244) 150. Tc 25 43 75 61 93 Bh Re Np Mn Pm 107 (98) Periodic (262) (145) 54.938 186.21 237.05 U W Cr 24 42 74 Sg 60 92 Mo Nd 106 (263) 52.00 93.94 183.85 144.24 238.03 ) 4 1 5 V Pr Ta 23 41 73 59 91 Pa Nb Db 10 (262 50.9 92.9 180.95 140.91 231.04 ) 0 2 4 Ti Zr Hf Rf 22 40 72 58 90 Th Ce 10 (261 47.9 91.2 178.49 140.12 232.04 : : Y 21 39 57 89 La Sc Ac * † 44.96 88.91 138.91 227.03 0 8 2 4 Sr 12 20 38 56 88 Be Ba Ca Ra Mg 9.012 24.3 40.0 87.6 137.33 226.02 ) 1 †Actinide Series 1 3 K H Li Fr 11 19 37 55 87 *Lanthanide Series Cs Na Rb (223 6.94 22.99 39.10 85.47 1.0079 132.91 ADVANCEDADVANCED PLACEMENTPLACEMENT CHEMISTRYCHEMISTRY EQUATIONSEQUATIONS ANDAND CONSTANTSCONSTANTS ThroughoutThroughout thethe testtest thethe followingfollowing symbolssymbols havehave ththee definitionsdefinitions specifiedspecified unlessunless otherwiseotherwise noted.noted. L,L, mLmL == liter(s),liter(s), milliliter(s)milliliter(s) mmmm HgHg == millimetersmillimeters ofof mercurymercury gg == gram(s)gram(s) J,J, kJkJ == joule(s),joule(s), kilojoule(s)kilojoule(s) nmnm == nanometer(s)nanometer(s) VV == volt(s)volt(s) atmatm == atmosphere(s)atmosphere(s) molmol == mole(s)mole(s) ATOMICATOMIC STRUCTURESTRUCTURE EE == energyenergy EE == hhνν νν == frequencyfrequency cc == λνλν λλ == wavelengthwavelength Planck’sPlanck’s constant,constant, hh == 6.6266.626 × × 1010−−3434 JJ s s SpeedSpeed ofof light,light, cc == 2.9982.998 ×× 101088 msms−−11 Avogadro’sAvogadro’s numbernumber == 6.0226.022 ×× 10102323 molmol−−11 ElectronElectron charge,charge, e e == −−1.6021.602 ××1010−−1919 coulombcoulomb EQUILIBRIUMEQUILIBRIUM [C][C]cdcd [D] [D] KKcc == ,, wherewhere aa AA ++ bb BB RR cc CC ++ dd DD EquilibriumEquilibrium Constants Constants [A][A]abab [B] [B] cdcd KKcc (molar(molar concentrations)concentrations) ((PPPPCC )( )(DD ) ) KKpp == abab KKpp (gas(gas pressures)pressures) ((PPPPABAB )( )( ) ) KKaa (weak(weak acid)acid) [H[H+-+- ][A ][A ] ] KK == K (weak base) aa [HA][HA] Kbb (weak base) KK (water)(water) [OH-+-+ ][HB ] ww KK == [OH ][HB ] bb [B][B] ++ −− −−1414 KKww == [H[H ][OH][OH ]] == 1.01.0 ××1010 atat 2525°°CC == KKaa ×× KKbb pHpH == −−log[Hlog[H++]] , , pOHpOH == −−log[OHlog[OH−−]] 1414 == pHpH ++ pOHpOH -- pHpH == ppKK ++ loglog[A[A ] ] aa [HA][HA] ppKKaa == −−loglogKKaa,, ppKKbb == −−loglogKKbb KINETICSKINETICS ln[A]ln[A] −− ln[A]ln[A] == −−ktkt tt 00 kk == raterate constantconstant 1111 tt == timetime -- == ktkt AAAA [][]tt [][]00 tt½½ == half-lifehalf-life 0.6930.693 tt == ½½ kk GASES, LIQUIDS, AND SOLUTIONS P = pressure V = volume PV = nRT T = temperature moles A n = number of moles PA = Ptotal × XA, where XA = total moles m = mass M = molar mass Ptotal = PA + PB + PC + . D = density n = m KE = kinetic energy M Ã = velocity K = °C + 273 A = absorbance a = molarabsorptivity m D = b = path length V c = concentration 1 KE per molecule = mv2 2 Gas constant, R = 8.314 J mol--11 K Molarity, M = moles of solute per liter of solution --11 = 0.08206 L atm mol K --11 A = abc = 62.36 L torr mol K 1atm= 760 mm Hg = 760 torr STP= 0.00D C and 1.000 atm THERMOCHEMISTRY/ ELECTROCHEMISTRY q heat = m = mass q= mcD T c = specific heat capacity T temperature DSSDD= ÂÂ products- S D reactants = SD = standard entropy DDHHDD= ÂÂproducts- D HfD reactants f HD = standard enthalpy GD standard free energy DDGGDD= ÂÂf products- D GfD reactants = n = number of moles D DDGD = HDD - TS D E = standard reduction potential I = current (amperes) = -RTln K q = charge (coulombs) t = time (seconds) = -nFED q Faraday’s constant , F = 96,485 coulombs per mole I ϭ t of electrons 1 joule 1volt = 1coulomb THERMODYNAMICS Enthalpy, Entropy, Free Energy, & Equilibrium What I Absolutely Have to Know to Survive the AP Exam The following might indicate the question deals with thermochemistry and thermodynamics: calorimeter; enthalpy (ΔH); specific heat (Cp); endothermic; exothermic; heat (q); heat capacity (C); heat transfer; bond energy; entropy (ΔS); Gibb’s free energy (ΔG); spontaneous; state function Laws of Thermodynamics Zeroth Law: Heat flows from hot to cold First Law: Energy and matter are conserved Second Law: Matter tends towards chaos Third Law: Entropy of a pure crystal at 0K is zero Thermodynamic Terms What does each term tell us? Enthalpy (ΔH) Energy content + endothermic − exothermic + increase in the dispersal of − decrease in the dispersal of Entropy (ΔS) Disorder matter matter Thermodynamically Free energy (ΔG) favored or not + not thermodynamically favored − thermodynamically favored favored Equilibrium (K ) Extent of reaction >1 reaction favors products <1 reaction favors reactants Internal Energy (ΔE) and Heat Flow § Refers to all of the energy contained within a chemical system. § Heat flow between the system and its surroundings involves changes in the internal energy of the system. It will either increase or decrease § Increases in internal energy may result in a § temperature increase § chemical reaction starting § phase change § Decreases in internal energy may result in a § a decrease in temperature § phase change § Note: even though the change in internal energy can assume several different forms, the amount of energy exchanged between the system and the surroundings can be accounted for ONLY by heat (q) and work (w) § ΔE = q + w § Work (w) refers to a force acting on an object; in chemical processes this acting force is done by a gas through expansion or to a gas by compression. § This is referred to as “pressure/volume” work § Thus, w = − PΔV § Where P is constant external pressure (atm) and ΔV (L) is the change in volume of the system Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 1 Thermodynamics Calculating Heat (q) § Heat (q) gained or lost by a specific amount of a known substance can be calculated using the heat capacity of the substance and the change in temperature the system undergoes. § Calorimetry § The process of experimentally measuring heat by determining the temperature change when a body absorbs or releases energy as heat. § Coffee-cup calorimetry –use a Styrofoam cup, mix reactants that begin at the same temperature and look for change in temperature; the heat transfer is calculated from the change in temp. q = mCΔT § q = quantity of heat (Joules) § ΔT is the change in temperature § ΔT = Tf − Ti (final – initial) § watch the sign; if the system loses heat to the surroundings then the ΔT = − § Cp = specific heat capacity (J/g°C) § m = mass in grams § the specific heat of water (liquid) = 4.184 J/g°C Also note: § q = −ΔH at constant pressure Enthalpy (ΔH) Enthalpy § Heat content; ∆H § Endothermic (+) or Exothermic (−) § Calculating Enthalpy 5 Ways § Calorimetry (see above) ° § Enthalpy of formation, ΔHf (using table of standard values) § Hess’s Law § Stoichiometry § Bond Energies ° ΔHf − Enthalpy of Formation § Production of ONE MOLE of compound FROM ITS ELEMENTS in their standard states (°) § Zero (0) for ELEMENTS in standard states: 25°C (298 K), 1 atm, 1M Big Mamma Equation: ΔH°rxn = Σ ΔH°f (products) − Σ ΔH°f (reactants) 3 Al(s) + 3 NH4ClO4(s) → Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g) ° Substance ΔHf (kJ/mol) NH4ClO4(s) −295 Al2O3(s) −1676 AlCl3(s) −704 NO(g) 90.0 H2O(g) −242 Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 2 Thermodynamics Hess’s Law § Enthalpy is not dependent on the reaction pathway. If you can find a combination of chemical equations that add up to the desired overall equation, you can sum up the ΔHrxn’s for the individual reactions to get the overall ΔHrxn. § Remember this: § First decide how to rearrange equations so reactants and products are on appropriate sides of the arrows.
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