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Advanced Placement Chemistry Enthalpy, Entropy and Free Energy 2014 29 Cu 63.55 30 Zn 65.39 41 Nb 92.91 42 Mo 93.94 43 Tc (98) †Actinide Series: *Lanthanide Series: 105 Db (262) 106 Sg (263) 107 Bh (262) 33 As 74.92 34 Se 78.96 35 Br 79.90 109 Mt (266) 110 § (269) 111 § (272) 112 § (277) §Not yet named 79 80 78 77 84 83 82 81 Au Hg Pt Ir Po Bi Pb Tl 192.2 195.08 196.97 200.59 204.38 207.2 208.98 (209) 85 At (210) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) 62 63 64 65 66 67 68 69 70 71 Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 150.4 151.97 157.25 158.93 162.50 164.93 167.26 168.93 173.04 174.97 108 Hs (265) 91 92 93 94 90 Th Pa U Np Pu 232.04 231.04 238.03 237.05 (244) 59 60 61 58 Ce Pr Nd Pm 140.12 140.91 144.24 (145) 88 104 89 87 † Rf Ac Ra Fr (223) 226.02 227.03 (261) 32 Ge 72.59 36 Kr 83.80 17 16 15 14 18 Cl S P Si Ar 28.09 30.974 32.06 35.453 39.948 9 10 F Ne 19.00 20.179 2 He 4.0026 86 Rn (222) 48 46 47 45 44 53 52 51 50 49 54 Cd Pd Ag Rh Ru I Te Sb Sn In Xe 101.1 102.91 106.42 107.87 112.41 114.82 118.71 121.75 127.60 126.91 131.29 56 76 75 73 74 72 57 55 * Os Re Ta W Hf La Ba Cs 132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.2 40 Zr 91.22 39 Y 88.91 28 Ni 58.69 38 Sr 87.62 27 Co 58.93 37 Rb 85.47 26 25 24 Fe Mn Cr 52.00 54.938 55.85 31 Ga 69.72 23 V 50.94 21 Sc 44.96 20 Ca 40.08 19 K 39.10 22 Ti 47.90 13 Al 26.98 12 Mg 24.30 11 Na 22.99 8 7 6 5 O N C B 10.811 12.011 14.007 16.00 4 Be 9.012 3 Li 6.941 1 H 1.0079 Periodic Table of the Elements AP Chemistry Equations & Constants ADVANCEDPLACEMENT PLACEMENTCHEMISTRY CHEMISTRYEQUATIONS EQUATIONSAND ANDCONSTANTS CONSTANTS ADVANCED Throughoutthe thetest testthe thefollowing followingsymbols symbolshave havethe thedefinitions definitionsspecified specifiedunless unlessotherwise otherwisenoted. noted. Throughout L,mL mL L, gg nm nm atm atm == == == == liter(s),milliliter(s) milliliter(s) liter(s), gram(s) gram(s) nanometer(s) nanometer(s) atmosphere(s) atmosphere(s) ATOMICSTRUCTURE STRUCTURE ATOMIC EE == hhνν λν cc == λν mmHg Hg mm kJ J,J,kJ VV mol mol == == == == millimetersof ofmercury mercury millimeters joule(s),kilojoule(s) kilojoule(s) joule(s), volt(s) volt(s) mole(s) mole(s) energy EE == energy frequency νν == frequency wavelength λλ == wavelength −34 Planck’sconstant, constant,hh == 6.626 6.626××10 10−34 Planck’s JJss −1 mss−1 Speedof oflight, light,cc == 2.998 2.998××10 1088m Speed −1 Avogadro’snumber number == 6.022 6.022××10 102323mol mol−1 Avogadro’s −19 Electroncharge, charge,ee == −1.602 −1.602××10 10−19 coulomb Electron coulomb EQUILIBRIUM EQUILIBRIUM [C]cc [D]dd [A] [B] [B] [A] EquilibriumConstants Constants Equilibrium ((PPCC))cc((PPDD))dd (molarconcentrations) concentrations) KKcc (molar (gaspressures) pressures) KKpp (gas [C] [D] R ccCC++ddD where aaAA++bbBB R D ,,where KKcc == aa bb KKpp == ((PPAA))aa((PPBB))bb [H++ ][A-- ] KKaa == [H ][A ] [HA] [HA] KKbb == [OH--][HB ][HB++]] [OH [B] [B] (weakacid) acid) KKaa (weak (weakbase) base) KKbb (weak (water) KKww (water) −14 [H++][OH ][OH−−]] == 1.0 1.0 ××10 10−14 25°C atat25°C KKww == [H == KKaa××KKbb pOH==−log[OH −log[OH−−]] pH == −log[H −log[H++]],, pOH pH 14 == pH pH++pOH pOH 14 [A--]] pH == pK pKaa++log log[A pH [HA] [HA] pKaa==−logK −logKaa,, pK pKbb ==−logK −logKbb pK KINETICS KINETICS ln[A]t t−−ln[A] ln[A]00 == −−ktkt ln[A] 11 -- 11 [[AA]]t t [[AA]]00 == ktkt 0.693 0.693 tt½ ½ == kk rateconstant constant kk == rate time tt ==time tt½ half-life ½ == half-life GASES, LIQUIDS, AND SOLUTIONS PV = nRT PA = Ptotal × XA, where XA = moles A total moles Ptotal = PA + PB + PC + . . . n= m M K = °C + 273 D= m V KE per molecule = 1 mv 2 2 Molarity, M = moles of solute per liter of solution A = abc THERMOCHEMISTRY/ ELECTROCHEMISTRY q = mcDT = DSD Â SD products - Â SD reactants = DH D Â DHfD products - Â DH fD reactants = DGD Â DGfD products - Â DGfD reactants G D DH D - T D S D D= = - RT ln K = - n F ED I q t P V T n m M D KE Ã A a b c = = = = = = = = = = = = = pressure volume temperature number of moles mass molar mass density kinetic energy velocity absorbance molar absorptivity path length concentration Gas constant, R = 8.314 J mol -1 K -1 = 0.08206 L atm mol -1 K -1 = 62.36 L torr mol -1 K -1 1 atm = 760 mm Hg = 760 torr STP = 0.00 D C and 1.000 atm q m c T SD = = = = heat mass specific heat capacity temperature = standard entropy D H = standard enthalpy GD = standard free energy n = number of moles E D = standard reduction potential I = current (amperes) q = charge (coulombs) t = time (seconds) Faraday’s constant, F = 96,485 coulombs per mole of electrons 1 joule 1volt = 1 coulomb 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 Zeroth Law: First Law: Second Law: Third Law: Laws of Thermodynamics Heat flows from hot to cold Energy and matter are conserved Matter tends towards chaos Entropy of a pure crystal at 0K is zero Thermodynamic Terms What does each term tell us? Enthalpy (ΔH) Energy content + endothermic − exothermic Entropy (ΔS) Disorder + increase in the dispersal of matter − decrease in the dispersal of matter Free energy (ΔG) Thermodynamically favored or not favored + not thermodynamically favored − thermodynamically 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 NH4ClO4(s) Al2O3(s) AlCl3(s) NO(g) H2O(g) ΔHf° (kJ/mol) −295 −1676 −704 90.0 −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. § If equations had to be reversed, change the sign of ΔH § If equations had be multiplied to get a correct coefficient, multiply the ΔH by the coefficient § Check to ensure that everything cancels out to give you the correct equation. § Hint** It is often helpful to begin working backwards from the answer that you want! C2H6(g) + 2 C(s) + 3 H2(g) → C2H6(g) C(s) + O2(g) → CO2(g) 1 H2(g) + O2(g) → H2O(g) 2 7 O2(g) → 2 CO2(g) + 3 H2O(g) 2 ∆H˚ = −84.68 kJ mol−1 ∆H˚ = −394 kJ mol−1 ∆H˚ = −286 kJ mol−1 Using Stoichiometry to Calculate ΔH § Often questions are asked about the enthalpy change for specific quantities in a reaction § Use a little stoichiometry to solve these; just remember the ΔH°rxn is per mole and convert to the unit and quantity given How much heat is released when 1.00 g iron is reacted with excess O2? 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ΔH°rxn = −1652 kJ/mol rxn heat released = § § 1.00g −1652kJ × = −7.39 kJ per gram of Fe 4 mol Fe 55.85 g mol the (–) represents the LOSS or release of heat can also write 7.39 kJ released per gram of Fe Using Bond Energy to Calculate ΔH § Be able to use individual Bond Energy data to calculate the overall enthalpy change for a reaction ΔH°rxn = Sum of Bonds Broken – Sum of Bonds Formed H2(g) + F2(g) → 2 HF(g) Bond Type H−H F−F H−F Bond Energy 432 kJ/mol 154 kJ/mol 565 kJ/mol Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 3 Thermodynamics Entropy (ΔS) Entropy § § § Dispersal of matter Less dispersal (−) or More dispersal (+) Calculating Entropy § Table of standard values § Hess’s Law § Entropy increases when: § Gases are formed from solids or liquids (most important!!!!) § H2O(ℓ) → H2O(g) § § § § C(s) + CO2(g) ⇌ 2 CO(g) A solution is formed Volume is increased in a gaseous system (energy is more efficiently dispersed) More complex molecules are formed Big Mamma Equation II: ΔS°rxn = Σ ΔS°(products) − Σ ΔS°(reactants) 2 SO2(g) + O2(g) → 2 SO3(g) Substance S° (J K−1 mol−1) SO2(g) 248.1 O2(g) 205.3 SO3(g) 256.6 Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 4 Thermodynamics Free Energy (ΔG) Free Energy § § § Thermodynamic favorability of the reaction Thermodynamically favorable (−ΔG°) or thermodynamically unfavorable (+ΔG°) Calculate: § Table of standard values § Hess’s Law Big Mamma Equation III: ΔG°rxn = Σ ΔG°(products) − Σ ΔG°(reactants) § ΔH°, ΔS°, and ΔG° may all be calculated from tables of standard values, from Hess’ Law or from the Gibb’s equation: Connections to ΔH° and ΔS°: Granddaddy of Them All: ΔG° = ΔH° − TΔS° Caution on units: ΔH° and ΔG° are typically given in kJ mol−1 whereas ΔS° typically given as J K−1mol−1 Conditions of ΔG ΔH ΔS − + + + − − + − Free Energy, Equilibrium, and Cell Potential ΔG° 0 negative positive § Connecting ΔG° to K § ΔG° = −RT lnK § Connecting ΔG° to E § ΔG° = −n ℑ Ε° K at equilibrium >1, products favored <1, reactants favored ΔG Spontaneous (−) at all temp Spontaneous (−) at high temp Spontaneous (−) at low temp Non-spontaneous (+) at all temp E° 0 + − Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 5 Thermodynamics Endothermic v. Exothermic PROBLEM STRATEGY You should be able to determine if the reactions are endothermic or exothermic and possibly determine the value of the ΔH° if the diagram has energy values given. For kinetics you will be asked to label the activation energy, Ea. CAUTION. Make sure to read carefully as questions are often asked about the reverse reaction. Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 6 Thermodynamics Thermochemistry Cheat Sheet q = mCpΔT q = ΔH (when pressure is constant/coffee cup) (−) qlost = qgained (same value; opp. sign) ΔHrxn = ∑ΔHprod − ∑ΔHreact ΔHrxn = ∑bondsbroken − ∑bondsformed Relationships ΔG = ΔH − TΔS ΔSrxn = ∑ΔSprod − ∑ΔSreact ΔGrxn = ∑ΔGprod − ∑ΔGreact ΔG° = −RTlnK (use 8.31× 10-3 kJ/molK for R) and watch your units for ΔG: they will be in kJ ΔG° = −n ℑ Ε° (96,500 for ℑ ) − ΔH is exothermic; +ΔH is endothermic ΔS = ΔH at equilibrium (including phase change) T ΔG = 0 at equilibrium and direction change Be cautious of which system component is losing heat Use ΔG = ΔH − TΔS equation to justify and which is gaining heat. thermodynamic favorability. Discuss ΔH “overtaking” Assign +/− signs accordingly. the TΔS term and vice versa. Connections Electrochem: ΔG = −n ℑ Ε° Kinetics – reaction diagrams Stoichiometry – Energy values are usually kJ/mol so if you have other than 1 mole adjust accordingly Equilibrium: ΔG = −RTlnK Potential Pitfalls ΔS is in J/K not in kJ like ΔH and ΔG −-1 ΔHrxn is usually in kJ mol (that’s per mol of rxn) ΔG must be negative for thermodynamic favorability ΔHf is usually in kJ mol−-1 Cp = J/g°C (specific heat units) Watch your signs and know what they mean UNITS CAUTION: this calculation gives w in units of (L·atm) not Joules (or kJ)!!!! N and 1 L = 0.001 m3 1 atm = 101,325 2 m 1 L·atm = 101.3 N·m = 101.3 J ALL PΔV calculations for work need to be × 101.3 to convert to Joules, J Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 7 Thermodynamics NMSI Super Problem Magnesium flakes were added to an open polystyrene cup filled with 50.0 mL of 1.00 M HCl solution. Assume the specific heat of the solution to be 4.18 J/g°C. Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) ΔH°rxn = −316.0 kJ mol−1 (a) If 0.600 g of the magnesium were added, determine the total amount of heat released into the calorimeter. (b) Determine the temperature change in the calorimeter. (c) Draw an energy diagram and label the enthalpy change, ΔH, for the reaction. The hydrogen gas produced in the reaction of magnesium and HCl was captured and placed in a sealed container, which occupies a volume of 650 mL at a constant pressure of 1.0 atm. The temperature of the container and gas was changed by 15°C; the resulting volume of the gas in the container is 620 mL. (d) Is the temperature of the system increasing or decreasing. Justify your answer (e) Is the statement in the box below correct? Justify your answer. The gas collected in the container does work on the surroundings Answer the following questions about the oxidation of magnesium metal. Mg(s) + ½ O2(g) → MgO(s) (f) Determine the value of the standard enthalpy change for the reaction in the box above. Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) ΔH°rxn = −316.0 kJ mol−1 MgO(s) + 2 HCl(aq) → MgCl2(aq) + H2O(ℓ) ΔH°rxn = −45.7 kJ mol−1 H2(g) + ½ O2(g) → H2O(ℓ) ΔH°rxn = −286 kJ mol−1 (g) Determine the value of the standard entropy change, ΔS°rxn, for the oxidation of magnesium using the information in the following table. Substance Mg O2 MgO S° (J mol−1 K−1) 33 205 27 Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 8 Thermodynamics (h) Calculate ΔG° rxn for the oxidation of magnesium at 25°C (i) Indicate whether the reaction is thermodynamically favored at 25°C. Justify your answer The hydrogen gas collected and placed in the sealed container above is mixed with nitrogen gas to produce ammonia according to the Haber process shown below. ΔG°rxn = −34.1 kJ mol−1 ΔH°rxn = −92.2 kJ mol−1 N2(g) + 3 H2(g) ⇌ 2 NH3(g) (j) In terms of the equilibrium constant, K, for the above reaction at 25°C i. Predict whether K will be greater than, less than, or equal to one. Justify your choice. ii. Calculate its value. (k) In terms of the standard entropy change, ΔS° i. Predict the sign of ΔS° for the above reaction. Justify your answer. ii. Calculate the value of ΔS° rxn for the above reaction at 25°C. (l) Using the data in the table below and the enthalpy of reaction, ΔH°rxn, calculate the approximate bond energy of the nitrogen−hydrogen bond in ammonia. N H Approximate Bond Energy (kJ mol−1) ??? H H 430 N N 960 Bonds Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 9