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Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K) John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapters 5,8: Energy (E), Heat (q), Work (w), and Enthalpy (∆H) John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc. Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ) Energy (E) What is it? • ability to do work OR transfer heat Work (w): transfer of energy by applying a force over a distance. Heat (q): transfer of energy by DT (high to low) • unit of energy: joule (J) • an older unit still in widespread use is… calorie (cal) 1 Cal = 1000 cal 1 cal = 4.18 J 2000 Cal ≈ 8,000,000 J ≈ 8 MJ!!! System and Surroundings • System: molecules to be studied (reactants & products) • Surroundings: everything else (container, thermometer,…) 1st Law of Thermodynamics • Energy is neither created nor destroyed. • total energy of an isolated system is constant (universe) (no transfer matter/energy) (conserved) Internal Energy (E): E = KE + PE (motions) (attractions) (Thermal Energy) (calculating E is too complex a problem) DE = Efinal − Einitial released or absorbed Changes in Internal Energy • Energy is transferred between the system and surroundings, as either heat (q) or work (w). DE = q + w DE = ? DE = (–) + (+) Surroundings System DE = q + w q in (+) q out (–) w on (+) w by (–) DE = + Changes in Internal Energy Efinal > Einitial Efinal < Einitial absorbed energy released energy (endergonic) (exergonic) Work (w) The only work done by a gas at constant P is change in V by pushing on surroundings. PDV = −w ΔV “–” b/c work done BY system ON surroundings Zn + H+ Zn2+ + H2(g) Enthalpy Enthalpy (H) is: H = E + PV internal work done energy work done heat/work energy DH = DE + PDV by system in or out DE = q+w PDV = −w of system (change in) H or DH (at constant P) : DH = DE + PDV DH = (q+w) + (−w) DH = heat DH = q • (change in) enthalpy IS heat absorbed/released Enthalpy of Reaction enthalpy is… CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) DH = DE + PDV DH = q …the heat transfer in/out of a system (at constant P) DH = Hproducts − Hreactants exergonic endergonic exothermic endothermic Endothermic & Exothermic • Endothermic: DH > 0 (+) DH(+) = Hfinal − Hinitial products reactants • Exothermic: DH < 0 (–) DH(–) = Hfinal − products Hinitial reactants DH(–) is thermodynamically favorable Enthalpy of Reaction 2 H2(g) + O2(g) 2 H2O(g) DHrxn, is the enthalpy of reaction, or “heat” of reaction. Demo –242 kJ units: per 1 mol O2 kJ (OR) molrxn –242 kJ kJ/molrxn –1 kJ∙molrxn per 2 mol H2 (OR) –121 kJ per 1 mol H2 DH = –242 kJ/molrxn Enthalpy HW p. 207 #34,35,38,45 1. DH depends on amount (moles, coefficients) 2. DHreverse rxn = –DHforward rxn 3. DHrxn depends on the state (s, l, g) of products & reactants DH1 = –802 kJ if 2 H2O(g) because… 2 H2O(l) 2 H2O(g) DH = +88 kJ/molrxn CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) Chp. 5,8: Calculate ∆H (4 Ways) 1) Bond Energies 2) Hess’s Law 3) Standard Heats of Formation (Hf ) 4) Calorimetry (lab) Overlap and Bonding When bonds/attractions form, energy is _________. released – + ++ What repulsive forces? –+ What attractive forces? Where is energy being released? Where must energy be added? Potential Energy of Bonds High PE Low PE (energy released when bonds form) + High PE (energy absorbed when bonds break) + p.330 Bond Enthalpy (BE) BE: ∆H for the breaking of a bond (all +) aka… bond dissociation energy Enthalpy of Reaction (∆H) BE: ∆H for the breaking of a bond (all +) To determine DH for a reaction: • compare the BE of bonds broken (reactants) to the BE of bonds formed (products). DHrxn = (BEreactants) (BEproducts) (bonds broken) (stronger) DH(+) = BEreac − BEprod DH(–) = BEreac − BEprod (stronger) (bonds formed) (released) (NOT on equation sheet) Enthalpy of Reaction (∆H) CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) DHrxn = [4(C—H) + (Cl—Cl)] [3(C—H) + (C—Cl) + (H—Cl)] = (4x413 + 242) (3x413 + 328 + 431) = (655) (759) DHrxn = 104 kJ/molrxn HW p. 339 #66, 68 Chp. 5,8: Calculate ∆H (4 Ways) 1) Bond Energies (NOT DH rxn = (BEreactants) (BEproducts) given) (+ broken) (– formed) 2) Hess’s Law 3) Standard Heats of Formation (Hf ) 4) Calorimetry (lab) Hess’s Law DH = Hfinal − Hinitial prod. react. DHrxn is independent of route taken DHrxn = sum of DH of all steps DHoverall = DH1 + DH2 + DH3 … (NOT on equation sheet) Calculation of DH by Hess’s Law C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ∆Hcomb = ? Given: + 3 C(gr.) + 4 H2(g) C3H8(g) ∆H1= –104 kJ 3(C(gr.) + O2(g) CO2(g) ) 3( ∆H2= –394 kJ ) 4( H2(g) + ½ O2(g) H2O(l) ) 4( ∆H3= –286 kJ ) Used: C3H8(g) 3 C(gr.) + 4 H2(g) ∆H1= +104 kJ 3 C(gr.) + 3 O2(g) 3 CO2(g) ∆H2= –1182 kJ 4 H2(g) + 2 O2(g) 4 H2O(l) ∆H3= –1144 kJ C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ∆Hcomb = –2222 kJ Standard Enthalpy of Formation o standard (DHf ) Standard Enthalpy of Formation (DHof ): heat released or absorbed by the formation of a compound from its pure elements in their natural states. (25oC , 1 atm) DHf = 0 for all elements in natural state o 3 C(gr.) + 4 H2(g) C3H8(g) o o DHf = 0 DHf = 0 o DHf = –104 kJ ∆Hf = –104 kJ Recall… DH = Hfinal − Hinitial …therefore ---> prod. react. Calculation of DH by DHf’s o …we can use Hess’s law in this way: “sum” DH = nHf(products) – nHf(reactants) (on equation sheet) n (mol) is the stoichiometric coefficient. Calculation of DH by DHf’s o C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ∆Hcomb = ? DH = nHf(products) – nHf(reactants) Appendix C (p. 1123 ) DH = [3(DHf CO2) + 4(DHf H2O)] – [1(DHf C3H8) + 5(DHf O2)] DH = (3 ∙ –393.5 + 4 ∙ –285.83) – (–103.85 + 5 ∙ 0) DH = (–2323.7) – (–103.85) DH = –2219.9 kJ HW p. 209 #60,63,66, 72,73 Chp. 5,8: Calculate ∆H (4 Ways) 1) Bond Energies (NOT DH rxn = (BEreactants) (BEproducts) given) (+ broken) (– formed) 2) Hess’s Law (NOT DH overall = DHrxn1 + DHrxn2 + DHrxn3 … given) 3) Standard Heats of Formation (Hf ) (given) DH = nHf(products) – nHf(reactants) 4) Calorimetry (lab) Calorimetry We can’t know the exact enthalpy of reactants and products, so we calculate DH by calorimetry, the measurement of heat flow. Calorimeter nearly isolated By reacting (in solution) in a calorimeter, we indirectly determine DH of system by measuring ∆T & calculating q of the surroundings (calorimeter). heat (J) q = mcDT mass (g) [of sol’n] (on equation sheet) Tf – Ti (oC) [of surroundings] Specific Heat Capacity (c) • specific heat capacity,(c): (or specific heat) energy required to raise temp of 1 g by 1C. (for water) c = 4.18 J/goC Metals have much lower c’s b/c they transfer heat and change temp easily. + 4.18 J of heat HW p. 208 #49, 52, 54 Calorimetry (in J) of calorimeter q = mcDT or surroundings – q = DHrxn (in kJ/mol) of system When 4.50 g NaOH(s) is dissolved 200. g of water in a calorimeter, the temp. changes from 22.4oC to 28.3oC. Calculate the molar heat of solution, ∆Hsoln (in kJ/mol NaOH). q = (4.50 + 200)(4.18)(28.3–22.4) qsurr = 5040 J DH = –5.04 kJ DHsys = –5.04 kJ 0.1125 mol 4.50 g NaOH x 1 mol = 0.1125 mol NaOH = –44.8 kJ mol 40.00 g Chp. 5,8: Calculate ∆H (4 Ways) 1) Bond Energies (NOT DH rxn = (BEreactants) (BEproducts) given) (+ broken) (– formed) 2) Hess’s Law (NOT DH overall = DHrxn1 + DHrxn2 + DHrxn3 … given) 3) Standard Heats of Formation (Hf ) (given) DH = nHf(products) – nHf(reactants) 4) Calorimetry (lab) (given) q = mc∆T (surroundings or thermometer) –q = ∆H ∆H/mol = kJ/mol (molar enthalpy) Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 19: Thermodynamics (∆H, ∆S, ∆G, K) John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc. Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ) ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w) ΔE = q + w PΔV = –w (at constant P) ΔH = ? q (heat) ΔS = ? ΔG = ? Big Idea #5: Thermodynamics Bonds break and form to lower free energy (∆G). Chemical and physical processes are driven by: • a decrease in enthalpy (–∆H), or • an increase in entropy (+∆S), or • both. 1st Law of Thermodynamics • Energy cannot be created nor destroyed (is conserved) • or…total energy of the universe is constant. DHsystem = –DHsurroundings OR DHuniv = DHsystem + DHsurroundings = 0 if (+) then (–) = 0 if (–) then (+) = 0 Thermodynamically Favorable • Thermodynamically Favorable (spontaneous) processes are those that can proceed without any outside intervention. • The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously Thermodynamically Favorable • Processes that are thermodynamically favorable (spontaneous) in one direction are NOT in the reverse direction. Thermodynamically Favorable • Processes that are favorable (spontaneous) at one temperature… …may not be at other temperatures. HW p. 837 #7, 11 melting freezing Entropy (S) •(okay but oversimplified) disorder/randomness “The energy of the •(more correct) dispersal of matter & energy universe is constant.” “The entropy of the among various motions of universe tends particles in space at a toward a maximum.” (ratio of heat to temp) temperature in J/K. ∆H DS = Sfinal Sinitial DS = T (more dispersal) DS = + therm fav DS = – not therm fav (less dispersal) (structure/organization) Entropy (S) ∆H DS = T (a part) (the rest) height AND weight • change in entropy (DS) depends on… …heat (∆H)…AND…temp. (T) System A 50 J (100 K) Surroundings (100 K) same ∆H diff. ∆S System B 50 J (25 K) Surroundings (25 K) ∆S = ____J/K +0.5 ∆S = ____J/K +2.0 (loud restaurant, less disturbed) (quiet library, more disturbed) Cough! Entropy • Example: melting 1 mol of ice at 0oC. DShand = ? DHhand = –6000 J 6000 J DHice = +6000 J DSice = ? ∆H +DH = –DH DH = 0 ice hand univ DS = Entropy T +DSice > –DShand DSuniv = + • The melting of 1 mol of ice at 0oC. DHfusion DSice = T (1 mol)(6000 J/mol) = = +22.0 J/K 273 K (gained by ice) • Assume the ice melted in your hand at 37oC. DHfusion (1 mol)(–6000 J/mol) –19.4 J/K DShand = = = 310 K T (lost by hand) DSuniv = DSsystem + DSsurroundings (hand) (ice) DSuniv DSuniv = (22.0 J/K) + (–19.4 J/K) = +2.6 J/K DHuniv = 0 2nd Law: +22 J/K > –19 J/K DSuniv = + 1st Law: 6000 J = 6000 J (usable E) 6000 J (usable E) (dispersed E) (in hand) (in ice) 5290 J + 710 J x Universe (isolated system) (∆Suniv) = (T) “dispersed” energy (unusable) Free energy (useful for work) 2nd Law of Thermodynamics DSuniv = DSsystem + DSsurroundings For thermodynamically favorable (spontaneous) processes… … +∆S gained always greater than –∆S lost, so… DSuniv = DSsystem + DSsurroundings > 0 2nd Law of Thermodynamics (formally stated): All favorable processes increase the entropy of the universe (DSuniv > 0) HW p. 837 #20, 21 Entropy (Molecular Scale) • Ludwig Boltzmann described entropy with molecular motion. Motion: Translational , Vibrational, Rotational • He envisioned the molecular motions of a sample of matter at a single instant in time (like a snapshot) called a microstate. Entropy (Molecular Scale) S = k lnW Boltzmann constant 1.38 1023 J/K microstates (max number possible) Entropy increases (+∆S) with the number of microstates in the system. < < < Entropy (Molecular Scale) S : dispersal of matter & energy at T • The number of microstates and, therefore, the entropy tends to increase with… ↑Temperature (motion as KEavg) ↑Volume (motion in space) ↑Number of particles (motion as KEtotal) ↑Size of particles (motion of bond vibrations) ↑Types of particles (mixing) Entropy (Molecular Scale) S : dispersal of matter & energy at T Maxwell-Boltzmann distribution curve: ∆S > 0 by adding heat as… …distribution of KEavg increases Entropy (Molecular Scale) S : dispersal of matter & energy at T Entropy increases with the freedom of motion. S(s) < S(l) < S(g) solid H2O(g) S(s) < S(l) < S(aq) < S(g) gas T V more microstates (s) + (l) (aq) H2O(g) Standard Entropy (So) • Standard entropies tend to increase with increasing molecular size. larger molecules have more microstates Entropy Changes (DS) • In general, entropy increases when gases form from liquids and solids liquids or solutions form from solids moles of gas molecules increase total moles increase Predict the sign of DS in these reactions: 1. Pb(s) + 2 HI(aq) PbI2(s) + H2(g) DS = + 2. NH3(g) + H2O(l) NH4OH(aq) DS = – 3rd Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. (not possible) S = k lnW S = k ln(1) S=0 increase only 1 microstate 0K S=0 Temp. >0K S>0 Standard Entropy Changes (∆So) HW p. 838 #29, 31, 40, 42, 48 Standard entropies, S. (Appendix C) DSo = nSo(products) – nSo(reactants) (on equation sheet) n (mol) is the stoichiometric coefficient. Energy (E) Enthalpy (H) (kJ) ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w) ΔE = q + w PΔV = –w Entropy (S) (J/K) (disorder) microstates dispersal of matter & energy at T (at constant P) ∆Suniv = + ΔH = q ΔS ΔS==ΔH ? T (heat) Free Energy (G) (kJ) ΔG = ? Big Idea #5: Thermodynamics Bonds break and form to lower free energy (∆G). Chemical and physical processes are driven by: • a decrease in enthalpy (–∆H), or • an increase in entropy (+∆S), or • both. Thermodynamically Favorable Chemical and physical processes are driven by: • decrease in enthalpy (–∆Hsys) causes (+∆Ssurr) • increase in entropy (+∆Ssys) (+) + DS (+) DSuniv = DSsystem surroundings > 0 • Thermodynamically Favorable: (defined as) increasing entropy of the universe (∆Suniv > 0) ∆Suniv > 0 (+Entropy Change of the Universe) (∆Suniv) ↔ (∆Gsys) For all thermodynamically favorable reactions: DSuniverse = DSsystem + DSsurroundings > 0 DSuniverse = DSsystem + (Boltzmann) DHsystem (Clausius) T multiplying each term by T: –TDSuniverse = –TDSsystem + DHsystem rearrange terms: –TDSuniverse = DHsystem – TDSsystem DGsystem = DHsystem – TDSsystem (Gibbs free energy equation) (∆Suniv) & (∆Gsys) –TDSuniv = DHsys – TDSsys DGsys = DHsys – TDSsys (Gibbs free energy equation) • Gibbs defined TDSuniv as the change in free energy of a system (DGsys) or DG. • Free Energy (DG) is more useful than DSuniv b/c all terms focus on the system. • If –DGsys , then +DSuniverse . Therefore… –DG is thermodynamically favorable. “Bonds break & form to lower free energy (∆G).” Gibbs Free Energy (∆G) ∆G : free energy transfer of system as work –∆G : work done by system (–w) favorably +∆G : work done on system (+w) to cause rxn (not react to completion) –DG +DG Q & ∆G (not ∆Go) Q<K [P] R P Q= [R] (not react to completion) can cause with electricity/light Q>K +DG –DG Gmin 0 –DG Q=K –DG (release), therm. fav. +DG (absorb), not therm. fav. DG = 0, system at equilibrium. (Q = K) DG = 0 DGo (1 M, 1 atm, 25oC) Q = 1 = K (rare) o (∆G ) Standard Free Energy and Temperature (T) (on equation sheet) (consists DG = DH – TDS of 2 terms) free enthalpy entropy units energy term term convert (kJ/mol) (kJ/mol) (J/mol∙K) to kJ!!! max energy energy energy used for transferred dispersed work as heat as disorder The temperature dependence of free energy comes from the entropy term (–TDS). o (∆G ) Standard Free Energy and Temperature (T) DG Thermodynamic o ∆G Favorability (fav. at high T) (high T) – (unfav. at low T) (low T) + (unfav. at ALL T) + (fav. at ALL T) – (unfav. at high T) (high T) + (fav. at low T) (low T) – = DH TDS = (∆Ho) – T(∆So) ( + ) –T( + ) = ( + ) – T( + ) = ( + ) – T( – ) = ( – ) – T( + ) – ) –T( – ) ( = ( – ) – T( – ) Energy (E) Enthalpy (H) (kJ) + ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w) ΔE = q + w PΔV = –w Entropy (S) (J/K) (disorder) microstates –T∆Suniv as: ΔHsys & ΔSsys at T dispersal of matter & energy at T max work done by favorable rxn (at constant P) ∆Suniv = + ΔH = q ΔS = ΔH T (heat) = Free Energy (G) (kJ) ΔGΔG = ΔH = ?– TΔS Calculating ∆Go (4 ways) 1) Standard free energies of formation, Gf : DG = nG f (products) – mG f (reactants) (given equation) 2) Gibbs Free Energy equation: HW p. 840 #52, 54, 56, 60 DG = DH – TDS (given equation) (may need to calc. ∆Ho & ∆So first) 3) From K value (next few slides) (given equation) 4) From voltage, Eo (next Unit) (given equation) Free Energy (∆G) & Equilibrium (K) Under any conditions, standard or nonstandard, the free energy change can be found by: [P] DG = DG + RT lnQ Q= [R] At equilibrium: Q=K DG = 0 therefore: rearrange: RT is “thermal energy” RT = (0.008314 kJ)(298) = 2.5 kJ at 25oC 0 = DG + RT lnK DG = –RT lnK Free Energy (∆G) & Equilibrium (K) DG = –RT ln K (on equation sheet) –1∙K–1 R = 8.314 J∙mol If DG in kJ, then R in kJ……… = 0.008314 kJ∙mol–1∙K–1 –∆Go = ln K RT Solved for K : –∆Go RT K = e^ (NOT on equation sheet) Free Energy (∆G) & Equilibrium (K) DG = –RT ln K ∆Go = –RT(ln K) – = –RT ( + ) K @ Equilibrium > 1 product favored (favorable forward) + = –RT ( – ) < 1 reactant favored (unfavorable forward) Energy (E) Enthalpy (H) (kJ) + ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w) ΔE = q + w PΔV = –w Entropy (S) (J/K) (disorder) microstates –T∆Suniv as: ΔHsys & ΔSsys at a T dispersal of matter & energy at T max work done by favorable rxn (at constant P) ΔH = q (heat) = Free Energy (G) (kJ) ΔS = ΔH T K > 1 means –∆Gsys & +∆Suniv ΔG = ΔH – TΔS p. 837 #6 G of Reactants What does x quantify? What is significant at this point? ΔGo G of Products ΔG = 0 (at equilibrium) p. 837 #4 What happens at 300 K? In what T range is this favorable? ΔG = – ΔG = ΔH – TΔS T > 300 K ΔH = TΔS so… ΔG = ΔH – TΔS ΔG = 0 (at equilibrium) HW p. 841 #62, 63, 76 ∆Go & Rxn Coupling Rxn Coupling: Unfav. rxns (+∆Go) combine with Fav. rxns (–∆Go) to make a Fav. overall (–∆Gooverall ). (zinc ore) (zinc metal) ZnS(s) Zn(s) + S(s) S(s) + O2(g) SO2(g) goes up if coupled (NOT therm.fav.) ∆Go = +198 kJ/mol ∆Go = –300 kJ/mol ZnS(s) + O2(g) Zn(s) + SO2(g) ∆Go = –102 kJ/mol (therm.fav.) ∆Go & Biochemical Rxn Coupling (weak bond broken, stronger bonds formed) ATP ADP ATP + H2O ADP + H3PO4 ∆Go = –31 kJ/mol Alanine + Glycine Alanylglycine ∆Go = +29 kJ/mol (amino acids) (peptide/proteins) ATP + H2O + Ala + Gly ADP + H3PO4 + Alanylglycine ∆Go = –2 kJ/mol ∆Go & Biochemical Rxn Coupling Rxn 1: Glu + Pi Glu-6-P ATP ADP + Pi Rxn 2: Overall Rxn: Glu + ATP Glu-6-P + ADP +14 (not fav) –31 (fav) –17 (fav) Overall Reaction: ∆Govr ∆Govr = ∆G1 + ∆G2 ∆Go & Biochemical Rxn Coupling Free Energy (G) Glucose (C6H12O6) Proteins ATP + O2 (oxidation) –∆G (fav) CO2 + H2O +∆G (not fav) ADP –∆G (fav) +∆G (not fav) Amino Acids Thermodynamic vs Kinetic Control Kinetic Control: (path 2: A C ) A thermodynamically favored process (–ΔGo) with no measurable product or rate while not at equilibrium, must have a very high Ea . ∆Go = +10 Ea = +20 (kinetic product) Free Energy (G) A B (initially pure reactant A) path 1 A B +10 kJ –50 kJ (low Ea , Temp , time) ∆Go = –50 path 2 Ea = +50 C (thermodynamic product) A C (–∆Go, Temp, Q<<K, time) Thermodynamic vs Kinetic Control Thermodynamic Product: ___ E Kinetic Product: ___ D Pain Rxn A E will be under ______________ kinetic (high Ea) control at low temp and Q > K .