Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K) 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.

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 (w) moving (q) due to DT Energy is neither created nor destroyed. total energy of an isolated system is constant (universe) (no transfer matter/energy) (conserved) System: molecules to be studied (reactants & products) Surroundings: everything else (container, thermometer) E = q + w on/by +/– in/out +/–

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) Enthalpy of Reaction CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) enthalpy is… …the heat transfer in/out of a system (at constant P) H = q exergonic exothermic endergonic endothermic H = Hproducts − Hreactants o Hf = 0 for all elements in standard state

Endothermic & Exothermic Endothermic: H > 0 (+) H(+) = Hfinal − Hinitial products reactants Exothermic: H < 0 (–) H(–) = Hfinal − Hinitial reactants products H(–) is thermodynamically favorable

Potential Energy of Bonds High PE chemical bond Low PE (energy released when bonds form) High PE (energy absorbed when bonds break) + +

Enthalpies of Reaction (∆H) BE: ∆H for the breaking of a bond (all +) To determine H for a reaction: compare the BE of bonds broken (reactants) to the BE of bonds formed (products). Hrxn = (BEreactants)  (BEproducts) (bonds broken) (bonds formed) (released) (stronger) H(+) = BEreac − BEprod (NOT on equation sheet) H(–) = BEreac − BEprod (stronger)

Calorimeter nearly isolated Calorimetry We can’t know the exact enthalpy of reactants and products, so we calculate H by calorimetry, the measurement of heat flow. By reacting (in solution) in a calorimeter, we indirectly determine H of system by measuring ∆T & calculating q of the surroundings (calorimeter). Calorimeter nearly isolated (on equation sheet) heat (J) q = mcT Tf – Ti (oC) [of surroundings] (thermometer) mass (g) [of sol’n]

Specific Heat Capacity (c) (or specific heat) energy required to raise temp of 1 g by 1C. (for water) c = 4.18 J/goC + 4.18 J of heat Metals have much lower c’s b/c they transfer heat and change temp easily.

(in J) of calorimeter or surroundings Calorimetry (in J) of calorimeter or surroundings q = mcT – q = Hrxn (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 H = –5.04 kJ 0.1125 mol Hsys = –5.04 kJ 4.50 g NaOH x 1 mol = 0.1125 mol NaOH 40.00 g = –44.8 kJ mol

Chp. 5,8: Calculate ∆H (4 Ways) 1) Bond Energies Hrxn = (BEreactants)  (BEproducts) 2) Hess’s Law Hoverall = Hrxn1 + Hrxn2 + Hrxn3 … 3) Standard Heats of Formation (Hf ) H = nHf(products) – nHf(reactants) 4) Calorimetry (lab) q = mc∆T (surroundings or thermometer) –q = ∆H ∆H/mol = kJ/mol (molar enthalpy) (NOT given) (+ broken) (– formed) (NOT given)     (given) (given)

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). Thermodynamically Favorable processes proceed without any outside intervention. (spontaneous)

1st Law of Thermodynamics total energy of the universe is constant. Hsystem = –Hsurroundings Huniv = Hsystem + Hsurroundings = 0 (Huniv = 0) 2nd Law of Thermodynamics All favorable processes increase the entropy of the universe. Suniv = Ssystem + Ssurroundings > 0 (Suniv > 0)

S : dispersal of matter & energy at T Entropy (J/K) S : dispersal of matter & energy at T S = + therm fav S = – not therm fav entropy increases as number of microstates: ↑Temperature (motion as KEavg) ↑Volume (motion in space) ↑Number of particles (motion as KEtotal) ↑Size of particles (motion of bond vibrations) ↑Types of particles (slaqg)

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 Temp. only 1 microstate 0 K S = 0 > 0 K S > 0

Thermodynamically Favorable Chemical and physical processes are driven by: decrease in enthalpy (–∆Hsys) increase in entropy (+∆Ssys) causes (+∆Ssurr) (+) (+) Suniv = Ssystem + Ssurroundings > 0 Thermodynamically Favorable: (defined as) increasing entropy of the universe (∆Suniv > 0) ∆Suniv > 0 (+Entropy Change of the Universe)

(∆Suniv) & (∆Gsys) –TSuniv = Hsys – TSsys Gsys = Hsys – TSsys (Gibbs free energy equation) Gibbs defined TDSuniv as the change in free energy of a system (Gsys) or G. Free Energy (G) is more useful than Suniv b/c all terms focus on the system. If –Gsys , then +Suniverse . Therefore… –G is thermodynamically favorable. “Bonds break & form to lower free energy (∆G).”

Standard Free Energy (∆Go) and Temperature (T) (on equation sheet) (consists of 2 terms) DG = DH – TS free energy (kJ/mol) enthalpy term (kJ/mol) entropy term (J/mol∙K) units convert to kJ!!! max energy used for work energy transferred as heat energy dispersed as disorder The temperature dependence of free energy comes from the entropy term (–TS).

Standard Free Energy (∆Go) and Temperature (T) DG = DH  TS Thermodynamic Favorability ∆Go = (∆Ho) ∆So – T( ) ( ) –T( ) (high T) – (low T) + (fav. at high T) (unfav. at low T) + + = ( ) – T ( ) + + – (unfav. at ALL T) + = ( ) – T( ) + (fav. at ALL T) – = ( ) – T( ) – + (high T) + (low T) – ( ) –T( ) (unfav. at high T) (fav. at low T) – – = ( ) – T ( ) – –

Calculating ∆Go (4 ways) Standard free energies of formation, Gf : Gibbs Free Energy equation: From K value From voltage, Eo (next Unit) DG = SnG(products) – SmG(reactants) f (given equation) DG = DH – TS (given equation) G = –RT ln K (given equation)

Free Energy (∆G) & Equilibrium (K) G = –RT ln K (on equation sheet) If G in kJ, then R in kJ……… R = 8.314 J∙mol–1∙K–1 = 0.008314 kJ∙mol–1∙K–1 –∆Go RT = ln K –∆Go RT Solved for K : (NOT on equation sheet) K = e^

Free Energy (∆G) & Equilibrium (K) G = –RT ln K ∆Go = –RT(ln K) K @ Equilibrium – + = –RT ( ) > 1 product favored (favorable forward) + – = –RT ( ) < 1 reactant favored (unfavorable forward)

∆Go & Rxn Coupling Rxn Coupling: Unfav. rxns (+∆Go) combine with Fav. rxns (–∆Go) to make a Fav. overall (–∆Gooverall ). goes up if coupled (zinc ore)  (zinc metal) (NOT therm.fav.) ZnS(s)  Zn(s) + S(s) ∆Go = +198 kJ/mol S(s) + O2(g)  SO2(g) ∆Go = –300 kJ/mol ZnS(s) + O2(g)  Zn(s) + SO2(g) ∆Go = –102 kJ/mol (therm.fav.)

Thermodynamic vs Kinetic Control Kinetic Control: (path 2: A  C ) A very high Ea causes a thermodynamically favored process (–ΔGo) to have no product. A  B ∆Go = +10 Ea = +20 (kinetic product) (initially pure reactant A) path 1 (low Ea , Temp , time) B A A  C ∆Go = –50 Ea = +50 (thermodynamic product) Free Energy (G)  +10 kJ path 2 –50 kJ C (–∆Go, Temp, Q<<K, time)

+ = Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ) ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w) (disorder) microstates –T∆Suniv as: ΔHsys & ΔSsys at a T dispersal of matter & energy at T max work done by favorable rxn ΔE = q + w PΔV = –w (at constant P) K > 1 means –∆Gsys & +∆Suniv ∆Suniv = + ΔH = q (heat) ΔS = ΔH T ΔG = ΔH – TΔS sys sys –T∆Suniv