3 laws of thermodynamics, Cheat Sheet of Physics

Download 3 laws of thermodynamics and more Cheat Sheet Physics in PDF only on Docsity! = =] ss =] = =] = =] = =] Ss =] = =] Chapter 17 1 The Three Laws of Thermodynamics 2 Spontaneous Processes 3. Entropy 4 The Second Law of Thermodynamics : 6 Free Energy and Chemical Equilibrium 7 Thermodynamics in Living Systems SPONTANEOUS” REACTION as time elapses y "Oh what's | cal PRM 8S Whal de you mean do NOT harm Kumeans 2 Gibbs Free Energy DG = DHsys - TDSsys Gibbs free energy (DG)- Can be used to predict spontaneity. For a constant temperature and constant pressure process: DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium. -DG = -T(+DSuniv) +DG = -T(-DSuniv) DSuniv > 0 -TDSuniv = DHsys - TDSsys DSuniv < 0 DG = -T(DSuniv) = 0 DSuniv = 0 2 The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD DG0 rxn nDG0 (products)f= S mDG0 (reactants)fS- Free Energy and Equilibrium DG° < 0 favors products spontaneously DG° > 0 favors reactants spontaneously Does not tell you it will go to completion! aA + bB cC + dD DG° fwd DG° rev DG° fwd = -DG° rev The value of G° calculated under the standard conditions characterizes the “driving force” of the reaction towards equilibrium. 5 DG° = -R T lnK Free Energy and Equilibrium standard free-energy (kJ/mol) gas constant (8.314 J/Kmol) temperature (K) equilibrium constant (Kp, Kc, Ka, Ksp, etc.) • Arguably most important equation in chemical thermodynamics! • It allows us to calculate the extent of a chemical reaction if its enthalpy and entropy changes are known. • The changes in enthalpy and entropy can be evaluated by measuring the variation of the equilibrium constant with temperature. • This relationship is only valid for the standard conditions, i.e. when the activities of all reactants and products are equal to 1. 6 Derivation http://en.wikipedia.org/wiki/Chemical_equilibrium#Thermodynamics At constant temperature and pressure, the Gibbs free energy, G, for the reaction depends only on the extent of reaction: § (Greek letter xi), and can only decrease according to the second law of thermodynamics. It means that the derivative of G with € must be negative if the reaction happens; at the equilibrium the derivative being equal to zero. (@) = =0: equilibrium de } Pp In general an equilibrium system is defined by writing an equilibrium equation for the reaction aA+8BoS4+7T In order to meet the thermodynamic condition for equilibrium, the Gibbs energy must be stationary, meaning that the derivative of G with respect to the extent of reaction: €, must be zero. It can be shown that in this case, the sum of chemical potentials of the products is equal to the sum of those corresponding to the reactants. Therefore, the sum of the Gibbs energies of the reactants must be the equal to the sum of the Gibbs energies of the products. Aftat+ Sg =ops+ Ter where J is in this case a partial molar Gibbs energy, a chemical potential. The chemical potential of a reagent Ais a function of the activity, {A} of that reagent. a= ue + RT In{ A}. ¢ eG is the standard chemical potential ). Substituting expressions like this into the Gibbs energy equation: k dG = Vdp —SdT + > pid; in the case of a closed system. i=1 Now dN; = v,d€ (Vj; corresponds to the Stoichiometric coefficient and d€ is the differential of the extent of reaction ). Using DG° and K Using the table of standard free energies, calculate the equilibrium constant, KP, for the following reaction at 25 C. 2HCl(g) H2(g) + Cl2(g) From appendix 3: H2(g) DGf = 0 kJ/mol Cl2(s) DGf = 0 kJ/mol HCl(g) DGf = -95.3 kJ/mol DG0 rxn nDG0 (products) f= S mDG0 (reactants) f S- DG0 rxn = [1(0) + 1(0)] - [2(95.3 kJ/mol)] DG0 rxn = 190.6 kJ/mol Non-spontaneous! Favors reactants! DG° = -R T ln K 10 Using DG° and K Using the table of standard free energies, calculate the equilibrium constant, KP, for the following reaction at 25 C. 2HCl(g) H2(g) + Cl2(g) DG0 rxn nDG0 (products) f= S mDG0 (reactants) f S- DG0 rxn = [1(0) + 1(0)] - [2(95.3 kJ/mol)] DG0 rxn = 190.6 kJ/mol Non-spontaneous! Favors reactants! DG° = -R T ln K 190.6 kJ/mol =  (8.314 J/K·mol)(25C) ln KP correct units 190.6 kJ/mol =  (8.314 x 103 kJ/K·mol)(298 K) ln KP KP = 3.98 x 1034 Favors reactants! 11 Example 12 17.6 Using data listed in Appendix 3, calculate the equilibrium constant (KP) for the following reaction at 25°C: 2H2O(l) 2H2(g) + O2(g) ΔG°rxn = [2ΔG°f(H2) + ΔG°f(O2)] - [2ΔG°f(H2O)] = [(2)(0 kJ/mol) + (0 kJ/mol)] - [(2)(-237.2 kJ/mol)] = 474.4 kJ/mol o rxn p p p p Δ = - ln 1000 J 474.4 kJ/mol × = -(8.314J/K mol)(298 K)ln 1 kJ ln = -191.5 = . -847 × 10   G RT K K K K e 1915 Bonus: What is the Kp for the reverse reaction? 2H2(g) + O2(g) 2H2O(l) ΔG°rxn = -474.4 kJ/mol or 1/Kp(fwd) = Kp(rev) lnKp(fwd) = -lnKp(rev) ( ) Free Energy and Equilibrium DG° = -R T lnK ln K = - DG° R T Substitution: Rearrange: DG = DH - TDS ln K = - DH - TDS R T ln K = - DH TDS R T R T Rearrange: ln K = - DH R R + + DS T 1 y = m • x + b Measure equilibrium with respect to temperature: 15 Free Energy and Equilibrium rateAB = kobs [A] rateBA = kf [B] A B At equilibrium: rateAB = rateBA kobs [A] = kf [B] [B] = kobs [A]kf = Ku kobs kf Find the DS and DH of the following: 16 Free Energy and Equilibrium kobs kf Slope = -7261.1 K DH° = 60 kJ/mol DS° = 200 J/KmolR DS - DH R Find the DS and DH of the following: 17 If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (DG < 0) If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (DG > 0) If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (DG = 0) Free Energy and Equilibrium standard free-energy (kJ/mol) gas constant (8.314 J/Kmol) temperature (K) reaction quotientDG = DG° + R T ln Q non-standard free-energy (kJ/mol) 20 21 Another Example H2(g) + Cl2(g)  2 HCl(g) For the following reaction at 298 K: H2 = 0.25 atm Cl2 = 0.45 atm HCl = 0.30 atm Given: From appendix 3: H2(g) DGf = 0 kJ/mol Cl2(s) DGf = 0 kJ/mol HCl(g) DGf = -95.3 kJ/mol Which way will the reaction shift to reach equilibrium? DG = DG° + R T ln Q constant given calculate calculate 22 Another Example H2(g) + Cl2(g)  2 HCl(g) For the following reaction at 298 K: H2 = 0.25 atm Cl2 = 0.45 atm HCl = 0.30 atm Given: From appendix 3: H2(g) DGf = 0 kJ/mol Cl2(s) DGf = 0 kJ/mol HCl(g) DGf = -95.3 kJ/mol Which way will the reaction shift to reach equilibrium? DG = DG° + R T ln Q G° = [2(95.27 kJ/mol)]  [0 + 0] = 190.54 kJ/mol 2 2( ) (0.30)HCl 0.80 ( ) ( ) (0.25) (0.45) H Cl 2 2 P QP P P    25 DG0 < 0 DG0 > 0 Free Energy and Equilibrium DG° = -R T ln KAt equilibrium: DG = DG° + R T ln QAt any time: Chapter 17 The Three Laws of Thermodynamics Spontaneous Processes Entropy The Second Law of Thermodynamics Gibbs Free Energy Free Energy and Chemical Equilibrium Thermodynamics in Living Systems Whal de you mean do NOT harm Kumeans 2 SPONTANEOUS” REACTION as time elapses y ORGANIZED EFFORT RECUIRING ENERGY INPUT 26 Synthesis of proteins: (first step) alanine + glycine  alanylglycine G° = 29 kJ/mol “Uphill” Reactions Because ΔG > 0, the reaction is non-spontaneous. alanine glycine No reaction! Need to couple two reactions! 27 Coupled Reactions in Biology glucose + Pi → glucose-6-phosphate ATP + H2O → ADP + Pi glucose + ATP → glucose-6-phosphate + ADP 30 G° < 0 G° > 0 Coupled Reactions in Biology Food Structural motion and maintenance Fats and Carbohydrates ATP and NADPH Coupled reactions ? Chemical Batteries for the Body Stored bond energy 31 Coupled Reactions in Biology Digestion/respiration: Generation of ATP: Burning Glucose Low Energy Higher Energy 32 Coupled reactions to drive the synthesis of: Aminoacids Ribose Nucleic acids Polypeptides DNA Phospholipids This is why we eat! …and why plants absorb light. 35 = =] ss =] = =] = =] = =] Ss =] = =] Chapter 17 1 The Three Laws of Thermodynamics 2 Spontaneous Processes 3. Entropy 4 The Second Law of Thermodynamics 5 Gibbs Free Energy 6 Free Energy and Chemical Equilibrium 7 Thermodynamics in Living Systems SPONTANEOUS” REACTION as time elapses y "Oh what's | cal PRM 8S Whal de you mean do NOT harm Kumeans 2