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Chemical Thermodynamics Chapter 19 A continuation of Chapter 5 Watkins Chem 1422, Chapter 19 1 Review Chapter 5 1st Law of Thermodynamics DEuniv = 0 Euniv = Esys + Esurr DEsys = q + w w = –PDV (expansion work) For DV = 0, w = 0 qv = DEsys For DP = 0, DEsys = qp – PDV qp = DEsys + PDV = DHsys Watkins Chem 1422, Chapter 19 DE and DH are state functions q and w are not state functions 2 Spontaneous Processes Some reactions occur spontaneously without outside intervention. For example: H2O(s, 25 oC) → H2O(l, 25 oC) ice always melts at room temperature The reverse of a spontaneous reaction is never spontaneous: H2O(l, 25oC) → H2O(s, 25oC) water never freezes at room temperature Thermodynamics tries to answer the question: Under what circumstances is a reaction spontaneous? Watkins Chem 1422, Chapter 19 3 Spontaneous Processes Every system tends spontaneously toward its lowest energy Watkins 1 spontaneous 1 2 Dge < 0 2 Gravitational Energy Gravitational Energy A ball rolls down a hill spontaneously because its gravitational energy at the bottom is less than its gravitational energy at the top: Dge < 0 A ball never rolls uphill all by itself; it has to be given extra energy Dge to climb up the (gravity) hill. Chem 1422, Chapter 19 2 non-spontaneous 2 1 D ge > 0 1 4 Spontaneous Processes Every system tends spontaneously toward its lowest energy Watkins R spontaneous R P Dce < 0 P Chemical Energy Chemical Energy A chemical reaction rolls down a chemical energy hill spontaneously if the chemical energy of the products is lower than the chemical energy of the reactants: Dce < 0 A non-spontaneous reaction can be forced to proceed by giving it extra energy Dce. P non-spontaneous P R Dce > 0 Chem 1422, Chapter 19 R 5 Thermodynamically Reversible Reactions Chemical Energy A chemical reaction at equilibrium is on a flat chemical energy surface (Dce = 0). A system at equilibrium is not spontaneous; in thermodynamics, a system at equilibrium is called reversible. Watkins R equilibrium P R P Dce = 0 Chem 1422, Chapter 19 6 ThermodynamicallyReversible Reactions A reversible reaction goes forward or backward just by adding or subtracting heat: H2O(l, 1 atm, 273K) ⇌ H2O(s, 1 atm, 273K) To freeze 1 mol of water to form 1 mol of ice, -q = -DHfus of heat is removed from the system. To reverse the process, +q = +DHfus of heat must be added to 1 mol of ice to form 1 mol of water. Therefore, converting between 1 mol of ice and 1 mol of water at this T and P is a reversible process. Watkins Chem 1422, Chapter 19 7 Reversible and Irreversible Reactions A reaction at equilibrium (Q = K) is reversible and therefore non-spontaneous. A reaction which is not at equilibrium (Q K) is irreversible; if Q < K, the reaction is spontaneous in the forward direction; if Q > K, the reaction is spontaneous in the reverse direction. Watkins Chem 1422, Chapter 19 8 Spontaneous Reactions Watkins R spontaneous R P Dce < 0 P Chemical Energy Chemical Energy We predict the spontaneity and direction of a chemical reaction by comparing the values of Q and K; thermodynamics cannot predict the speed at which the reaction will occur (that's kinetics!). So what is “Chemical Energy”? The first thing to consider is DH. P non-spontaneous P R Dce > 0 Chem 1422, Chapter 19 R 9 Spontaneous Reactions We predict the spontaneity and direction of a chemical reaction by comparing the values of Q and K; thermodynamics cannot predict the speed at which the reaction will occur (that's kinetics!). So what is “Chemical Energy”? The first thing to consider is DH. Many spontaneous reactions are exothermic; they seem to roll down an enthalpy hill. (Dce = DH?) But some spontaneous reactions are endothermic, so Dce DH. But is Dce = DE? There are some spontaneous processes for which there is no apparent energy change at all! Watkins Chem 1422, Chapter 19 10 Spontaneous Processes Expansion of an Ideal Gas Initial state: two 1 L flasks are connected by a closed stopcock; one flask is evacuated and the other contains 1 atm of an ideal gas. Watkins Chem 1422, Chapter 19 11 Spontaneous Processes Expansion of an Ideal Gas Process: when the stopcock is opened, the gas spontaneously expands to fill both flasks. Watkins Chem 1422, Chapter 19 12 Spontaneous Processes Expansion of an Ideal Gas Final state: the two flasks are connected by an open stopcock; each flask contains gas at 0.5 atm. Watkins Chem 1422, Chapter 19 13 Spontaneous Processes Expansion of an Ideal Gas The expansion of an ideal gas is isothermal (DT = 0) so no heat is transferred (q = 0; adiabatic). The flasks are rigid so there is no expansion work (w = 0). Thus, DE = q + w = 0 Watkins Chem 1422, Chapter 19 14 Spontaneous Processes Expansion of an Ideal Gas If DE = 0 and DH = 0, why does the gas expand spontaneously (Dce < 0)? The spontaneous expansion of an ideal gas rolls down a chemical energy hill created by ENTROPY Watkins Chem 1422, Chapter 19 15 Entropy - S Entropy is a measure of the probability and disorder of a system. Analogy: unwrap a new deck of cards; it is arranged in suits and is highly ordered – the new deck has low entropy. Now throw the deck into the air... the cards will probably land in a disordered jumble: high S. the cards will (probably) never land in a neat stack with the cards in suits: low S. high S = high probability = disorder low S = low probability = order Watkins Chem 1422, Chapter 19 16 Entropy - S high S = high probability = disorder low S = low probability = order Example: adiabatic expansion of a gas... Gas molecules are in constant and random motion. In a small volume, the molecules have less room for their random motion, but in a larger volume, the molecules can move in a more random (disordered) way. Thus... Gas in a smaller volume low S & low probability Watkins DS > 0 Gas in a larger volume high S & high probability Chem 1422, Chapter 19 17 Entropy - S high S = high probability = disorder low S = low probability = order Example: ice has low S because the molecules are held in ordered positions by H-bonds. Ice spontaneously melts at room temperature because liquid water is composed of molecules in random motion (high S). H2O(s,298K) → H2O(l,298K) DS > 0 Solid = low S Liquid = higher S Gas = highest S Watkins Chem 1422, Chapter 19 18 Entropy - S high S = high probability = disorder low S = low probability = order Example: an ionic solid dissolves in water: the ions leave the crystal and float off into the solution (increasing S) the water organizes into hydrates about the ions (decreasing S) DSdiss, the total entropy of dissolution is the sum (usually > 0) Watkins Chem 1422, Chapter 19 19 Measuring DS The entropy of a system is a state function: DSsys = Sfinal – Sinitial It can be measured in a reversible system. For water and ice in equilibrium at 0 oC, the two states can be reversibly exchanged by adding or subtracting heat (DHfus = qrev = 6010 J/mol) at constant T = 273K: qrev DHfus DSrev = T = T ice → water @ 273K, qrev > 0: DSrev = +22.0 J/mol-K water → ice @ 273K, qrev < 0: DSrev = -22.0 J/mol-K Watkins Chem 1422, Chapter 19 20 Measuring DS Any chemical reaction, if carried out in a closed system, will come to equilibrium at the T and P specified by the experimenter: LHR(states,T,P) ⇌ RHR(states,T,P) Furthermore, every reaction has a heat of reaction at temperature T, DHrx,T , which can be calculated using Hess’s Law and a table of Heats of Formation. Therefore the entropy change for the reversible reaction at temperature T, DSrx,T , can be computed: DHrx,T DSrev = DSrx,T = T Note that only values at T = 298K can be computed from the data in Appendix C. Watkins Chem 1422, Chapter 19 21 Measuring DS Now consider an irreversible (spontaneous) reaction (e.g., a mixture of water and ice at room temperature). It has been found experimentally that for all spontaneous (non-equilibrium) reactions: DSirrev > DSrev The concept of entropy, and the rules governing its behavior, have been formulated as the 2nd Law of Thermodynamics Watkins Chem 1422, Chapter 19 22 Second Law of Thermodynamics DSsys may be + or –, but DSuniv is always positive! The change in entropy of the universe is the sum of the change in entropy of the system and the change in entropy of the surroundings DSuniv = DSsys + DSsurr For every reversible (equilibrium) process DSsys = -DSsurr DSuniv = 0 For every irreversible (spontaneous ) process DSsys > -DSsurr DSuniv > 0 Watkins Chem 1422, Chapter 19 23 Second Law of Thermodynamics DSsys may be + or –, but DSuniv is always positive! What does this mean? TDS is the energy associated with entropy; TDS is waste energy which cannot be used for any purpose! But the total amount of energy in the universe is constant (First Law)! Every spontaneous process that occurs at T > 0 converts some of the energy of the universe into waste energy. Once all of the energy in the universe is converted to waste energy, there can be no more spontaneous processes. The universe is running down! Watkins Chem 1422, Chapter 19 24 Molecular Interpretation of Entropy Absolute Entropy of State high S g l low S s Adding heat increases both entropy and temperature ... except at a phase change, where added heat increases only entropy. The entropy of a perfect crystal at 0 K is 0 (3rd Law of Thermodynamics) Watkins Chem 1422, Chapter 19 25 Molecular Interpretation of Entropy Entropy Change vs. Amount of Gas Any reaction that increases the number of gas molecules leads to a increase in entropy 2NO2(g) → 2NO(g) + O2(g) 2 moles gas lower S 3 moles gas higher S DS > 0 Entropy Change vs. Volume of Gas Entropy increases as a gas expands Watkins Chem 1422, Chapter 19 26 Spontaneous Reactions Watkins R spontaneous R P Dce < 0 P Chemical Energy Chemical Energy A spontaneous (non-equilibrium) reaction always rolls down a “chemical energy” hill which clearly must involve entropy. So what is “Chemical Energy”? We are now ready to answer this question. P non-spontaneous P R Dce > 0 Chem 1422, Chapter 19 R 27 Spontaneous Reactions For DP = 0, the total heat energy (DH) is divided into useful heat energy (DG) and waste heat energy (TDS): DH = DG + TDS DG = DH - TDS G is Gibbs Free Energy G is free to cause spontaneous reactions G is the chemical energy that drives a spontaneous chemical reaction. Watkins Chem 1422, Chapter 19 28 Thermodynamic Energies aA + bB ⇌ cC + dD EA HA SA GA EB HB SB GB EC HC SC GC ED HD SD GD DErx DHrx DSrx DGrx Only these three are used in this course (see Appendix C) Each reagent in a reversible reaction contains four specific molar energy quantities called total internal energy E kJ/mol “heat content” or enthalpy H kJ/mol entropy S J/mol.K (the energy quantity is TS J/mol) Gibbs free energy G kJ/mol The energy changes for the reaction are (Hess’s Law) DErx = cEC + dED – aEA – bEB Watkins Chem 1422, Chapter 19 29 Calculation of DHrx The heat of a chemical reaction which produces n moles of products from m moles of reactants at constant P and T is computed using Hess's Law: DHorx(kJ) = SnDHof(products) - SmDHof(reactants) Standard heat of formation DHof is the enthalpy change when 1 mole of a substance is formed from its elements at constant P and T (Appendix C). DHof for an element is zero. For example: DHof (kJ/mol): H2(g) + ½O2(g) → H2O(g) at 298K 0 0 -241.82 DHorx = –241.82 kJ/mol Watkins Chem 1422, Chapter 19 30 Calculation of DGrx The free energy change of a chemical reaction which produces n moles of products from m moles of reactants at constant P and T is computed using Hess's Law: DGorx(kJ) = SnDGof(products) - SmDGof(reactants) Standard free energy of formation DGof is the free energy change when 1 mole of a substance is formed from its elements at constant P and T (Appendix C). DGof for an element is zero. For example: DGof (kJ/mol): H2(g) + ½O2(g) → H2O(g) 0 0 at 298K –228.57 DGorx = –228.57 kJ/mol Watkins Chem 1422, Chapter 19 31 Calculation of DSrx The entropy change for a chemical reaction which produces n moles of products from m moles of reactants at constant P and T is computed using Hess's Law: DSoT(J/K) = SnSoT(products) - SmSoT(reactants) Standard molar entropy SoT is the absolute entropy of 1 mole of a substance in its standard state at constant P and T (Appendix C, units J/mol.K) SoT for an element is not zero. H2(g) + ½O2(g) → H2O(g) For example: So298 (J/mol.K) 130.6 205.0 at 298K 188.8 DSorx = –44.3 J/K Watkins Chem 1422, Chapter 19 32 DG = DH - TDS Gibbs Free Energy G is the chemical energy that drives a reaction. RHR (G ) DG = G – Glhr < 0 The forward rxn is driven harder than the reverse rxn; the forward rxn is spontaneous. LHR (Glhr) LHR (G ) lhr rhr rhr RHR (Grhr) DG = Grhr – G > 0 lhr The reverse rxn is driven harder than the forward rxn; the reverse rxn is spontaneous. LHR (Glhr) RHR (Grhr) DG = Grhr – Glhr = 0 The forward and reverse reactions are driven equally, so neither rxn is spontaneous: equilibrium! Watkins Chem 1422, Chapter 19 33 DGof Standard Free Energies of Formation, DGf Subscript f: specifies a formation reaction: Elements are the only reactants, One mole of a pure compound is the only product: 1/ N (g) + 3/ H (g) → NH (g) 2 2 2 2 3 Superscript o: Standard states: Every solid or liquid is pure (except aq); Every gas is at 1 atm (partial) pressure All dissolved substances are 1 M Watkins Chem 1422, Chapter 19 34 DGof Standard Free Energies of Formation, DGf Temperature is not part of the standard state specification. DGf values at 298K are tabulated in App. C for many substances; these values are different at other temperatures. Note that for an element DGof = 0 at all T. DGof {O2(g)} = 0 kJ/mol at any T DGof {O(g)} = 230 kJ/mol at 298K Watkins Chem 1422, Chapter 19 35 Temperature, DGo, DHo, DSo For any reaction under standard conditions and at any temperature DGorx,T = DHorx,T – TDSorx,T The only values we can calculate from App. C are for T = 298K: DGorx,298, DHorx,298 and DSorx,298. But it has been found that DHorx and DSorx are almost constant for temperatures between about 200K and 400K, and only change slowly outside this range. Therefore, we will approximate, at all temperatures: DHoT ≈ DHo298 DSoT ≈ DSo298 Watkins Chem 1422, Chapter 19 36 Temperature, DGo, DHo, DSo Let's rearrange the defining equation: DGorx,T = DHorx,T - TDSorx,T DGorx,T = (-DSorx,T)T + DHorx,T y = m x + b (a stright line) x =T: the independent variable y =DGorx,T: the dependent variable m = -DSorx,T -DSorx,298: the slope b =DHorx,T DHorx,298: the intercept Watkins Chem 1422, Chapter 19 37 Temperature, DGo, DHo, DSo There are four kinds of reactions DGoT = (-DSo298)T + DHo298 4. (DHo > 0) (DSo < 0) 3. (DHo > 0) (DSo > 0) (+) 4 Non-Spontaneous 2 DGo 0 (-) (DHo (DSo 2. < 0) < 0) 1. (DHo < 0) (DSo > 0) Watkins T Spontaneous 3 1 Chem 1422, Chapter 19 38 Temperature, DGo, DHo, DSo 1. Exothermic (DHo < 0) with increasing entropy (DSo > 0) 2O3(g) → 3O2(g) spontaneous at all T O3 is "unstable" with respect to O2 at all temperatures. (but the reaction is kinetically slow) DGoT = (-DSo298)T + DHo298 4 Non-Spontaneous 2 T Spontaneous 3 1 Watkins Chem 1422, Chapter 19 39 Temperature, DGo, DHo, DSo 2. Exothermic (DHo < 0) with decreasing entropy (DSo < 0) DGoT = (-DSo298)T + DHo298 CaO(s) + CO2(g) → CaCO3(s) Non-Spontaneous spontaneous at T < To nonspontaneous at T > To To is the temperature at which equilibrium is established under standard conditions To = DHo /DSo Watkins 4 2 T To Spontaneous 3 1 Chem 1422, Chapter 19 40 Temperature, DGo, DHo, DSo 3. Endothermic (DHo > 0) with increasing entropy (DSo > 0) DGoT = (-DSo298)T + DHo298 CaCO3(s) → CaO(s) + CO2(g) Non-Spontaneous spontaneous at T > To nonspontaneous at T < To To = DHo/DSo 4 2 T To Spontaneous 3 1 Watkins Chem 1422, Chapter 19 41 Temperature, DGo, DHo, DSo 4. Endothermic (DHo > 0) with decreasing entropy (DSo < 0) DGoT = (-DSo298)T + DHo298 3O2(g) → 2O3(g) Non-Spontaneous nonspontaneous at all T (this reaction has to be “pushed uphill” by the input of energy – UV photons from the sun) Watkins 4 2 T Spontaneous 3 1 Chem 1422, Chapter 19 42 DGo and Equilibrium Equilibrium at To under standard conditions (DGo = 0): 2. Exothermic (DHo < 0) with decreasing entropy (DSo < 0) or 3. Endothermic (DHo > 0) with increasing entropy (DSo > 0) To = DHo /DSo DGoT = (-DSo298)T + DHo298 4 Non-Spontaneous 2 T To Spontaneous 3 1 Watkins Chem 1422, Chapter 19 43 DGo and Equilibrium DGo = DHo -TDSo aA + bB ⇌ cC + dD Reactions of type 2 & 3 can reach equilibrium under standard conditions: [A] = [B] = [C] = [D] = 1 so Kc = 1 p(A) = p(B) = p(c) = p(D) = 1 so Kp =1 DGo = 0 at To = DHo/DSo Most reactions reach equilibrium under nonstandard conditions (the concentrations are not 1 M). Are there any common reactions which do reach equilibrium under standard conditions? Watkins Chem 1422, Chapter 19 44 DGo and Equilibrium DGo = DHo -TDSo Are there any common reactions that do reach equilibrium under standard conditions? Yes! Normal Phase Changes. Melting (fusion): a pure solid is in equilibrium with its pure liquid under 1 atm pressure at temperature To (the normal melting point) : A(s) ⇌ A(l) K = 1 (DHofus > 0, DSofus > 0) DG DGofus,T = DHofus,T – TDSofus,T T To = DHofus,T/DSofus,T ≈ DHofus,298/DSofus,298 0 o o when T < T , DG fus,T > 0 (liquid freezes) when T > To, DGofus,T < 0 (solid melts) when T = To, DGofus,T = 0, (no spontaneous reaction) o o Watkins Chem 1422, Chapter 19 45 T DGo and Equilibrium DGo = DHo -TDSo Normal melting point of Cs(s): Cs(s) ⇌ Cs(l) DHof 298K kJ/mol Cs(s) 0 Cs(l) 2.09 DHofus = +2.09 kJ DGof kJ/mol 0 0.03 So J/mol.K 85.15 92.07 DSofus = +6.92 J/K o DH 2090 J To ≈ DSo fus = = 302K = 29 oC 6.92 J/K fus (experimental value: 28.4 oC) Watkins Chem 1422, Chapter 19 46 DGo and Equilibrium DGo = DHo -TDSo Are there any common reactions that do reach equilibrium under standard conditions? Yes! Normal Phase Changes. Boiling (vaporization): a pure liquid is in equilibrium with its pure vapor at 1 atm pressure at temperature To (the normal boiling point) : A(l) ⇌ A(g) K = 1 (DHovap > 0, DSovap > 0) DG DGovap,T = DHovap,T – TDSovap,T T To = DHovap,T/DSovap,T ≈ DHovap,298/DSovap,298 0 o o when T < T , DG vap,T > 0 (vapor condenses) when T > To, DGovap,T < 0 (liquid vaporizes) when T = To, DGovap,T = 0, (no spontaneous reaction) o o Watkins Chem 1422, Chapter 19 47 T DGo and Equilibrium DGo = DHo -TDSo Normal boiling point of methanol: CH3OH(l) ⇌ CH3OH(g) 298K CH3OH(l) CH3OH(g) DHof kJ/mol -238.6 -201.2 DHovap = +37.4 kJ DGof kJ/mol -166.2 -161.9 So J/mol.K 126.8 237.6 DSovap = +110.8 J/K o DH 37.4 kJ vap o T ≈ DSo = = 337K = 64 oC 0.111 kJ/K vap (experimental value: 64.6 oC) Watkins Chem 1422, Chapter 19 48 DGrx,T DGof,298 is for a formation reaction at 298 K under standard conditions (appendix C) DGorx,298 is for any reaction at 298 K under standard conditions (Hess’s Law + appendix C) DGorx,T is for any reaction at any temperature under standard conditions DGorx,T = DHoT - TDSoT ≈ DHo298 – TDSo298 DGrx,T is for any reaction at any temperature under any conditions. Watkins Chem 1422, Chapter 19 49 DGrx,T aA + bB ⇌ cC + dD DGrx,T is the driving force for this reaction under the specified conditions of T, P, molarities, etc.: DGrx,T = DGorx,T + RT ln Q Q specifies concentrations and/or gas pressures before, during or after the reaction. To match the units of G (kJ), the required value of R is 0.008314 kJ/mol.K DGrx,T = DHorx,T - TDSorx,T + RT ln Q DGrx,T DHorx,298 - TDSorx,298 + RT ln Q Watkins Chem 1422, Chapter 19 50 DGrx,T aA + bB ⇌ cC + dD DGrx,T 0: the reaction is driven forward (< 0) or backward (> 0) spontaneously. In a closed system, as the reaction proceeds, DGrx,T approaches zero. In an open system, reactants and/or products are lost and DGrx,T never reaches zero. DGrx,T = 0: the reaction is at equilibrium (only in a closed system). Watkins Chem 1422, Chapter 19 51 Equilibrium and Thermodynamics C D Q a b A B c aA + bB cC + dD Initially [A]i, [B]i, [C]i, [D]i DGi,T = DGoT + RT ln Qi d If DGi,T < 0, reaction proceeds to the right to reach equilibrium. As reaction proceeds ... Q increases and Qi → Kc DG increases until ... DGi → 0 DG = 0 and Qeq = Kc forward reaction is spontaneous, reverse reaction is non-spontaneous Watkins Chem 1422, Chapter 19 52 Equilibrium and Thermodynamics C D Q a b A B c aA + bB cC + dD Initially [A]i, [B]i, [C]i, [D]i DGi,T = DGoT + RT ln Qi d If DGi > 0, reaction proceeds to the left to reach equilibrium. As reaction proceeds ... Q decreases and Qi → Kc DG decreases until ... DGi → 0 DG = 0 and Qeq = Kc forward reaction is non-spontaneous, reverse reaction is spontaneous Watkins Chem 1422, Chapter 19 53 Equilibrium and Thermodynamics C D Q a b A B c aA + bB cC + dD When DG = 0, Q ≡ Kc Under standard conditions: Only two kinds of reaction can reach equilibrium! DG And at equilibrium: d o K=1 T = DHo/DSo Watkins Chem 1422, Chapter 19 54 Equilibrium and Thermodynamics C D Q a b A B c aA + bB cC + dD When DG = 0, Q ≡ Kc d Under non-standard conditions: DGT = DGoT + RT ln Q DHo298 - TDSo298 + RT ln Q Any forward reaction is spontaneous (DGT < 0) if Q is small enough: lim Q → 0 => lim lnQ → -∞ Watkins Chem 1422, Chapter 19 55 Equilibrium and Thermodynamics C D Q a b A B c aA + bB cC + dD When DG = 0, Q ≡ Kc d Under non-standard conditions: DGT = DGoT + RT ln Q At equilibrium, DGT = 0 and Q = KT 0 = DGoT + RT ln KT o /RT -DG T KT = e Watkins Chem 1422, Chapter 19 56 2SO2(g) + O2(g) ⇌ 2SO3(g) Calculate DHo298, DSo298, and DGo298 for this reaction. Standard conditions: all gases at 1 atm Gas 298K SO2 DHof kJ/mol -296.9 DGof kJ/mol -300.4 So J/mol.K 248.5 O2 SO3 0 -395.2 0 -370.4 205.0 256.2 Using Hess's Law: DHo298 = [2(-395.2)] – [2(-296.9)+(0)] = -196.6 kJ DGo298 = [2(-370.4)] – [2(-300.4)+(0)] = -140.0 kJ DSo298 = [2(+256.2)] – [2(+248.5)+(+205.0)] = -189.6 J/K Watkins Chem 1422, Chapter 19 57 2SO2(g) + O2(g) ⇌ 2SO3(g) Calculate DGo298 from its defining equation. DGo298 = DHo298 – TDSo298 DHo298 = -196.6 kJ DGo298 = -140.0 kJ DSo298 = -189.6 J/K DGo298 = (-196.6 kJ)-(298K)(-0.1896 kJ/K) = -140.1 kJ Match energy units! Watkins Chem 1422, Chapter 19 58 2SO2(g) + O2(g) ⇌ 2SO3(g) Plot DGo as a function of temperature, and find the equilibrium temperature. DHo298 = -196.6 kJ o o o o DG DG T = DH T – TDS T 298 = -140.0 kJ DSo298 = -189.6 J/K DGoT ≈ DHo298 – TDSo298 Intercept (T = 0): -197 kJ Slope: +0.19 kJ/K The equilibrium temperature at which DGoT = 0 and KT = 1 DGoT = 0 DHo298 – TDSo298 T DHo298/DSo298 = (-196.6 kJ)/(-0.1896 kJ/K) = 1037 K = 764 oC Watkins Chem 1422, Chapter 19 59 2SO2(g) + O2(g) ⇌ 2SO3(g) Calculate the equilibrium constant at 298K and 350K o -DG 298/RT DHo298 = -196.6 kJ K298 = e o 140.0/(0.008314×298) DG 298 = -140.0 kJ K298 = e DSo298 = -189.6 J/K 24 K298 = 3.47×10 K350 R = 8.314 J/mol.K R = 0.008314 kJ/mol.K o -DG 350/RT =e DGo350 = DHo350 – (350)DSo350 DHo298 – (350)DSo298 (-196.6 kJ) - (350 K)(-0.1896 kJ/K) = -130.2 kJ K350 = e 130.2/(0.008314×350) = 2.74×1019 Watkins Chem 1422, Chapter 19 60 2SO2(g) + O2(g) ⇌ 2SO3(g) Calculate DG and K under the following conditions: T = 315K, p(SO2)=0.1 atm, DHo298 = -196.6 kJ p(O2) = 0.2 atm, p(SO3) = 10 atm. DGo298 = -140.0 kJ DGo DG315 = 315 + RT ln Q DGo315 (-196.6) - (315)(-0.1896) = -136.9 kJ Q = (10)2/(0.1)2(0.2) = 50000 DSo298 = -189.6 J/K p(SO 3 )2 Q 2 p(SO 2 ) p(O 2 ) DG315 = -136.9 + (0.008314)(315) ln (50000) = -108.5 kJ K315 = e Watkins 136.9/(0.008314*315) = 4.99×1022 Chem 1422, Chapter 19 61 2SO2(g) + O2(g) ⇌ 2SO3(g) Summary T K 298 DHoT kJ -197 DSoT J/K -190 DGoT kJ -140 315 ~ -197 ~ -190 ~ -137 ~ -109 5.0×1022 350 ~ -197 ~ -190 ~ -130 - 2.7×1019 1037 ~ -197 ~ -190 - ~ 1.0 ~0 DGT kJ - KT (none) 3.5×1024 Le Chatelier: as the temperature of this exothermic reaction increases, K decreases. Watkins Chem 1422, Chapter 19 62 C2H5OH(l) The forward reaction is vaporization. The reverse reaction is condensation. C2H5OH(g) C2H5OH liquid DHof kJ/mol -278 DGof kJ/mol -175 So J/mol.K 161 vapor -235 -169 283 (all values at 298K) DHovap,298 = (-235) – (-278) = +43 kJ/mol vaporization is endothermic DSovap,298 = (283) – (161) = +122 J/mol.K vaporization increases disorder DGovap,298 = (-169) – (-175) = +6 kJ/mol vaporization is not spontaneous at 298K condensation is spontaneous at 298K Watkins Chem 1422, Chapter 19 63 C2H5OH(l) The forward reaction is vaporization. The reverse reaction is condensation. C2H5OH(g) DHovap,298 = +43 kJ/mol DSovap,298 = +122 J/mol.K DGovap,298 = +6 kJ/mol Normal Boiling Point: the temperature T at which a pure liquid is in equilibrium with its vapor at 1 atm. Standard conditions: Kp,T = pvap = 1 DGovap,T = 0 ≈ DHovap,298 – TDSovap,298 T ≈ 43/0.122 = 352K = 79 oC experimental Tb = 78.29 oC Watkins Chem 1422, Chapter 19 64 C2H5OH(l) The forward reaction is vaporization. The reverse reaction is condensation. C2H5OH(g) DHovap,298 = +43 kJ/mol DSovap,298 = +122 J/mol.K DGovap,298 = +6 kJ/mol Kp = pvap At its normal boiling point (352K), the vapor pressure of ethanol is 1 atm. What is the equilibrium vapor pressure pvap of ethanol at 298K? Kp = pvap = e Watkins -DGovap,298/RT = 0.089 atm (67 torr) Chem 1422, Chapter 19 65 Summary Data in Appendix C for a given substance (at 298K): DHof,298, DGof,298, So298 Calculate for any reaction at 298K: DHorx,298 = SmprodDHof,298 – SnreactDHof,298 DGorx,298 = SmprodDGof,298 – SnreactDGof,298 DSorx,298 = SmprodSo298 – SnreactSo298 Watkins Chem 1422, Chapter 19 66 Summary Calculated for any reaction at any temperature: DGorx,298 = DHorx,298 – 298DSorx,298 DGorx,T = DHorx,T – TDSorx,T ≈ DHorx,298 – TDSorx,298 DGorx,T = 0 and KT = 1 at T = DHorx,T/DSorx,T ≈ DHorx,298/DSorx,298 DGrx,T = DGorx,T + RT ln QT Calculated for any reaction at equilibrium: DGrx,T = 0 KT = e -DGorx,T/RT ≈ e DSorx,298/R e -DHorx,298/RT Watkins Chem 1422, Chapter 19 67