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Chapter 19 Chemical Thermodynamics Grace Ngubeni Office: C106 Email: [email protected] © 2017 Pearson Education, Inc. 2 What is Thermodynamics? THERMODYNAMICS Heat Temperature Energy “Not static” from the Greek Word for “power” Thermodynamics is the study of energy and its transformation. 3 Recap FIRST LAW OF THERMODYNAMICS • You will recall that energy cannot be created nor destroyed. • Therefore, the total energy of the universe is a constant. • Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa. 4 Recap FIRST LAW OF THERMODYNAMICS Two ways for energy to be transferred Heat lost by the system Heat gained by the system q + q SYSTEM w − w Work done on the system Work done by the system or ΔE = q + w 5 Recap Are q and w State Functions? No, they are Path Functions Whether the battery is shorted out or is discharged by running the fan, its E is the same. 6 SPONTANEOUS PROCESSES • Spontaneous processes are those that can proceed without any outside intervention, i.e. without ongoing input of energy from outside the system. • The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B. © 2017 Pearson Education, Inc. 7 SPONTANEOUS PROCESSES • Processes that are spontaneous in one direction are nonspontaneous in the reverse direction. • Spontaneity does not predict the rate (speed) of the reaction. 8 SPONTANEOUS DOESN’T MEAN FAST • Rusting of an iron nail is spontaneous and slow. Fe + O2 → Fe2O3 © 2017 Pearson Education, Inc. 9 SPONTANEOUS PROCESSES • Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. – Above 0 oC, it is spontaneous for ice to melt. – Below 0 oC, the reverse process is spontaneous. 10 QUESTION Melting a cube of water ice at 10°C and atmospheric pressure is a ________ process. a) Spontaneous b) Nonspontaneous Attempt the question first. The solution will be provided in a separate document. 11 REVERSIBLE PROCESSES System at higher T than Surroundings In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process. System at lower T than Surroundings The direction of heat flow can be changed by an Infinitesimal change in T or another property. 12 IRREVERSIBLE PROCESSES • Irreversible processes cannot be undone by exactly reversing the change to the system. • Spontaneous processes are irreversible. 13 SPONTANEOUS PROCESSES Reversible Irreversible • Chemical systems in equilibrium are reversible. • Both the system and its surroundings can be restored to their original state. • In an spontaneous process, the path between reactants and products is irreversible. • The system cannot return to its original state without there being a permanent change in the surroundings. Thermodynamics gives us the direction of a process. It cannot predict the speed at which the process will occur. Direction of a process can be exothermic or endothermic. 14 ENTROPY • Entropy (S) is a term coined by Rudolph Clausius in the 19th century. • Clausius was convinced of the significance of the ratio of heat (q) delivered and the temperature (T) at which it is delivered. 𝑞 𝑆= 𝑇 ENTROPY 15 gas increasing disorder • Entropy – measure of the disorder/randomness of a system. • It is related to the various modes of motion in molecules. liquid solid Entropy increases with the freedom of motion of molecules. Therefore, S(g) > S(l) > S(s) 16 ENTROPY • Like internal energy, E, and enthalpy, H, entropy (S) is a state function. • Therefore, S = Sfinal Sinitial 17 ENTROPY • For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature: q rev S = T Heat transferred under reversible conditions The units of entropy are J K-1 Temp at which process takes place 18 The sign of entropy change, ΔS, associated with the boiling of water is _______. a) Positive b) Negative c) Zero Attempt the question first. The solution will be provided in a separate document. QUESTION H2O (l) → H2O (g) 19 S FOR PHASE CHANGES Melting of a substance at its melting point and vaporization at its boiling point are isothermal processes. Consider melting of ice at 0 oC, q = Hfusion for a small change in heat by raising the temp of the surrounding infinitely slowly, qrev = Hfusion 20 S FOR PHASE CHANGES • For the system, qrev = Hfusion q rev Sfusion = (constant T) T Hfusion = (units: T J/K) 21 EXAMPLE • Consider ΔHfusion for H2O is 6.01 kJ/mol (take note: the values is positive because melting is an endothermic process). Therefore, calculate the ΔSfusion for melting 1 mol of ice at 0°C. 𝑞𝑟𝑒𝑣 ∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛 ∆𝑆𝑓𝑢𝑠𝑖𝑜𝑛 = = 𝑇 𝑇 1 𝑚𝑜𝑙 (6.01 × 103 𝐽. 𝑚𝑜𝑙−1 ) = = 22.0 𝐽. 𝐾 −1 273 𝐾 22 SECOND LAW OF THERMODYNAMICS States: • the entropy of the universe increases for spontaneous processes, and • the entropy of the universe does not change for reversible processes. 23 SECOND LAW OF THERMODYNAMICS In other words: For reversible processes: Suniv = Ssystem + Ssurroundings = 0 For irreversible processes: Suniv = Ssystem + Ssurroundings > 0 These last truths mean that as a result of all spontaneous processes, the entropy of the universe increases. ENTROPY IS NOT CONSERVED 24 EXAMPLE 1 The normal melting point of ice is 0 oC and its molar enthalpy of fusion is ∆Hfus = 6.01 kJ/mol. What is the entropy of the system when 1 mol of ice melts at the normal freezing point? 25 • ∆S is a state function so we can calculate the heat transferred if the process is reversible • ∆Ssys = qrev/T = ∆Hfus /T ={(1 mol)(6.01x103 J/mol)} / 273K = 22.0 J/K • ∆Suniv= ∆Ssys + ∆Ssurr = 22.0 J/K + (-22.0 J/K) =0 J/K 26 EXAMPLE 2 Calculate the change in entropy of the system and the change in entropy of the surrounding, when our system is 1 mol of ice melting in the palm of our hand, which is part of the surrounding. 27 • Note: The process is not reversible because the system and surroundings are at different temperatures. ∆Ssys = qrev/T= ∆Hfus /T ={(1 mol)(6.01x103 J/mol)} / 273 K = 22.0 J/K The surroundings immediately in contact with the ice is your hand. (Assume 370C = 310K) ∆Ssurr = qrev/T= ∆Hfus /T ={(1 mol)(-6.01x103 J/mol)} / 310 K = -19.4 J/K 28 ∆Suniv= ∆Ssys + ∆Ssurr = 22.0J/K + (-19.4J/K) =2.6 J/K • The entropy of the universe increases for any spontaneous process. • This can be used to predict whether a process is spontaneous or not. Reversible: Suniv = Ssystem + Ssurroundings = 0Example 1 Irreversible: Suniv = Ssystem + Ssurroundings > 0Example 2 29 Which of these properties are state functions? a) i) and ii) only b) all of the above c) ii) iii) iv) d) iii) and iv) only QUESTION Consider the following thermodynamic properties. i) Work done on a system ii) Heat absorbed iii) Entropy iv) Enthalpy Attempt the question first. The solution will be provided in a separate document. 30 ENTROPY ON A MOLECULAR SCALE • • Ludwig Boltzmann described the concept of entropy on the molecular level. Temperature is a measure of the average kinetic energy of the molecules in a sample. 31 ENTROPY ON A MOLECULAR SCALE • Molecules exhibit several types of motion: – Translational: Movement of the entire molecule from one place to another. – Vibrational: Periodic motion of atoms within a molecule. – Rotational: Rotation of the molecule about an axis or rotation about the bonds. 32 ENTROPY ON A MOLECULAR SCALE • Boltzmann envisioned the motions of a sample of molecules, at a particular instant in time. • This would be akin to taking a snapshot of all the molecules. • He referred to this sampling as a microstate of the thermodynamic system. • Microstate is a single possible arrangement of the positions and kinetic energies of the molecules when the molecules are in a specific thermodynamic state. 33 ENTROPY ON A MOLECULAR SCALE • Each thermodynamic state has a specific number of microstates, W, associated with it. • Entropy is S = k ln W where k is the Boltzmann constant, 1.38 1023 J/K. Entropy is a measure of how many microstates are associated with a specific macroscopic state. 34 S = k ln W R 8.3154 J K -1 mol1 -23 -1 k 1.38 10 J K N A 6.022 1023 mol1 The Boltzmann constant (k) is in effect the equivalent of the universal gas constant, R, but per molecule rather than per mole. 35 ENTROPY ON A MOLECULAR SCALE Expansion of a Gas at the Molecular Level The two molecules are colored red and blue to keep track of them Four possible arrangements (microstates) once the stopcock is opened Possible arrangements of two gas molecules in two flasks. (a) Before the stopcock is opened, both molecules are in the left of the flask. (b) After the stopcock is opened, there are four possible arrangements of the two molecules. 36 ENTROPY ON A MOLECULAR SCALE Expansion of a Gas at the Molecular Level • Probability of red molecule being in the left flask is two in four (2/4) or ½. • Probability of both molecules being in the left flask is (½)2 = 1/4. • For three gas molecules, the probability of all three molecules being in the left flask is (1/2)3 = 1/8. 37 ENTROPY ON A MOLECULAR SCALE Expansion of a Gas at the Molecular Level • For one mole of gas, the probability of all molecules being in the left flask at the same time is (1/2)N where N = 6.022 x 1023. • The gas spontaneously expands to fill both the left and right flasks, and it does not spontaneously all go back into the left flask. • With the stopcock opened, the arrangement of gas molecules is more random of dispersed than when the molecules are confined in the left flask. 38 ENTROPY ON A MOLECULAR SCALE • The change in entropy for a process, then, is S = k ln Wfinal k ln Winitial S = k ln • Wfinal Winitial Entropy increases with the number of microstates in the system. 39 ENTROPY ON A MOLECULAR SCALE • The number of microstates and, therefore, the entropy tends to increase with increases in – temperature – volume – the number of molecules 40 QUESTION Predict the sign of ΔS for each the following processes: a) Dissolving a solute in a solvent to produce a solution. b) Freezing a liquid. c) Condensing vapour. d) Evaporating a liquid. Attempt the question first. The solution will be provided in a separate document. 41 ENTROPY AND PHYSICAL STATES Entropy increases with the freedom of motion of molecules. Therefore, S(s) < S(l) < S(g) 42 INCREASE vs. DECREASE IN ENTROPY 2NO(g) +O2(g)2NO2(g) Generally, when a solid is dissolved in a solvent, entropy increases. Entropy for the reaction? Entropy decreases when NO(g) is oxidized by O2(g) to NO2(g) 43 CHANGES IN ENTROPY • In general, entropy increases when – Gases are formed from liquids and solids. – Liquids or solutions are formed from solids. – The number of gas molecules increases. – The number of moles increases. 44 QUESTION If a process is exothermic (loss of energy), does the entropy of the surroundings: A) always increase B) always decrease C) sometimes increase and sometimes decrease, depending on the process? Attempt the question first. The solution will be provided in a separate document. 45 Predict whether change in entropy (ΔS) is positive or negative for each of the processes assuming each occurs at constant temperature. CO2(g) (ii) HCl (g) + NH3(g) a) b) c) d) NH4Cl(s) (i) Negative (ii) Positive (i) Negative (ii) Negative (i) Positive (ii) Negative (i) Positive (ii) Positive QUESTION (i) CO2(s) Attempt the exercise first. The solution will be provided in a separate document. 46 EXERCISE Choose the sample of matter that has greater entropy in each pair and explain your choice: (i) 1 mole of NaCl(s) or 1 mole of HCl (g) at 25oC. (ii) 1 mole of HCl (g) or 1 mole of Ar (g) at 25oC. Attempt the exercise first. The solution will be provided in a separate document. 47 THIRD LAW OF THERMODYNAMICS The entropy of a pure crystalline substance at absolute zero is 0. S (0K) = 0 At T = 0 K there is only one microstate; W = 1 So S = k ln W = 0 48 THIRD LAW OF THERMODYNAMICS Since, the entropy of a pure crystalline substance at absolute zero is 0. Remember that S = k ln W k = 1.38 1023 J/K When W = 1, S = 1.38 1023 J/K ln 1 =0 49 ENTROPY CHANGES IN CHEMICAL REACTIONS Entropy vs Temperature graphs can be obtained by carefully measuring how the heat capacity (C) of a substance varies with temperature and we can use the data to obtain absolute entropies at different temperatures. 50 STANDARD MOLAR ENTROPIES • These are molar entropy values of substances in their standard states. • Standard entropies tend to increase with increasing molar mass. • Unlike enthalpies of formation, standard molar entropies of elements at the reference temperature of 298K are not zero. 51 STANDARD MOLAR ENTROPIES • Standard molar entropies of gases are generally greater than those of liquids and solids. • Standard entropies tend to increase with increasing number of atoms in the formula of a substance. 52 STANDARD ENTROPIES Larger and more complex molecules have greater entropies. 53 Predict whether ΔS is positive or negative, assuming each process occurs at constant temperature. 1. 2. 3. 4. 5. 6. 7. 8. H2O(l) H2O(g) Ag+(aq) + Cl-(aq) AgCl(s) 4Fe(s)+ 3O2(g) 2Fe2O3(s) CO2(l) CO2(g) 2SO2(g) + O2(g) 2SO3(g) N2(g) + O2(g) 2NO(g) NaCl(s) Na+(aq) + Cl-(aq) CaSO4(s) Ca2+(aq) + SO42-(aq) QUESTION • Attempt the question first. The solution will be provided in a separate document. 54 EXAMPLE Calculate the standard entropy change of reaction at 25oC for the synthesis of ammonia. N2(g)+ 3H2(g) → 2NH3(g) ΔSo= ∑ nSo (Products) – ∑ mSo (Reactants) ΔSo= 2So(NH3) – { So(N2) + 3So(H2)} = {2 mol x 192.5 J/K mol} – {(1mol x 191.5 J/K mol) + (3 mol x (130.6 J/K mol)} = -198.3 J/K 55 QUESTION Calculate the standard entropy change for the following reactions: a. H2O(l) H2O(g) b. H2(g) + 1/2 O2(g) H2O(l) c. CH3OH(l) CH3OH(g) Attempt the question first. The solution will be provided in a separate document. 56 ENTROPY CHANGES IN SURROUNDINGS • Heat that flows into or out of the system changes the entropy of the surroundings. • For an isothermal process: Ssurr = qsys T At constant pressure, qsys is simply H for the system. ∴ Ssurr = Hsys T 57 ENTROPY CHANGE IN THE UNIVERSE • The universe is composed of the system and the surroundings. • Therefore, Suniverse = Ssystem + Ssurroundings • For spontaneous processes Suniverse > 0 58 ENTROPY CHANGE IN THE UNIVERSE • It is preferable to predict the entropy on the basis of the property of the system • This becomes: Suniverse = Ssystem + Hsystem T Multiplying both sides by (T), TSuniverse = TSsystem + Hsystem TSuniverse is defined as the Gibbs energy (also known as the “free energy” or Gibbs free energy), G 59 GIBBS FREE ENERGY • Endothermic process that is spontaneous e.g. dissolution of ammonium nitrate in water. – must always be accompanied by an increase in entropy of the system. • Also processes that are spontaneous yet proceed with a decrease in entropy of the system e.g. formation of NaCl from its constituent elements. – always exothermic. • Spontaneity therefore involves two thermodynamic concepts – enthalpy and entropy. 60 GIBBS FREE ENERGY G = H – TS (G = Gibb’s free energy or just free energy) For an isothermal process, the change in free energy of the system, G, is G = H - TS Under standard conditions, G° = H TS 61 GIBBS FREE ENERGY • TSuniverse is defined as the Gibbs free energy, G. G = TSuniverse • When Suniverse is positive, G is negative. • Therefore, when G is negative, a process is spontaneous. 62 GIBBS FREE ENERGY • If both T and P are constant, the relationship between the sign of G and the spontaneity of a reaction is: 1. If G < 0 (negative), the forward reaction is spontaneous. 2. If G = 0, the system is at equilibrium. 3. If G > 0 (positive), the reaction is spontaneous in the reverse direction. It is more convenient to use G as a criterion for spontaneity than to use Suniv because G relates to the system alone and avoids the complication of having to examine the surroundings. 63 GIBBS FREE ENERGY Potential energy Free energy Analogy between the gravitational potential energy change of a boulder rolling down a hill and the free energy change in a spontaneous reaction. In any spontaneous process carried out at constant temperature and pressure, the free energy always decreases. 64 GIBBS FREE ENERGY N2 (g) + 3H2 (g) ⇄ 2NH3 (g) 1. If G < 0, Q < K and the forward reaction is spontaneous. 2. If G = 0, Q = K and the system is at equilibrium. 3. If G > 0, Q > K and the reverse reaction is spontaneous 65 FREE ENERGY CHANGES At temperature of 25°C, G° = H TS How does G change with temperature? 66 FREE ENERGY AND TEMPERATURE G° = H TS • There are two parts to the free energy equation: H— the enthalpy term TS — the entropy term • The temperature dependence of free energy, then comes from the entropy term. 67 FREE ENERGY AND TEMPERATURE G° = H TS 68 STANDARD FREE ENERGY CHANGES Analogous to standard enthalpies of formation are standard free energies of formation, G. G = SnGf (products) SmGf (reactants) where n and m are the stoichiometric coefficients. 69 Thermodynamic Quantities for Selected Substances at 298.15 K (25°C) Substance Carbon C (s, diamond) C (s, graphite) C2H2 (g) C2H4 (g) C2H6 (g) CO (g) CO2 (g) Hydrogen H2( g) Oxygen O2 (g) H2O (l) ΔH°f (kJ/mol) 1.88 0 226.7 52.30 -84.68 -110.5 -393.5 0 0 -285.83 ΔG°f (kJ/mol) S° (J/K-mol) 2.84 0 209.2 68.11 - 32.89 -137.2 -394.4 2.43 5.69 200.8 219.4 229.5 197.9 213.6 0 130.58 0 -237.13 205.0 69.91 70 EXAMPLE Calculate ΔG° (in kJ) for the following reaction at 1 atm and 25°C: 2C2H6 (g) +7O2 (g) → 4CO2 (g)+6H2O (l) ΔGf° C2H6 (g) = -32.89 kJ/mol ΔGf° CO2 (g) = -394.4 kJ/mol ΔGf° H2O (l) = -237.2 kJ/mol What is the value of O2? 71 G = SnGf (products) SmGf (reactants) ΔGo={4ΔGf°(CO2(g))+6ΔGf°(H2O(l))} – {2ΔGf°(C2H6 (g))+ 7ΔGf°(O2(g))} ={4×(-394.4) + 6× (-237.2)} – {(-2×32.89)+(7×0)} = -2935.0kJ 72 EXAMPLE Calculate the standard free-energy change for ammonia synthesis at 298K. N2 + 3H2 2NH3 Given ΔHo = -92.2kJ and ΔSo = -198.7J/K G° = H TS = (-92.2 × 103 J) - (298 K) (-198.7 J/K) = (-92.2 × 103 J) - (-59.2 × 103 J) = -33 000 J = -33.0 kJ 73 EXAMPLE Calculate the value of ΔS° for the catalytic hydrogenation of acetylene to ethene, C2H2 (g) + H2 (g) → C2H4 (g) S° = SnS°(products) - SmS°(reactants) S° = {1mol × 219.4 J/mol-K} – {(1mol × 200.8 J/mol-K) + (1mol × 130.58 J/mol-K)} = 219.4 J/K - 331.38 J/K = -112.0 J/K 74 EXAMPLE Calculate ΔG° (in kJ) for the following reaction at 1 atm and 25°C: 2C2H6 (g) + 7O2 (g) →4CO2 (g) +6 H2O (l) • ΔHf° C2H6 (g) = -84.7 kJ/mol S° C2H6 (g) = 229.5 J/K·mol • ΔHf° CO2 (g) = -393.5 kJ/mol; S° CO2 (g) = 213.6 J/K·mol • ΔHf° H2O (l) = -285.8 kJ/mol; S° H2O (l) = 69.9 J/K · mol • S° O2 (g) = 205.0 J/K · mol LONG METHOD ΔH°=Sum Products – Sum of Reactants ΔS°=Sum Products – Sum of Reactants G° = H TS Different manifestation of earlier example Answer: -2935.0 kJ. 75 EXAMPLE Iron metal can be produced by reducing iron(III)oxide with hydrogen. Fe2O3(s) +3 H2(g) 2Fe(s) +3 H2O(g) ΔHo = 98.8kJ; ΔSo = 141.5J/K • Is the reaction spontaneous at 25oC? • At what temperature will the reaction become spontaneous? 76 ΔGo= ΔHo – TΔSo = 98.8 kJ- (298 K)(0.1415 kJ/K) = 56.6 kJ Therefore, reaction is not spontaneous because ΔG>0. We can estimate the temperature at which ΔG changes from a positive to a negative value by setting ΔGo= ΔHo – TΔSo=0 Therefore, this will give us reaction at equilibrium. 77 T=ΔHo/ΔSo =(98.8 kJ)/(0.1415 kJ/K) = 698 K Note: that this calculation assumes that the value of ΔH and ΔS remains unchanged over the temperature range. Therefore, the reaction will be spontaneous at a temperature just above 698 K. ΔGo= ΔHo – TΔSo 78 Find the temperature above which a reaction with a ΔHo of 123.0 kJ and a ΔSo of 90.00 J/K becomes spontaneous. QUESTION Attempt the question first. The solution will be provided in a separate document. 79 FREE ENERGY AND EQUILIBRIUM Under any conditions, standard or nonstandard, the free energy change can be found this way: G = G + RT ln Q (R = universal gas constant = 8.314Jmol-1K-1) (Under standard conditions, all concentrations are 1 M, so Q = 1 and ln Q = 0; the last term drops out.) 80 FREE ENERGY AND EQUILIBRIUM Gf° = 0 81 THE REACTION QUOTIENT, Q • The reaction quotient, Q, is obtained by substituting the reactant and product concentrations or partial pressures at any point during a reaction into an equilibrium constant expression. aA + bB • dD + eE The reaction quotient expression for this reaction would be Qc = [D]d[E]e [A]a[B]b OR PDdPEe Qp = PAaPBb 82 FREE ENERGY AND EQUILIBRIUM At equilibrium, Q = K, and G = 0. The equation {G = G + RT ln Q} becomes 0 = G + RT ln K Rearranging, this becomes G = RT ln K ln K = G/-RT or, K = eG/RT 83 If Q = K, the system is at equilibrium. 84 If Q > K, there is too much product and the equilibrium shifts to the left. 85 If Q < K, there is too much reactant, and the equilibrium shifts to the right. 86 EXAMPLE Calculate the free-energy change for ammonia synthesis at 298K given the following sets of partial pressures. a. 1.00atm N2, 3.0 atm H2, 0.020 atm NH3 N2+ 3H2 2 NH3 Go=-33.0 kJ Qp =(0.020)2/{(1.0)(3.0)3} =1.5x10-5 aA + bB PDdPEe Qp = PAaPBb dD + eE 87 EXAMPLE CONT’D G can be obtained from data tables. Value obtained from earlier calculation & it is given in the question. G = G + RT ln Q =(-33.0 X103Jmol-1) + (8.314J.K-1mol-1)(298K)(ln 1.5x0-5) = -60 519.5 J/mol = -60.5 kJ/mol 88 QUESTION Methanol an important alcohol used in the manufacturing of adhesives, fibers and plastics is synthesized industrially by the reaction. CO(g) + 2 H2(g) CH3OH(g) Calculate the equilibrium constant for this reaction using the data for standard conditions: ΔGo = ΔGof(CH3OH) - {ΔGof (CO)+2ΔGof(H2)} 89 RELATING Kp and Kc The equilibrium constant obtained by this procedure is Kp because the reaction and products are gases and their standard states are defined in terms of pressure. If we want the value of Kc we must calculate it from the relation Kp=Kc(RT)Δn where R must be expressed in the proper units 0.08206(L.atm)/(K.mol) 90 EXAMPLE Given: Kp = 3.5; Δn = -2; Temperature = 474.15 K Calculate Kc Kp=Kc(RT)Δn 3.5 = Kc (0.08206 x 474.15 K)-2 Kc = 5299 91 EXAMPLE The equilibrium constant for a reaction is 0.48 at 25 °C. What is the value of ΔG° (kJ/mol) at this temperature? ΔGo = -RT ln K Ans: 1.8 kJ/mol 92 EXAMPLE The value of ΔG° at 25 °C for a reaction is 1.8 kJ/mol. Calculate the equilibrium constant for this reaction at 25 °C ΔGo = -RT ln K ln K = -ΔGo/RT K = eG/RT K = e-1.8x10^3/(8.314)(298) (Ans: K=0.48) 93 DRIVING NON SPONTANEOUS REACTIONS • Consider extraction of the copper metal from the mineral chalcocite, which contains Cu2S. The decomposition is nonspontaneous Cu2S(s)2 Cu(s) + S(s) ΔGo=+86.2kJ Cu cannot be obtained directly since ΔGo is positive. The reaction can be coupled so that the overall reaction is spontaneous S(s) +O2(g)SO2(g) ΔGo=-300.4kJ 94 By coupling these reactions Cu can be extracted via a spontaneous reaction. CuS(s)+O2(g)2Cu(s)+SO2(g)ΔGo=-214.2kJ. Biological reactions also use spontaneous reactions to drive non spontaneous reactions. The metabolism of food is the usual source of free energy needed to do work of maintaining biological system. C6H12O6(s)+6O2(g)6CO2(g)+6H2O(l) ΔGo=-2880 kJ. Free energy of oxidation used to convert ADP to ATP. The conversion of ATP to ADP releases free energy that can be used to drive other reactions (Converts simple molecules to more complex ones). 95 COUPLED REACTIONS 96 QUESTION Calculate ΔG (the free-energy change under non standard condition) at 298K for a reaction mixture that consists of 1.0atm N2, 3.0 atmH2 and 0.50 atm NH3. 97 QUESTION Consider the decomposition of N2O4. N2O4 (g) 2NO2 (g) ΔHo=57.1kJ ΔSo=175.8 J/K. • Is the reaction spontaneous at 25oC? • Estimate the temperature at which the reaction becomes spontaneous? Solution • Reaction is not spontaneous. ΔGo= 4.71 kJ. • Estimated T = 325 K 98 EXAMPLE • Calculate ∆G at 298K for a reaction mixture that consists of 0.50 atm N2, 0.075 atm H2 and 2.0atm NH3 • Remember : • ∆Go from earlier calculation = -33.3 kJ/mol. 99 QUESTION Calculate the standard free energy change,ΔGo, and the equilibrium constant K, at 298K for the reaction: H2(g)+Br2(l) 2HBr(g) 100 101 Feel free to contact me via email should you have any questions and/or queries. Email address: [email protected]