Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Thermomechanical analysis wikipedia , lookup
Determination of equilibrium constants wikipedia , lookup
Electrolysis of water wikipedia , lookup
Bioorthogonal chemistry wikipedia , lookup
Marcus theory wikipedia , lookup
Photosynthetic reaction centre wikipedia , lookup
Self-assembly of nanoparticles wikipedia , lookup
Internal energy wikipedia , lookup
Equilibrium chemistry wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Thermodynamics wikipedia , lookup
Chapter 19 Spontaneous Change: Entropy and Free Energy Dr. Peter Warburton [email protected] http://www.chem.mun.ca/zcourses/1051.php Spontaneous processes We have a general idea of what we consider spontaneous to mean: A spontaneous process WILL OCCUR in a system WITHOUT any outside action being performed on the system. 2 Spontaneous processes Ice melting above zero Celcius is spontatneous Object falling to earth is spontaneous 3 Spontaneous processes Ice will melt above zero Celcius. We don’t have to DO anything! Objects will fall to earth. We don’t have to DO anything! 4 Spontaneous processes Since we DON’T have to DO anything for these spontaneous processes to occur it APPEARS that an overall energy change from potential energy to kinetic energy IS SPONTANEOUS 5 Non-spontaneous processes We have a general idea of what we consider non-spontaneous to mean: A non-spontaneous process WILL NOT OCCUR in a system UNTIL an outside action is performed on the system. 6 Non-spontaneous processes Ice freezing above zero Celcius is non-spontatneous Object rising from earth is non-spontaneous 7 Non-spontaneous processes We can make water freeze above zero Celcius by increasing the pressure. We can make an object rise from the earth by picking it up. 8 Non-spontaneous processes Since we DO have to ACT for these non-spontaneous processes to occur it APPEARS that an overall energy change from kinetic energy to potential energy IS NON-SPONTANEOUS 9 Chemistry and spontaneity We know there are chemical processes that are spontaneous because we can put the chemical system together and reactants become products without us having to do anything. H3O+ (aq) + OH- (aq) 2 H2O (l) 10 Chemistry and non-spontaneity We know there are chemical processes that are non-spontaneous because we can put the chemical system together and reactants DO NOT become products. The system we put together stays like it is UNTIL WE CHANGE SOMETHING! 2 H2O (l) 2 H2 (g) + O2 (g) 11 Spontaneous vs. non-spontaneous It is obvious by the examples we’ve looked at that the opposite of every spontaneous process is a non-spontaneous process. In chemical systems we’ve seen that if we put a chemical system together a reaction occurs until the system reaches equilibrium. Whether the forward reaction or the reverse reaction dominates depends on which of the two reactions is spontaneous at those conditions! 12 Equilibrium and spontaneity are related! 13 Spontaneity and energy In our examples it APPEARED that the spontaneous process ALWAYS takes a system to a lower potential energy. 14 Spontaneity and energy If this were true all exothermic processes would be spontaneous and all endothermic processes would be nonspontaneous. THIS ISN’T TRUE! NH4NO3 (s) NH4+ (aq) + NO3- (aq) is spontaneous even though H2O DH = +25.7 kJ 15 Recall the First Law The First Law of thermodynamics stated that the energy of an ISOLATED system is constant. What’s the largest ISOLATED system we can think of? It’s the UNIVERSE! The energy of the universe is constant! 16 Recall the First Law On the universal scale, there is no overall change in energy, and so lower energy CANNOT be the only requirement for spontaneity. There must be something else as well! 17 Further proof lower energy isn’t enough If an ideal gas expands into a vacuum at a constant temperature, then no work is done and no heat is transferred 18 Further proof lower energy isn’t enough No work done and no heat transferred means NO OVERALL CHANGE in energy of the system 19 Further proof lower energy isn’t enough This spontaneous process has no overall change in energy! 20 Entropy Entropy (from Greek, meaning “in transformation”) is a thermodynamic property that relates the distribution of the total energy of the system to the available energy levels of the particles. 21 Entropy A general way to envision entropy is “differing ways to move” Consider mountain climbers on a mountain. Two factors affect the distribution of mountain climbers on a mountain: Total energy of all the climbers and how many places can you stop on the mountain 22 Consider hungry mountain climbers Hungry mountain climbers have little total energy amongst themselves to climb a mountain, so most of them are near the bottom, while some are distributed on the lower parts of the mountain. Few energy levels can be reached! 23 Consider well fed mountain climbers Well-fed mountain climbers have more total energy amongst themselves to climb a mountain, so the climbers will be more spread out on the whole mountain. More energy levels can be reached! 24 Temperature and total energy The total energy shared by molecules is related to the temperature. A given number of molecules at a low temperature (less total energy) have less “differing ways to move” than the same number of molecules at a high temperature (more total energy). 25 Higher mountain means more places to stop If a well-fed mountain climber tries to climb Signal Hill, they will most likely reach the top. They have only a few places to stop (levels) because Signal Hill is a small mountain. The same well-fed mountain climber on Mount Everest has a greater number of places to stop (levels) because it is a larger mountain. 26 Volume and energy levels The number of levels of energy distribution of molecules is related to the volume. A given number of molecules in a small volume have less “differing ways to move” than the same number of molecules in a larger volume. 27 Entropy The greater the number of “differing ways to move” molecules can take amongst the available energy levels of a system of a given state (defined by temperature and volume, and number of molecules), the greater the entropy of the system. 28 Expansion into vacuum A gas expands into a vacuum because the increased volume allows for a greater number of “differing ways to move” for the molecules, even if the temperature is the same. 29 Expansion into vacuum That is, the entropy increases when the gas is allowed to expand into a vacuum. Entropy increase plays a role in spontaneity! 30 Entropy is a state function Entropy, S, is a state function like enthalpy or internal energy. The entropy of a system DEPENDS ONLY on the current state (n, T, V, etc.) of the system, and NOT how the system GOT TO BE in that state. 31 Boltzmann equation and entropy More available energy levels when the size of a box increases – like expanding a gas into a vacuum – ENTROPY INCREASES! More energy levels are accessible when the temperature increases – ENTROPY INCREASES! 32 Change in entropy is a state function Because entropy is a state function, then change in entropy DS is ALSO a state function. The difference in entropy between two states ONLY depends on the entropy of the initial and final states, and NOT the path taken to get there. 33 Hess’s Law Recall Hess’s Law – as long as we get from the same initial state to the same final state then DH will be the same regardless of the steps we add together. Change in entropy DS will work exactly the same way! As long as we get from the same initial state to the same final state then DS will be the same regardless of the steps we add together. 34 Boltzmann equation and entropy n, T, V help define the number of states (number of available energy levels) the system can have. The many “different ways to move” of molecules in a particular state are called microstates. Hopefully it makes sense that more total states should automatically mean more total microstates. The number of microstates is often symbolized by W. 35 Playing cards Say we have a deck of 52 playing cards. Choosing one playing card is a state. If we choose the first card out of the pack, there are 52 microstates for this first state. The second card (second state) we choose has 51 microstates, and so on. 36 Playing cards Overall there are W = 52! 8 x 1067 possible distributions (total microstates) for 52 playing cards! 37 Playing cards and coin flips If we flip 52 coins (a coin is one state), with two possible microstates (heads or tails) each, there are W = 252 = 4.5 x 1015 possible distributions. (total microstates) A deck of 52 playing cards has greater entropy than 52 coins! 38 Boltzmann equation and entropy Ludwig Boltzmann formulated the relationship between the number of microstates (W) and the entropy (S). S = k ln W The constant k is the Boltzmann constant which has a value equal to the gas constant R divided by Avagadro’s number NA 39 Boltzmann equation and entropy S = k ln W where k = R / NA k = (8.3145 JK-1mol-1) / (6.022 x 1023 mol-1) k = 1.381 x 10-23 JK-1 We can see the units for entropy will be Joules per Kelvin (JK-1) 40 Measuring entropy change From the units for entropy (JK-1) we get an idea of how we might measure entropy change DS It must involve some sort of energy change relative to the temperature change! 41 Measuring entropy change DS = qrev / T The change in entropy is the heat involved in a reversible process at a constant temperature. 42 Heat IS NOT a state function Since heat IS NOT a state function we need a reversible process to make it ACT LIKE a state function. 43 Reversible processes In a reversible process a change in one direction is exactly equal and opposite to the change we see if we do the change in the reverse direction. In reality it is impossible to make a reversible process without making an infinite number of infinitesimally small changes. 44 Reversible processes We can however imagine the process is done reversibly and calculate the heat involved in it, so we can calculate the reversible entropy change that could be involved in a process. DSrev = qrev / T 45 Endothermic increases in entropy In these three processes the molecules gain greater “differing ability to move.” The molecules occupy more available microstates at the given temperature, and so the entropy increases in all three processes! 46 Generally entropy increases when… …we go from solid to liquid. …we go from solid or liquid to gas. …we increase the amount of gas in a reaction. …we increase the temperature. …we allow gas to expand against a vacuum. …we mix gases, liquids, or otherwise make solutions of most types. 47 Problem Predict whether entropy increases, decreases, or we’re uncertain for the following processes or reactions: a) 2 H 2 S (g) SO 2 (g) 3 S (s) 2 H 2 O (g) b) 2 HgO (s) 2 Hg (l) O 2 (g) c) Zn (s) 2 Ag 2 O (s) ZnO (s) 2 Ag (s) is d) 2 Cl - (aq) 2 H 2 O (l) electrolys 2 OH - (aq) H 2 (g) Cl 2 (g) Answers: a) decreases b) increases c) uncertain d) increases 48 Evaluating entropy and entropy changes Phase transitions – In phase transitions the heat change does occur reversibly, so we can use the formula DSrev = qrev / T to calculate the entropy change. In this case the heat is the enthalpy of the phase transition and the temperature is the transition temperature DS = DHtr / Ttr 49 Evaluating entropy and entropy changes Phase transitions – For water going from ice (solid) to liquid, DHfus = 6.02 kJmol-1 at the melting point (transition temp.) of 273.15 K (0 C) DSfus = DHfus / Tmp DSfus = 6.02 kJmol-1 / 273.15 K DSfus = 22.0 JK-1mol-1 50 Evaluating entropy and entropy changes Phase transitions – For water going from liquid to gas, DHvap = 40.7 kJmol-1 at the boiling point (transition temp.) of 373.15 K (100 C) DSvap = DHvap / Tbp DSvap = 40.7 kJmol-1 / 373.15 K DSvap = 109 JK-1mol-1 51 Problem What is the standard molar entropy of vapourisation DSvap for CCl2F2 if its boiling point is -29.79 C and DHvap = 20.2 kJmol-1? Answer: DSvap = 83.0 JK-1mol-1 52 Problem The entropy change for the transition from solid rhombic sulphur to solid monoclinic sulphur at 95.5 C is DStr = 1.09 JK-1mol-1. What is the standard molar enthalpy change DHtr for this transition? Answer: DHtr = 402 Jmol-1 53 Absolute entropies Say we imagine a system of molecules that has no total energy. At this zeropoint energy there can ONLY be ONE possible distribution of microstates, as no molecule has the energy to occupy a higher energy level. The entropy CAN NEVER get smaller than its value in this situation, so we define the entropy S of this situation as ZERO. 54 Absolute entropies This kind of imagining is the Third Law of Thermodynamics which states that The entropy of a pure perfect crystal at 0K is zero. At conditions other than at absolute zero, our entropy is that of the perfect system (zero) PLUS any entropy changes that come changing temperature and/or volume. These are absolute entropies! 55 Methyl chloride entropy as a function of temperature 56 Standard molar entropies One mole of a substance in its standard state will have an absolute entropy that we often call the standard molar entropy S. These are usually tabulated at 298.15 K In a chemical process we can then use these standard molar entropies to calculate the entropy change in the process. 57 Standard molar entropies DS = [SnpS(products) - SnrS(reactants)] Hopefully this looks somewhat familiar! We have seen a special treatment of Hess’s Law in Chem 1050 where DH = [SnpDHf(products) - SnrDHf(reactants)] We can do something similar with ANY thermodynamic property that IS A STATE FUNCTION! 58 Standard molar entropies DS = [SnpS(products) – SnrS(reactants)] DH = [SnpDHf(products) – SnrDHf(reactants)] Enthalpies of formation ARE NOT absolute! 59 Problem Use the data given to calculate the standard molar entropy change for the synthesis of ammonia from its elements. N2 (g) + 3 H2 (g) 2 NH3 (g) S298 for N2 = 191.6 JK-1mol-1 S298 for H2 = 130.7 JK-1mol-1 S298 for NH3 = 192.5 JK-1mol-1 Answer: -198.7 JK-1 mol-1 (per mole of rxn) 60 Problem N2O3 is an unstable oxide that readily decomposes. The decomposition of 1.00 mol N2O3 to nitrogen monoxide and nitrogen dioxide at 25 C is accompanied by the entropy change DS = 138.5 JK-1mol-1. What is the standard molar entropy of N2O3 (g) at 25 C? S298 for NO (g) = 210.8 JK-1mol-1 S298 for NO2 (g) = 240.1 JK-1mol-1 Answer: 312.4 JK-1 mol-1 61 The second law of thermodynamics We’ve seen that entropy MUST play a role in spontaneity, because the total energy of the universe doesn’t change. We could say that an entropy increase leads to spontaneity, but we have to be careful. 62 The second law of thermodynamics Ice freezing below 0 Celcius is spontaneous, but the entropy of the water decreases in the process! DSfreeze = -DHfus / Tmp Since DHfus and Tmp are +ve, then DSfreeze is –ve! -ve since “freezing” is the reverse of “fusion” (like we do in Hess’s Law) 63 The second law of thermodynamics The water is ONLY the system. The rest of the universe (the surroundings) must experience an opposite heat change as it takes the heat the freezing water gave off (a +ve DH for the surroundings), which means the entropy of the REST OF THE UNIVERSE INCREASES in the process of water freezing! 64 The second law of thermodynamics The total entropy change in any process is the entropy change for the system PLUS the entropy change for the surroundings DSuniverse = DStotal = DSsys + DSsurr Now we can connect entropy and spontaneity! 65 The second law of thermodynamics In any spontaneous process the entropy of the universe INCREASES. DSuniverse = DSsys + DSsurr > 0 This is the Second Law of Thermodynamics! 66 Water freezing So while water freezing below zero Celcius decreases the entropy of the system, the heat given off to the surroundings increases the entropy of the surroundings to a greater extent. The total entropy change of the universe is positive and the process of water freezing below 0 Celcius is spontaneous! 67 Free energy We’ve seen entropy increases when molecules have more ways to distribute themselves amongst the energy levels. 68 Free energy However, some of the energy a molecule uses to put itself at a higher energy level CAN NO LONGER be used to to do work because doing work would put the molecule back at a lower energy level, which would automatically decrease entropy. The energy is NOT free (or available) to be used! 69 Free energy DSuniverse = DSsys + DSsurr TDSuniverse = TDSsys + TDSsurr TDSuniverse = TDSsys + DHsurr DS = DHrev / T then Reversible since the rest of the universe is SO BIG TDS = DH 70 Free energy TDSuniverse = TDSsys + DHsurr TDSuniverse = TDSsys - DHsys Since DHsys = -DHsurr 71 Free energy TDSuniverse = TDSsys – DHsys -TDSuniverse = DHsys - TDSsys DG = DHsys – TDSsys DG is the free energy (Gibbs free energy) 72 Free energy DG = -TDSuniv For a spontaneous process DSuniv > 0, which means for a spontaneous process DG < 0! 73 Free energy DG < 0 is spontaneous DG > 0 is non-spontaneous DG = 0 is at equilibrium 74 75 Problem Predict the spontaneity at low and high temperatures for: N2 (g) + 3 H2 (g) 2 NH3 (g) DH = -92.22 kJ 2 C (s) + 2 H2 (g) C2H4 (g) DH = 52.26 kJ 76 Problem N2 (g) + 3 H2 (g) 2 NH3 (g) DH = -92.22 kJ Spontaneous @ low T and nonspontaneous @ high T 2 C (s) + 2 H2 (g) C2H4 (g) DH = 52.26 kJ Nonspontaneous @ all T 77 Standard free energy change DG Just like we can have a standard enthalpy change DH for chemicals, we can also define the standard free energy change… DG = DH - TDS 78 Standard free energy of formation DGf Standard enthalpies of formation DHf of elements in their standard states are zero: DHrxn = [Snp DHf(products) – Snr DHf(reactants)] Standard free energies of formation DGf of elements in their standard states are zero: DG = [Snp DGf(products) – Snr DGf(reactants)] 79 Problem Calculate DG at 298.15 K for the reaction 4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s) by using the two following sets of data. Compare your answers. a) DH = -1648 kJ mol-1 and DS = -549.3 JK-1 mol-1 b) DGf (Fe) = 0 kJmol-1 DGf (O2) = 0 kJmol-1 DGf (Fe2O3) = -742.2 kJmol-1 80 Problem answer a) DG = -1484 kJmol-1 b) DG = -1484.4 kJmol-1 The answers are the same because free energy is a state function. 81 Free energy and equilibrium We’ve already seen that DG < 0 is spontaneous DG > 0 is non-spontaneous One process in a written reaction is spontaneous while the reverse is not, as long as DG ≠ 0. Therefore DG = 0 is at equilibrium 82 Water and steam H2O (l, 1 atm) H2O (g, 1 atm) DG373.15 K = 0 The system is at equilibrium at 1 atm (standard conditions) and at the boiling point temperature! -1 kJmol 83 Water and steam H2O (l, 1 atm) H2O (g, 1 atm) DG298.15 K = 8.590 The system is not at equilibrium at 1 atm (standard conditions) and at the room temperature! -1 kJmol 84 Water and steam H2O (l, 1 atm) H2O (g, 1 atm) DG298.15 K = 8.590 kJmol-1 The forward process is nonspontaneous (DG > 0) so the reverse process is spontaneous and condensation occurs. 85 Water and steam H2O (l, 0.03126 atm) H2O (g, 0.03126 atm) DG298.15 K = 0 kJ The system is at equilibrium at 0.03126 atm (non-standard conditions) and at the room temperature! 86 Water and steam H2O (l, 0.03126 atm) H2O (g, 0.03126 atm) DG298.15 K = 0 kJmol-1 Water CAN evaporate at room temperature, just not to give an equilibrium pressure of 1 atm! 87 Non-standard conditions As we’ve seen with the previous water example, our interest in an equilibrium system is often at non-standard conditions, so knowing DG is usually not as useful as knowing DG. 88 Non-standard conditions For an ideal gas DH does not change if pressure changes, so at all nonstandard conditions DH = DH. For an ideal gas DS does change if pressure changes (expansion into vacuum shows us this!), so at all non-standard conditions DS ≠ DS. 89 Non-standard free energy Because of these facts, the nonstandard free energy change is DG = DH - TDS But the standard free energy change is DG = DH - TDS 90 Non-standard free energy The difference between standard and non-standard free energy is totally due to the difference in entropy change between the standard and non-standard conditions DG - DG = - T(DS-DS) DG = DG + T(DS-DS) 91 Boltzmann distribution By the Boltzmann distribution S = R ln W for one mole of particles DG = DG + T(R ln W - R ln W) DG = DG + T(R ln W / W) We are comparing a real system with a standard one. We are dealing with activities! 92 Boltzmann distribution DG = DG + RT ln Qeq In the comparison, we are looking at a reaction quotient! In other words, we are looking at how our non-standard system is different from the system we have at standard conditions! 93 Boltzmann distribution DG = DG + RT ln Qeq If our system is at equilibrium then Qeq = Keq and DG = 0 94 Boltzmann distribution DG = DG + RT ln Keq = 0 which means DG = -RT ln Keq 95 Equilibrium constant The thermodynamic equilibrium constant for a reaction is directly related to the standard free energy change! DG = -RT ln Keq -DG/RT Keq = e 96 Equilibrium constant DG 0 Keq 1 DG >> 0 then Keq is very small DG << 0 then Keq is very large 97 Equilibrium constant at 298 K DG Keq Meaning Keq = e-DG/RT 98 Predicting reaction direction DG < 0 means the forward reaction is spontaneous at the given non-standard conditions DG < 0 means the forward reaction is spontaneous at standard conditions and so K > 1 99 Predicting reaction direction DG = 0 means the system is at equilibrium at the given nonstandard conditions DG = 0 means the system is at equilibrium at standard conditions and so Keq = 1 which only occurs at one specific T! 100 Predicting reaction direction DG > 0 means the forward reaction is non-spontaneous at the given non-standard conditions DG > 0 means the forward reaction is non-spontaneous at standard conditions and so K < 1 101 Predicting reaction direction DG = DG ONLY at standard conditions! 102 Thermodynamic equilibrium constant Keq The thermodynamic equilibrium constant Keq is expressed in terms of activities, which are unitless quantities. Activities relate properties like concentration or pressure compared to a standard property value, like 1 M for concentration or 1 bar for pressure. 103 Thermodynamic equilibrium constant Keq aA+bBcC+dD a C a D a b a A a B c K eq d where ax = [X] / c0 (c0 is a standard concentration of 1 M) or ax = Px / P0 (P0 is a standard pressure of 1 atm) Note that ax = 1 for pure solids and liquids 104 Free energy and equilibrium constants DG = -RT ln Keq IS ALWAYS TRUE DG = -RT ln Kc and DG = -RT ln Kp DO NOT have to be true! 105 Problem Write thermodynamic equilibrium constant expressions for each of the following reactions and relate them to Kc and Kp where appropriate: a) Si (s) + 2 Cl2 (g) SiCl4 (g) b) Cl2 (g) + H2O (l) HOCl (aq) + H+ (aq) + Cl- (aq) 106 Problem answer a) Si (s) + 2 Cl2 (g) SiCl4 (g) K eq a a a SiCl 4 2 Si Cl 2 PSiCl 4 PSiCl 4 P Kp 2 2 PCl 2 PCl 2 1 P b) Cl2 (g) + H2O (l) HOCl (aq) + H+ (aq) + Cl- (aq) K eq [HOCl] [H ] [Cl ] a HOCl a H a Cl- c c c [HOCl][H ][Cl - ] P a Cl 2 a H 2O PCl 2 Cl 2 1 P 107 Problem Use the given data to determine if the following reaction is spontaneous at standard conditions at 298.15 K: N2O4 (g) 2 NO2 (g) DGf (N2O4) = 97.89 kJmol-1 DGf (NO2) = 51.31 kJmol-1 Answer: DG = 4.73 kJmol-1, not spontaneous at standard conditions. 108 Problem Based on the problem of the previous slide, determine which direction the reaction will go in if 0.5 bar of each gas is placed in an evacuated container: N2O4 (g) 2 NO2 (g) Answer: Since DG = 4.73 kJmol-1, then Keq = 0.15 = Kp. Since Qp = 0.5 the reaction should go from right to left. 109 Problem Determine the equilibrium constant at 298.15 K for the following reaction using the given data: AgI (s) Ag+ (aq) + I- (aq) DGf (AgI) = -66.19 kJmol-1 DGf (Ag+) = 77.11 kJmol-1 DGf (I-) = -51.57 kJmol-1 110 Problem answer Since DG = 91.73 -17 then Keq = 8.5 x 10 = Kc = Ksp. If we compare to the Ksp value in Table 18.1 (8.5 x 10-17) we see we are definitely in the right ballpark! -1 kJmol , 111 DG and Keq are functions of T We’ve already seen that equilibrium constants change with temperature. Why? -DG/RT e Keq = and DG = DH - TDS 112 DG and Keq are functions of T DG = -RT ln Keq = DH - TDS ln Keq= -(DH/RT) + (DS/R) 113 Problem At what temperature will the following reaction have Kp = 1.50 x 102? 2 NO (g) + O2 (g) 2 NO2 (g) DH = -114.1 kJmol-1 and DS = -146.5 J K-1mol-1 Answer: T = 606 K 114 DG and Keq are functions of T DG = -RT ln Keq = DH - TDS ln Keq= -(DH/RT) + (DS/R) If DH and DS are constant over a temperature range then 115 DG and Keq are functions of T ln K1= -(DH/RT1) + (DS/R) minus ln K2= -(DH/RT2) + (DS/R) ln K1 - ln K2 = -(DH/RT1) - (-DH/RT2) 116 DG and Keq are functions of T ln K1 - ln K2 = [-DH/R] [(1/T1) - (1/T2)] OR ln [K1/K2] = [-DH/R] [(1/T1) - (1/T2)] 117 van’t Hoff equation The van’t Hoff equation relates equilibrium constants to temperatures ln [K1/K2] = [-DH/R] [(1/T1) - (1/T2)] It looks very similar to the Arrhenius equation and so a plot of ln K versus 1/T should give a straight line with a slope of [-DH/R] 118 Plot of ln Kp versus 1/T for the reaction. 2 SO2 (g) + O2 (g) 2 SO3 (g) slope = -DH / R, so for this reaction DH = -180 kJmol-1 119 Problem The following equilibrium constant data have been determined for the reaction H2 (g) + I2 (g) 2 HI (g) Kp = 50.0 @ 448 C Kp = 66.9 @ 350 C Estimate DH for the reaction. Answer: DH = -11.1 kJmol-1. 120 Coupled reactions If we have a non-spontaneous (DG > 0) reaction with a product that appears as a reactant in a different reaction that is spontaneous (DG < 0) then we can couple the two reactions (do them in the same container at the same time) to drive the non-spontaneous reaction! 121 Coupled reactions This should make sense because if we have two equilibria in one container, one with a small K (like Ksp) and one with a large K (like Kf for complex formation), then the coupled reaction that is the sum of the two reactions has a K value that is larger than Ksp! We have made the non-spontaneous reaction occur! 122