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15 Thermodynamics: What thermo tells us: Whether a reaction happens or not. By the time we finish this chapter, you will be able to look at reactions like the ones in the table below and tell me all kinds of interesting things about the reaction, including, most importantly, whether the reaction happens (whether the reaction is spontaneous.) And since this is the start of getting really quantitative about chemistry, you will be able to tell not only whether it happened, but also the extent to which a reaction occurs. Pretty exciting stuff from just a single chapter. Reaction spontaneity: it’s not just for thermo anymore: What is even more interesting is that except for Kinetics in Chapter 16, the rest of the semester will be spent asking the question, did the reaction happen? And we will use three different approaches to answering the question: Did the reaction happen? Thermodynamics CH 15 DG < 0 Did the reaction happen? Equilibria CH 17-20 K >1 Did the reaction happen? Electrochemistry CH 21 E >0 Thermodynamic reaction worksheet Predict the vales for DH, Dngas, w, DS and DG. At the least provide a sign, but if you can offer a guess of the magnitude without using a calculator, that is even better. reaction DH Dngas w DS DG CH4(g) + O2(g) ‡ CO2(g) + 2H2O(g) 2H2(g) + O2(g) ‡ 2H2O(g) 2H2O(g) ‡ 2H2(g) + O2(g) C2H5OH(l) + 3O2(g) ‡ 2CO2(g) + 3H2O(g) C2H5OH(l) + 3O2(g) ‡ 2CO2(g) + 3H2O(l) C3H8(g) + 5O2(g) ‡ 3CO2(g) + 4H2O(g) 4H2O(g) + 3CO2(g) ‡ C3H8(g) + 5O2(g) CCl4(l) ‡ C(s) + 2Cl2(g) Ba(OH)2(H2O)8(s) + 2NH4NO3(s) ‡ Ba(NO3)2(s) + 2NH3(g) + 10H2O(l) 3O2(g) ‡ 2O3(g) H2O(s) ‡ H2O(l) CO2(g) ‡ CO2(s) NH3(g) + HCl(g) ‡ NH4Cl(s) 2H2O2(l) ‡ 2H2O(l) + O2 (g) (You should be able to do this worksheet by the time we finish this chapter (if you really understand the material.) Course Notes on Thermodynamics A timeout before we get started: If you compare the notes for Chapter 15 to the review sheet on thermodynamics (entitled “Thoughts in Thermodynamics”), you will see some differences. In particular, the review sheet first presents the concept of DG, the free energy of the system, while the notes begin with a discussion of DH, the heat in the system. This is because the notes follow the textbook, while the review sheet follows the way thermo should be taught (IMHO). Plan on me providing an overview of thermo using the review sheet in the first lecture, and taking a more detailed look at the course notes in the next two lectures. Thoughts on Thermodynamics: A Study Guide for Chapter 15 Nine Content Areas You Should Know In organizing your study plans for thermo, it is essential to be able to distinguish the seemingly endless stream of very similar concepts from each other. This is best done by creating a hierarchy of important facts about thermodynamics—note that the relative importance of this material doesn’t run from front to back, it is more like from back to front. Thus we start with DG, the first of none topic areas: 1. DG, The Gibb’s Free Energy. The most important thing about thermodynamics is this: thermodynamics tells you whether a reaction is spontaneous or not, whether it occurs or not. To do this we look at the Gibb’s Free Energy, DG and ask the following: If DG positive (+) the reaction is not spontaneous—it doesn’t happen If DG negative (-) the reaction is spontaneous—it happens 2. So what is DG? It is the free energy of a system. The free energy is a combination of three kinds of energy: • heat energy, q used to calculate the enthalpy, D H • work, usually a change in P or V (PDV) for a reaction involving gases • TDS, the entropy, which is a measure of the disorder in a system 3. We can combine these three forms of energy in a couple of ways: • DG = DH – TDS which is a really famous equation used to calculate free energy • DE = q + w = DH – PDV which describes the internal energy in a system 4. Enthalpy, DH: The heat of reaction: Very simply, this term provides a quantitative measure of how much heat is given off in a reaction. • Endothermic reaction: DH is +. The surroundings get colder as the system gets hotter. • Exothermic Reaction: DH is -. The surroundings get hotter as the system gets colder. 5. Calculating DH: There are two ways to do this. • Experimentally using a bomb calorimeter: Calculate DH = mCDT. This is how they measure the calories of food products. • Using tables and Hess’ Law. Hess’ Law Hrxno = S DHfo products – S DHfo reactants 6. Hess’ Law: its where you start and stop, not how you get there. No matter how we calculate a change in a state function, it isn’t the path that matters, it is where you start and where you end. Because of the first Law of Thermodynamics, we know that energy is conserved. So if we want to calculate the DH for a reaction, we can put together any combination of reactions that cancel out to leave the DHrxn we want. There are two ways we apply Hess’ Law in CH302 problems: heats of formation and bond energy: • Heat of formation, DHfo values in Appendix K. DHfo is the heat of reaction when elements in their standard state react to make an product. • Bond Energy: Rather than use Appendix K, we can apply Hess’ Law using average values for each and every broken and formed bond. We use tables of average covalent bond energies. 7. PV Work (bombs) (and why it often is ignored): Defining PV work. Some reactions don’t just give off heat, they produce and use up gases that change the pressure or volume of a system. For example, the pistons in your car do this with combusted gasoline. Very simply, this is PV work that occurs when either the pressure changes or the volume changes when a reaction of gases occurs. If a reaction involves only liquids and solids, or no net change in the number of moles of gas, there is no PV work and DE = q + w = DH – PDV or DE = DH Calculating PV work. If there is PV work it is easy to calculate from the ideal gas law: PDV = DnRT For example, for a one mole change in gas, PDV = DnRT = 2.5 kJ which is much smaller than most combustion reactions. A practical note about PV work: It is often approximately true that DE = q + w = DH + (something small) and again DE = DH 8. Entropy, DS: It takes a lot of energy to keep things neat. Systems tend to become messy. This is because there is energy in the world above absolute zero that makes bonds vibrate and molecules jump around. Before you know it, the entropy, S is larger. • When DS = -, the system is increasingly ordered. • When DS = +, the system is increasingly disordered. Examples of how to predict DS: On the exam you will be asked to predict system that are ordered or disordered. • Disorder occurs as phase changes happen and solids become liquids become gases • Disorder occurs when the volume increases, when temperature increases, when the number of particles increases. • The opposite conditions favor increased order and require that energy be removed from the system to the surroundings so all that jiggling will stop and things will freeze over. 9. Putting together S and H to make G: Look at the equation DG = DH – TDS Note that to make a reaction spontaneous, DG must be negative. This is done by making S more positive and H more negative. For any reaction, one of four possible combinations of DH and DS occur: • If DH is negative and DS is positive a reaction happens • If DH is positive and DS is negative a reaction can’t happen • If both DH and DS are positive or both are negative, it is possible for a reaction to occur and depends on the temperature. A Study Hint: Everything you need to study for thermo is founded on the 9 concept areas presented on the review sheet. When working thermo problems, before actually finding an answer, identify the material on the review sheet associated with the problem. That way you know where to go for help in figuring out how to do them. And finally, the course notes: Thermodynamics is the study of energy change, D, in chemical systems. Here are the kinds of things that change in a system. Some of them you have seen before. DH, DE, DS, DG, DT, DV, DP What thermodynamic changes can do for you: They will tell you if two things will react. They will tell you how much energy change occurs. They will tell you how far a reaction will go to completion. DH Reaction Categories: Exothermic and Endothermic For thermodynamic purposes, chemical reactions fall in two categories, the ones that happen and the ones that don’t. Spontaneity favored = exothermic = negative DH example: NaOH + HCl -----> H2O + Na+ + Cl- + heat (temperature of surroundings increases) Spontaneity not favored = endothermic = positive DH example: Ba(OH)2.8H2O + 2NH4NO3 + heat ----> Ba(NO3)2 + 2NH3 + 10H2O (temperature of surroundings decreases) Thermodynamic concepts that you should know: system the substances involved in the chemical system surroundings everything else in the universe universe system and surroundings first law of thermodynamics energy is conserved in chemical reactions second law of entropy in the universe is always increasing thermodynamics state functions parameters that define the current state of a chemical system: P, V, T, H, S, G, composition change in state functions all we care about in thermo: D = (what we end with - what we start with) change in enthalpy (DH) change in heat content of system calorimetry experimental technique to find DH Hess’ Law The path to a change does not matter, only where you start and where you stop B.E. Bond energy calculation Path through gaseous atoms to determine DHrxn DHfo Heat of formation Path through elements in std. states to determine DHrxn Time out for Specific Heat I know that everyone remembers specific heat and heat capacity since you learned about them in Chapter 1-13, but I thought we could use a breather before things get hard. specific heat: The amount of heat required to raise the temperature of a gram of substance one degree C. (amount of heat, J) specific heat has units of: C = (mass of substance, g)(DT) heat capacity: Amount of heat required to raise temperature of an object one degree C. (You get the heat capacity just by multiplying the mass of an object times the specific heat.) molar heat capacity: Of course everyone’s favorite mass is the mass of a mole of things which yields the idea of molar heat capacity A few constants for those who don’t want to peruse Appendix E : substance specific heat molar heat capacity aluminum 0.90 J/goC 24.3J/moloC iron 0.444 J/goC 24.5 J/moloC benzene 1.74 J/goC 136 J/moloC water 4.18 J/goC = 1.0 cal/ goC 75.3 J/moloC Of these, the one that matters for bomb calorimetry measurements is water. FAMOUS EQUATION FOR ALL PREMEDS!!!! Rearrange the definition of specific heat above to yield: DH = mCDT ﬂ the experimental way to find DHrxn Example: How much heat is necessary to raise the temperature of 205 g of H2O from 21.2oC to 91.4oC? DH = mCDT = (205g)(4.18J/goC)(70.2oC) = 60,154 J = 60.1 kJ Wait, this is how to do the problems back in Chapter 13!! How to calculate DH experimentally using a bomb calorimeter: Applying the Concept of Conservation of Energy. The problem. Those values for calories on the back of a candy bar were obtained by using a calorimeter. A calorimeter is a device that captures all of the heat produced in a combustion reaction. Now you might wonder why we would go to all of that trouble. After all, if DH = mCDTcandy bar then if we know the values for m, C, and DTcandy bar for the candy bar, we just calculate DH for the candy bar. The difficulty, of course is finding DT, I don’t know about you, but determining the DT for a candy bar when you catch it on fire is not an easy thing. A solution: In a closed system, the first law of thermodynamics states that: DHsystem = DHsurroundings So what about placing the candy bar into a container in the middle of a vat of water (a calorimeter). Then is must be true that DHsystem (candy bar) = mCDT candy bar = DH water (surroundings) = mCDT water And since it is much easier to measure temperature change for a container of water, we use values of m and C for water to get DH for the system (the burning candy bar) indirectly. Nothing is perfect. Of course, not all the heat from an exploding candy bar will end up in the water. Some will transfer to the calorimeter itself. So we need to add a correction for the lost heat and we get the true calorimeter equation: DHsystem (candy bar) = mCDT candy bar = DH water + DH container = mCDT water + mCDT container Working problems: Now with this big old equation, we can ask you to find just about any value by giving you all the other data. The big trick: keeping track of what is the system and what is the surroundings. Thermodynamic Equations A balanced equation that includes a value for DH is a thermochemical equation. Example, the combustion of pentane: C5H12(l)+ 8 O2(g) -----> 5CO2(g) + 6 H2O(l) DH = -3523kJ/mol rxn 1 mole 8 moles 5 moles 6 moles Things to note about pentane combustion: 1. This is an exothermic combustion reaction (the DH is negative). 2. The reaction could be made endothermic by reversing reactants and products. 5CO2(g) + 6 H2O(l) -----> C5H12(l)+ 8 O2(g) DH = +3523kJ/mol rxn 3. The physical state (gas, liquid or solid) of the materials matters in assigning DH. 4. The assumption is made that this is a “mole reaction” in which the amount of energy released is for one mole of reaction. If there is a change in the amount of material, then DH must be changed proportionally. Getting ready to work Thermo problems: Standard States thermochemical standard state: the most stable state of a substance under standard pressure (one atmosphere) and temperature (usually 25 oC) Examples of standard states: water clear liquid, H2O (l) mercury silvery liquid, Hg (l) hydrogen diatomic gas, H2 (g) limestone chalky solid, CaCO3 (s) o We denote that a change occurs under standard conditions with a superscript zero, standard enthalpy change, DHrxno : This is DH when reactants in standard states are converted to products in standard states A special case of DHrxno is: heat of formation, DHfo :This is the case where DH is for one mole of substance formed under standard conditions from its elements in their standard conditions. By definition, DHfo of an element is zero. Examples of DHfo: CH4 ,(g) -74.21 kJ/mole C 6H6 (l) 49.03 kJ/mole CO2 (g) -393.53 kJ/mole H2O (l) -285.8 kJ/mole Br2 (l) 0 kJ/mole Br2 (g) 30.91 kJ/mole There are plenty more in Appendix K. Example: Calculate the enthalpy change when for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere. 4Al(s) + 3O2(g) -----> 2 Al2O3(s) DH = -3352KJ/mol rxn kJ = (15g Al)(1mol Al/27g Al)(1mol rxn/4 mol Al)(-3352kJ/mol rxn) = -465 kJ released to surroundings Hess’ Law Hess’ Law: The DH for a reaction is the same no matter what the pathway is between reactants and products. or stated as a nice equation to be memorized for working dozens of problems: Hess’ Law: DHrxno = S n DHfoproducts - S n DHforeactants What this law allows us to do is combine thermodynamic equations any way we want to fine new thermodynamic equations for unknown reactions. Bond Energies bond energy: Energy necessary to break one mole of bonds in a gaseous substance. Note the substance has to be in the gas phase to use bond energies. Bond energies can be used in a special case example of Hess’ Law to give a rough estimate of DH rxno when all else fails. Consider a reaction in which the H-H bond for hydrogen in its standard elemental state is broken to form atoms. H2(g) ------> 2H (g) DHrxno = +435 kJ/mol This is the simplest example of a bond energy. Consider the more complicated case of successive cleavage of a methane CH4 ---> CH3(g) + H(g) DHo = +427kJ/mole CH3 ---> CH2(g) + H(g) DHo = +439kJ/mole CH2---> CH (g) + H(g) DHo = +452 kJ/mole CH ---> C (g) + H(g) DHo = +347kJ/mole If you think about it, you can cleave a C-H bond in any kind of organic molecule and the DHo will be a little different each time. For this reason we need to use an AVERAGE bond energy. The good thing is that we can include all the average bond energies in a single table. A nice table lifted from Davis, p561 gives us some single bond values. We can also construct a table of average multiple bond energies. Why go to this trouble? Because in a variation on Hess’ Law we can find the DHrxno for any reaction IN THE GAS PHASE by DHrxno = S B.E. reactants - S B.E. products Example: Use the bond energies in the two table above to estimate the DHrxno for CH4 (g) + 2O2 (g) ----> CO2 (g) + 2H2O(g) DHrxno = [4BEC-H + 2BEO=O ] - [2BEC=O + 4BEO-H] = [(4)(414) + (2)(498)] - [(2)(741) + (4)(464)] = -686 kJ/mole Notes about BE: • Using bond energies we found a value for DHrxno and we never had to access thermochemical data like other DHrxno or DHfo. • But... we could only look at gas phase reactions. • And...the answer is only approximate and is only close when all reactants and products are gases. There are huge errors when using BE for other phases. • The advantage: all the data you need for the calculation can be found in a small table of individual bond energies. • Internal Energy Change: An Intro to DE We’ve examined DH for a system and talked a lot about changes in energy. But we’ve not discussed DE. What is the relationship between the two? The fundamental equation relating DH and DE is H = E + PV PV is pressure-volume work. It is work in which during the course of a reaction, there is a change in pressure or in volume of the system. As we will see, on many occasions there is no DP or DV for a reaction and DH = DE. Very often even when DP or DV are non-zero they are negligible and the approximation that DE = DH holds. To understand the experimental conditions under which these cases occur, it is necessary to better understand PV work. DE defined: DE = Eproducts- E reactants = q + w where q is the heat absorbed or released by the system and w is work done on or by the system. We will see that w is PV work. But first, we have to get our signs straight: According to Davis (and therefore all the problems you will do) q is positive: when heat is absorbed by the system q is negative: when heat is released by the system w is positive: when work is done on the system w is negative: when work is done by the system In this convention whenever energy is added to the system, the energy of the system increases. PV work Why is PV work, work anyway? Look at the units: P x V F/d2 x d 3 = Fd = w !!! PDV Work: Our first example of PV work will be the case where we keep P constant and vary V. This is PDV work. Actually, most of the reactions you do will involve PDV work because they are done at atmospheric pressure. Thus DE = q - PDV But wait. Why is there a minus sign? Well consider what happens when we have a reaction in which the amount of gas in the system increases compared to a reaction in which the amount of gas decreases in the system. In the picture above, a reaction occurs in which the volume of gas decreases. Work is being done by the surroundings to the system. From our table of signs above, this means w is positive. But the volume decreased, i.e., DV is negative. To make the equality hold, + - - w = PDV we must add a negative sign to the equation. hence the general equation DE = q - PDV Note that if the volume of a reaction doesn’t change, then DE = qv The energy of the system is just the heat released or absorbed by the system. When does this happen? Well to a good approximation, in the following three cases. When can we ignore PDV work? 1. In reactions involving only solids and liquids 2. In reactions involving gases in which an equal number of moles of gas reactants and products are involved. 3. In a bomb calorimeter. That is right, our friend the bomb calorimeter measures only qv because the volume is held constant and no PDV work is done so DE = qv and we can do exactly the kind of calorimetry problems we learned about at the beginning of the chapter. A little more on DE and DH. Now that we better understand PV work, we can better understand the relationship between DE and DH. Remember by definition H = E + PV And at constant T and P DH = DE + PDV Example: Can we show that at constant pressure, DH = qp we know DH = DE + PDV and DE = q + w so, DH = q + w + PDV (constant T, P) at constant P, w = -PDV so, DH = q + (-PDV) + PDV = qp How does DH compare to w in magnitude? Thus the difference between DE and DH is PDV and if we have no gas present or if equal moles are exchanged then DH = DE. But what about when we do have a gas present. Assume an ideal gas law and PDV = DnRT so DH = DE + DnRT How significant is this DnRT term? Sample calculation of w Example: If a reaction occurs at standard temperature in a bomb calorimeter and one mole of gas is produced, how much PV work is done? w = -PDV = -DnRT w = -(1mol)(8.314J/mol K)(298K) = 2.5 x 103 J so, PDV work is a couple of kJ/mole at room temperature Note we get a value of a couple of kJ/mole. How does this compare with typical combustion reaction DHrxno? Consider for example the DHrxno of methane which is -890 kJ/mol, PDV work is much smaller. For many gas reactions, then, the PDV contribution to the energy of the system is negligible and again, DH ª DE Spontaneity and Entropy I lied. Actually I approximated. On page one of the notes it is suggested that an exothermic reaction, -DH, is synonymous with a spontaneous reaction while an endothermic reaction, +DH, is synonymous with a non- spontaneous reaction. This is not always true. Usually it is, but not always--because in addition to the energy change associated with a reaction, there is a second parameter to consider in determining spontaneity: the change in entropy, DS Entropy and the Second Law—the universe is getting messier Now you all know a bunch of jokes about entropy, and it may be kind of easy to tell you pat definitions about entropy like The second law of thermodynamics: In spontaneous changes in chemical systems, the universe tends to a state of greater disorder. But actually, entropy is quite difficult to discuss on a quantitative level. Thus whereas you will work hundred of problems involving energy changes, you will work only a few involving entropy changes. The problem is that to describe entropy mathematically, you have to understand probability theory and be a whiz at statistics. I would bet most of you don’t and aren’t. So we will stick to a kind of hazy overview in which we hand wave about entropy being an increase in disorder of a system--we’ll save the quantitative details for when you take a statistical mechanics course, that’s CH354L for those who want to plan ahead. Entropy and signs: When we talk about increasing disorder in a system, we are describing an increase in S. In other words, DS is positive. For example, consider a phase change as molecules go from solid to liquid to gas. There is increasing disorder so DS for the change in the system if positive. Spontaneity and entropy: According to the second law, if entropy increases, the spontaneity of the process is favored. But, this doesn’t mean the process will be spontaneous--remember, we have to consider DH as well. Entropy and perfection. Suppose we try to increase the order of a chemical system, make it perfectly ordered. This is the same as saying that entropy of the system is zero. Can this happen? No. This is the basis of the third law of thermodynamics. Third law of thermodynamics: The entropy of a perfectly ordered crystal is zero at absolute zero (0K). Note in the picture that were a perfect crystal to be created, then by adding heat to the system (raising the temperature), the bonds in the crystal will vibrate and entropy increases. We will see this at play when we look at kinetic theory in the study of gases, liquids and solids. A Hess’ Law calculation using S data from Appendix K Standard Entropy Changes. We can actually treat calculations of DSrxno as we would DHrxno and use a Hess’ Law equivalent to take known DS rxno values from Appendix K to find unknown DSrxno values. The equation we use is DSrxno = S n DSfoproducts - S n DSforeactants and the problems are worked identically to those for DHrxno. Predicting the sign of DS. We can’t do the math right now, but we all have an intuitive feel for whether disorder increases or decreases in a system. Can we predict the sign of entropy changes that occur in molecular systems? Phase changes: This is an easy one. As we move from solid to liquid to gas, molecules have increased freedom of motion and hence the opportunity for increased disorder, so DS is positive. Conversely, as molecules are cooled from gas to liquid to solid, DS is negative as order increases. Temperature changes: The argument for phase changes holds here. Adding heat to a system gives the molecule greater ability to vibrate, rotate or travel. Hence DS is positive. Volume changes: If we increase the volume of a system, the molecules have more places they can go, hence DS is positive. If the volume decreases, DS is negative. Increase in number of particles. Obviously the more particles we have in a system, the greater the opportunity for disorder, hence DS is positive. A particular application of this is to chemical reactions. Consider the dissociation of a diatomic molecule. H2 (g) ----> 2H (g) DSrxno = +98.0J/molK Example of entropy prediction In general we can look at a chemical reaction and evaluate whether there is an increase in the number of reactant species or an increase in the number of product species. If the number of products increases, DS is positive. If the number of products decreases, DS is negative. For example, 2H2(g) + O2(g) ----> 2H2O (g) Because Dn for the system goes down, there is more order and we would predict a decrease in entropy And sure enough, DSrxno = -89J/molK from App K data. Turning entropy into energy units: TDS Where do we go from here? Well it isn’t too much of a stretch to see that we will want to combine what we know about DS and DH. But before we do, let’s make a guess at how to relate the magnitude of D S to DH. Consider for example the 89 J/moleK just calculated for the diatomic molecule above. How do we convert it into something to compare to DH? Why not just multiply by T to get J/mol? At room temperature: TDS = (298K)(89J/moleK) = 26 kJ/mole Compare this value with some of the DHrxno in Appendix K. Note that in many cases, the entropy term for a reaction will be relatively insignificant in determining the spontaneity of a reaction. DG (Gibbs Free Energy) and Spontaneity Now it is time to put all we have learned about DH and DS together. As noted at the beginning of the chapter, spontaneity is actually determined from both DH and DS. This is because when energy, DH, evolves from a system some of it may do work, but some may also be used to alter the entropy of the system, DS. A state function was developed by Willard Gibbs to account for this. Gibbs Free Energy, DG: DG = DH - TDS (constant T,P) Note that DG is increasingly negative if : i. the system gives off more heat, larger negative DH ii. the system becomes more disordered, larger positive DS iii. T for the system increases If you think about it, these are the kinds of things that happen to make a reaction more likely to occur. In general we can determine spontaneity of a reaction from a determination of DG. DG spontaneity positive nonspontaneous 0 at equilibrium negative spontaneous DG and the Equalizer Bunny: As we will learn, later in the course, if DG is large negative then we have a system capable of doing a lot of work, for example, a brand new battery. If DG = 0 we are at a special place in the reaction called equilibrium where nothing happens--DG = 0 is the scientific definition of a dead battery. The special case of a standard reaction: Consider the case for a thermochemical reaction that occurs at 1 atmosphere pressure, at 25oC, and with stoichiometrically correct molar quantities of all the reactants and products. We will have created the special case of a standard reaction with a DG rxno value found right in Appendix K. We can create a Hess’ Equation equivalent for DG rxno standard reactions that allows us to mix and match other thermochemical reactions just like we did for DHrxno and DSrxno DGrxno = S n DGfoproducts - S n DGforeactants Simple calculation of DGrxno using Appendix K Example. Calculate DGrxno for the reaction below. Is the reaction spontaneous? C3H8(g) + 5O2 (g) -----> 3CO2 (g) + 4H2O (l) from App.K -23.49 0 3(-394) 4(-120.4) DGo = 4(-120.4) + 3(-394) - (-23.49) = -1638 kJ/mol In the example above, DGrxno is negative so the reaction is spontaneous (which you already knew, because you know that combustion reaction are spontaneous which is why your car works.) Note that the reverse reaction is endothermic because DGrxno would be positive. This also makes sense because you can’t make gasoline by blowing into a bowl of H2O. The Real World and DG. The thing is, DGrxno is valid only at 1 atm and 298K for a molar reaction. This is not a value we would use in real life where the concentrations are not often 1M and partial pressures are not often 1 atmosphere. In the real world we evaluate spontaneity by examining DGrxn (no tiny superscript o) which differs from DGrxno by an amount dependent on concentration DGrxn = DGrxno + something that accounts for concentrations not = 1M We will learn about this equation later. For now we will just play around with DGrxno. Temperature and spontaneity. Look at the Gibbs equation: DG = DH - TDS The temperature term is a nice easy experimental variable just begging to be manipulated. Try sticking various values of DH and DS into the equation while you vary T. You’ll note the following trends: class DH DS DG one - + spontaneous (-) at all T two - - spontaneous (-) below a specific T three + + spontaneous (-) above a specific T four + - nonspontaneous (+) at all T Combinations of DS and DS that determine DG Consider an example from each class. Try thinking about the reactions in terms of real world experiences. class Reaction Example DH DS sponta kJ/mol J/molK neity? 1 2H2O2 (l) --->2H2O (l) + O2 (g) -196 +126 all T 2 NH3 (g) + HCl (g) ----> NH4Cl (s) -176 -285 below 619K 3 CCl4 (l) ---> C (s) + 2Cl2(g) +136 +235 above 517K 4 3O2 (g) ---> 2 O3 (g) +285 -137 never Thinking for yourself. For each of these examples, think about the conditions under which a spontaneous reaction is to occur. Also, could you predict whether DH and DS would be + or - from what we’ve learned, and could you then use these values to look at any reaction and make a reasoned guess as to whether it is spontaneous? Rearranging an DG = DH – TDS to determine when spontaneity occurs Example. The answer is given above, but why don’t you calculate from thermochemical data the temperature at which the reaction written below becomes spontaneous. As a hint, remember, a reaction changes from spontaneous to nonspontaneous when DGrxno = 0. NH3 (g) + HCl (g) ----> NH4Cl (s) DG = DH - TDS = 0 DH -176 T = ææ = æææ = 618 K DS -.285 Rearranging an DG = DH – TDS to calculate boiling points To top it all off, use the same idea to determine the boiling point of water, thermochemically, when you don’t have a thermometer around. Example. Use the thermochemical data in Appendix K to estimate the boiling point of H2O Consider the reaction H2O (l) ----> H2O (g) This is the temperature at which liquid and gas exist in equilibrium at 1 atmosphere for standard conditions. This means DGo = 0. DHorxn = DHf H2O(g) - DHf H2O(l) = = -241 kJ/mole - (-285 kJ/mole) = 44 kJ/mole DSorxn = SoH2O(g) - SoH2O(l) = = 0.188 kJ/moleK - 0.69 kJ/moleK = 0.119 kJ/molK DH 44 kJ/mol DG = 0 = DH - TDS T = ææ = æææææ = 370K = 97oC DS 0.119 kJ/molK