Entropy and Standard Free Energy:

Transcription

1 To ΔG or to ΔG 0 : Improving Conceptual Understanding in Thermodynamics A Presentation of the Flinn AP Chemistry Symposium CHEMED 2005 Paul D. Price Trinity Valley School Fort Worth, TX pricep@trinityvalleyschool.org Introduction: The concepts of thermodynamics are frequently troubling for students. Although it is easy to quantify ideas such as heat and work, the theoretical underpinnings of entropy, free energy, and voltage are much more difficult to understand. Like many of us, I could see that my students were trying to memorize equations and plug and chug their way through problems, as opposed to obtaining the firm conceptual understanding of the first and second laws that make solving standard problems in thermodynamics so much easier. Ironically, the biggest problems I saw were in making the true connection between thermodynamics and equilibrium, with its application to electrochemistry. Although my students could frequently give me the numerical answer to a question, rarely could they, for example, explain why a galvanic cell s voltage changed with concentration. Thus, I set out on a different path to see if there was a better way to explain these concepts while getting my students to think more critically about thermodynamics and electrochemistry. What follows is an outline of some of the techniques I use to help make these connections more clear To motivate our discussion let us consider two AP style problems that illustrate some of the conceptual problems students have. The first is a modified electrochemistry problem from the 1996 AP exam (I have simply changed the chemical reaction and its associated equilibrium constant; the conclusions drawn can be generalized to any electrochemical reaction), while the second is a thermodynamics problem from the 1991 exam to which I have added a question (part e) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Consider the reaction represented above that occurs at 25 C. All reactants and products are in their standard states. The value of the equilibrium constant, K eq, for the reaction is 1.71 x at 25 C. (a) Predict the sign of the standard cell potential, E, for a cell based on the reaction. Explain your prediction. (b) Predict the sign of ΔG 0 for the reaction. Explain your prediction (c) If the reaction were carried out at 60 C instead of 25 C, how would the cell potential change? Justify your answer. (d) How would the cell potential change if the reaction were carried out at 25 C with a 1.0-molar solution of Zn(NO 3 ) 2 and a 0.10-molar solution of Cu(NO 3 ) 2? Explain. (e) When the cell reaction in (d) reaches equilibrium, what is the cell potential?

2 1991 BCl 3 (g) + NH 3 (g) Cl 3 BNH 3 (g) The reaction represented above is a reversible reaction. (a) Predict the sign of the entropy change, S, as the reaction proceeds to the right. Explain your prediction. (b) If the reaction spontaneously proceeds to the right, predict the sign of the enthalpy change, H. Explain your prediction. (c) The direction in which the reaction spontaneously proceeds changes as the temperature is increased above a specific temperature. Explain. (d) What is the value of the equilibrium constant at the temperature referred to in (c); that is, the specific temperature at which the direction of the spontaneous reaction changes? Explain. (e) If more Cl 3 BNH 3 (g) was added to the system at equilibrium, what would happen to the value of ΔG? Explain Note right away that neither question wants a numerical calculation. Indeed, both of the problems are expected to be answered without a calculator. Thus, to be successful in solving these types of questions, one must conceptually understand the principles of electrochemistry and equilibrium thermodynamics. Interestingly I expect that students would do better on the electrochem question than the standard free energy question, as electrochemistry frequently gets moved to the end of the year simply because it is typically the last chapter in most texts. However, since both questions are really two applications of the same base phenomena, is there a way to easily link the two in the mind of the student, particularly without resorting to calculations? To answer this question, I am going to present some examples I show to my students to help improve their critical thinking skills in thermodynamics while avoiding some of the pitfalls that can easily occur when working problems for the first time. Entropy and Standard Free Energy: I believe a lot of the problem students have in thermodynamics can be traced back to the immediate emphasis we all place on free energy because of the ease at which knowledge of free energy leads to information about reaction spontaneity. However, as the importance of free energy in chemistry is a specialized case of the second law of thermodynamics, it is instructive to deal with some examples utilizing the second law directly before defining and using free energy. Entropy itself is presented in many different ways depending on the current textbook you are using. Of course, most books promote the intuitive relationship between entropy and disorder and present arguments explaining why the entropy of gases > entropy of liquids > entropy of solids, as well as giving students a rationalization for the increase in entropy with temperature. There is a growing trend among texts to use a more probabilistic interpretation of entropy, which is in fact how I introduce entropy in my

3 classes. However, either method seems to give students a decent beginning grasp on the factors that can increase the entropy of a given chemical system. Interpreting the Second Law is the first critical step for students. With our frequent emphasis on the system, we often forget the entropy change of the universe must be positive for a spontaneous process.!s universe =!S system +!S surroundings > 0 To emphasize that both the system and the surroundings play a key role in determining the entropy change of the universe, I frequently do a calculation involving phase change. For example, in the table below, we see how both the system and the surroundings contribute to the entropy change of the universe for the vaporization of water at various temperatures: Entropy Change of the Universe for the Vaporization of Water at Various Temperatures Temp (deg C) ΔS 0 (system) J/K ΔS 0 (surr) ΔS 0 (univ) (J/K) J/K This table is interesting because it predicts that the entropy change of the universe is positive when the temperature is approximately 97 degrees C, which is below the expected value of 100 degrees C. The main problem here is the assumption we frequently make that values of enthalpy and entropy are nearly temperature independent, so that we may take values from 298 K and extrapolate them to, in this case, 373K. The small error in making this approximation (which is repaired when one takes physical chemistry) results in an estimated temperature slightly lower than the experimental value.

4 It is from an example such as this that one can introduce free energy. I define free energy for the class as G = H! TS and then have the class determine the standard formula for ΔG at constant temperature mathematically, obtaining!g =!H " T!S We then graph both ΔS universe and ΔG for the previous problem to prove a point: Figure I: Demonstrating how Entropy and Free Energy change as Water Vaporizes at Different Temperatures It is evident that at the boiling point for this data both the free energy change of the system and the entropy change of the universe are 0. However, where the entropy of the universe is positive, the free energy is negative, and vice versa. Thus, free energy changes in a reverse manner from entropy: If the entropy of the universe increases, the free energy of the system decreases. In addition, we see right away that several standard free energies exist, one for each given temperature. Thus we can immediately define the Standard Free Energy Change for a Process, ΔG 0, as the change in free energy that occurs if a set of reactants in their standard states is completely converted to products in their standard states. Note that the standard state makes no mention of temperature. In fact by definition, The standard state of a gas is the pure gas at a pressure of 1 atm 1. The standard state of a condensed phase is a sample of the pure substance The standard state of a solution is a sample of the solution at a concentration of 1 M. 1 Note that IUPAC has redefined the standard pressure to be 1 bar instead of 1 atm. However, both pressure values are still used regularly and the error in the use of one over the other is negligible in AP Chemistry.

5 Thus, there are an infinite number of ΔG 0 values that can be calculated, depending on the temperature at which the transformation takes place. Even if we assume that the values of enthalpy and entropy changes are temperature independent, the functional form of the definition of free energy makes it a function of temperature. Thus using free energy in place of entropy allows us to focus on the system and still discover if a physical or chemical process will increase the entropy of the universe. To ΔG or to ΔG 0? Although the sign of ΔG 0 for a physical or chemical transformation indicates the spontaneity of change, a little more is needed in order to relate free energy to equilibrium. For what follows let s consider the reaction A(g) B(g), which we know is spontaneous at 298 K. As many students have had significant experience with the ideas of equilibrium before discussing thermodynamics, it is not a shock to recognize that although the reaction happens, it does not go to completion. Based on discussions of K values as well as the entropy of mixtures as opposed to pure substances, it is no surprise that even if stoichiometric amounts of A and B react, there will still be at least a small amount of reactant in the container after equilibrium is reached. Let s use this information for a little thought experiment. Assume we have some amount of pure A and it begins to react to form B. As the reaction proceeds, what must happen to the entropy of the universe? In the figure below, three possible examples are given as the reaction proceeds from A to B. Entropy of Universe A B Figure II: Different Ways the Entropy of the Universe Could Change as the Reaction A(g) B(g) Proceeds

6 If we consider each path we can easily eliminate two of them. Since we know the reaction between A and B occurs, path 3 is not possible, because it violates the second law by allowing the entropy to decrease as the reaction proceeds. In addition path 2 cannot occur either because it says that the entropy would increase without bound until the reaction mixture only contained pure B. If this were the case the equilibrium constant would be infinite, which counters our experimental and problem solving evidence relating to chemical reactions. Thus, path 1 must give a general description of how the entropy of the universe changes with our spontaneous reaction; the reaction will proceed in the forward direction as long as the entropy of the universe increases. Once we reach that maximum, the reaction stops. More specifically, we have reached equilibrium! Since we have a general idea of how the entropy of the universe changes as the reaction proceeds, what happens to the free energy of the system during the course of the reaction? As free energy and entropy are inversely related, the graph must look something like Figure III. Figure III: Free Energy Profile of the Spontaneous Reaction A(g) B(g). Note the Position of Equilibrium as well as the Value of ΔG 0 for the Reaction. As the reaction proceeds from pure A towards pure B, the free energy must lower, eventually reaching a minimum when the system reaches equilibrium. The question becomes how can we use this graph to obtain experimental information about the equilibrium position? Consider Figure IV, which again shows the free energy profile of the reaction at which the slope of the free energy curve has been determined at three points.

7 Figure IV: The Green, Blue, and Red Arrows Illustrate the Slope of the Tangent Line to the Curve. Note the Slope is 0 at the equilibrium position of the reaction. From physical chemistry, the value of this slope can be shown to be: slope =!G 0 + RT lnq where ΔG 0 is the standard free energy change of the reaction, and Q is the reaction quotient commonly used to determine if a combination of amounts of reactants and products satisfy the equilibrium constant. As a reaction continues to proceed spontaneously, the value of the slope will be an increasing negative number, eventually reaching a value of 0 when the free energy function reaches its minimum, and the system reaches equilibrium. At this point, the ratio of the pressures of B to A will be at their equilibrium values. Thus, we arrive at perhaps the most important equation in thermodynamics:!g 0 = "RT ln K which allows the determination of equilibrium constants from free energy data and viceversa. Unfortunately, the equation for the slope above often gets a confusing label. Instead of denoting a differential change, this equation is frequently written as!g = slope =!G 0 + RT lnq As can be seen, one cannot tell that the right side of the equation equals the slope of the free energy vs. reaction coordinate graph, instead implying a physical difference between two unspecified values of the free energy. Thus, one can see why is it so important to

8 remember the superscript to clearly denote if a difference in standard free energies or a differential change is being reported. Note that we now have a powerful conceptual framework in which to answer questions like 1991 on page 2. Using the combination of the formula for the standard free energy change as well as having a mind s eye view of what the graph of free energy looks like as the reaction proceeds allows us to immediately answer questions like part e. Since the reaction composition is now past equilibrium, the slope of the curve, and hence ΔG, must be positive. In addition, if the value of ΔG 0 is 0, than the only way the slope of the free energy curve can also be 0 is if the equilibrium constant has a value of 1. This last point is one that can really confuse students because if both ΔG and ΔG 0 are interpreted as differences, it is hard to see how they can both be zero at the same time. But upon understanding the slope of the function depends both on the standard free energy change as well as the composition of the mixture makes these questions much easier. The Relationship to Electrochemistry To easily understand how the concepts of electrochemistry easily fit into the discussion, lets consider the reaction of a zinc bar with a solution of copper(ii) sulfate. The balanced chemical equation for the reaction is: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) If one places a zinc bar into a solution containing copper(ii) ion, one immediately sees the formation of copper on the bar. Thus this reaction must be spontaneous. Of course calculations back up the observation: Table I: Thermodynamic Parameters of the Reaction Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) At 298 K ΔH 0 (J) ΔS 0 (J/K) ΔG 0 (298 K) Consider for a minute why this reaction occurs in terms of the second law of thermodynamics. Not surprisingly, the entropy change of this reaction is small since we are essentially switching forms of solids and aqueous solutions. However, the change in entropy of the system is negative, hence the reaction must release some energy into the environment to increase the entropy of the surroundings. What is interesting is if you place the zinc bar in the copper solution, all the enthalpy of the reaction enters the environment as heat. Therefore the entropy change of the surroundings would be q/t = J / 298 K = 734 J/K. Thus number is much greater than what is required by the second law, as in the case of all of the enthalpy of the reaction creating entropy in the surroundings, the total entropy change of the universe would be 713 J/K. Of course not all of the available enthalpy needed to go into heat to satisfy the second law. In fact we can back calculate how much enthalpy must enter the environment to satisfy the universe.

9 q 20.98J / K = 298K q = heat needed to be released = 6252 J!H "left over" = J J = J But the ΔH left over is just the free energy! This calculation immediately gives a conceptual interpretation of ΔG 0 ; the energy is free because it is the energy available to do something useful after ensuring that the second law has been satisfied. The problem is extracting this available free energy, as simple combination of the reactants just releases all the available enthalpy as heat. Thus, a perfect lead-in to electrochemical cells results as providing a method by which this free energy can be extracted from the reaction. The fundamental equation describing the voltage (or the energy per electric charge) of a voltaic cell is the Nernst equation. E = E 0! RT nf lnq To gain some insight into this relationship, consider some of the things we know about the copper-zinc cell (which are applicable to any voltaic cell). When are species are at their standard states, the voltage of the cell is 1.10V When the cell reaches equilibrium, the voltage is 0 Let s use this information along with a hypothetical graph of the free energy profile for the zinc-copper cell to extract useful graphical information about the voltage. Figure V: Hypothetical Free Energy Profile of the Copper-Zinc Voltaic Cell. The slope of the graph at each point is proportional to the cell voltage at those concentrations.

10 Recall at any point, the slope of the graph = ΔG 0 + RTlnQ. Thus, when the system is in its standard state, the slope is simply the numerical value of ΔG 0, and the slope of the graph must be 0 when the system is in equilibrium. Thus, the slope of the graph itself is proportional to the voltage of the system. Indeed, taking the slope and dividing it by a constant (in this case -nf) yields the voltage of the galvanic cell at the specific concentrations of reactants and products represented by that point of the graph. This implies that the voltage of a cell is an indication of the distance that system is from equilibrium, distance here meaning that a high voltage implies much reactant will have to react to reach equilibrium. The graphical interpretation of voltage gives students an easy conceptual way to visualize the Nernst equation. If one knows that the voltage of a nonstandard state is greater than E 0, it is clear that the system is further from equilibrium than the standard system, or that the value of Q < 1. As this system reacts and more product is made, the voltage will continually drop until the system voltage reaches 0 at equilibrium. Thus, it is the voltage of the system that drives a galvanic cell to equilibrium. The further the system is from the equilibrium state the more free energy it is willing to release to get there. Returning to the 1996 AP problem, it is seen that parts a,b,d, and e are immediately obvious from the graph above. Part c, however, is more difficult. Many students would argue immediately that an increase in temperature would cause the voltage of the system to drop because of the T in the Nernst equation. However, we must remember that we are at the standard state, and the voltage, and thus the slope, is only dependent on ΔG 0 at that point. The problem is that ΔG 0 itself is of course temperature dependent, as we have seen with the discussion of the vaporization of water. The student must then be able to discern how ΔG 0 changes with temperature, which realistically requires knowledge of ΔH 0 and ΔS 0 for the reaction. Although one could surmise that this reaction must be exothermic (as it is spontaneous and one would expect the entropy of the reaction to be small considering the states of the reactants and products), this really is not a fair question given the information at hand. A calculation does show that the standard voltage of the system does decrease with increasing temperature. With the graphical ideas developed in this handout, conceptual problems in free energy, equilibrium and electrochemistry become much simpler for students as a common picture binds these ideas. Based on knowledge of the stoichiometry of the reaction plus the equilibrium constant or ΔG 0, students can easily make qualitative predictions on the voltage of systems of arbitrary concentration. This should place some of the more abstract concepts of thermodynamics on more solid ground, allowing students to gain more insight into the forces that ultimately drive chemistry as well as provide insightful answers to important conceptual questions.