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Helmholtz and Gibbs Energy Examples

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Helmholtz and Gibbs Energy Examples 1 INITIAL DEMONSTRATION 1 Helmholtz and Gibbs Energy Examples These are the examples to be used along with the powerpoint lecture slides. The problems are numbered to match the tags in the the lower left hand corner of the powerpoint slides. The numbers of the examples are # the in the HG EX# tags on the slides. 1 Initial Demonstration This first example takes liquid sodium acetate at room temp and with a small perturbation it crystallizes with a release of heat. Here is a nice demonstration of a clear increase in order (∆S < 0) but the reaction clearly proceeds spontaneously. Have a student make note of the temperature, which has increased. The example here is the building of a mountain of a solid from a stream of liquid and the hand warmers. You can also use 5 minute epoxy for an example, where you take a number of smaller molecules and make one very big molecule (seems like a loss of dissorder?), again the student should notice that the mixture warms up. Less order (entropy), but still goes spontaneously and there seems to be a release of heat to the surroundings... Perhaps there is a balance between the need to maximize disorder and minimize the energy. 2 Calculating the entropy of the universe? That's a lot of work. • The First Law gave us energy balance. • The Second Law gave us a criterion for a spontaneous process, ∆S ≥ 0. This is great, but the change in entropy here is the TOTAL change and this means we must determine the change for both the system and the surroundings. Calculating the change in a state function for the Universe is not always an easy task, so we would like to consider the criteria for a spontaneous process from only the properties of the system. In fact, we would like to define a thermodynamic state function in terms of the system alone that would provide this information. The total change in entropy is simply the sum of the entropy change for the system and the entropy change of the surroundings (system + surrounding = everything), dStot = dSsys + dSsur 2 CALCULATING THE ENTROPY OF THE UNIVERSE? THAT'S A LOT OF WORK.2 Consider a system at constant T . Since it is at a constant temperature it must be interacting with the environment via heat exchange to maintain the same T , ie. it is not isolated. Now consider a process that occurs in the system that results in the transfer of an infinitesimal amount of heat, δq, from the system to the surrounding. So the change in the entropy for δqsur the surrounding is defined as dSsur = T , and the total entropy change is now, δq dS = dS + sur tot sys T Our convention is that heat transferred from the system to the surrounding is negative, and we note that the amount of heat transferred to the surrounding must be equal and opposite to the amount of heat transferred from the system (it is the same heat), −δqsys = δqsur Using this relationship, δq dS = dS − sys tot sys T This is now an expression for the total change in the entropy (change in the Universe) in terms of only the system! Just what we wanted. Look at a constant pressure process, δqsys = dHsys From the Second law we know that, dStot ≥ 0 Substitute our result in terms of the system for the total entropy change, dH dS = dS − sys ≥ 0 tot sys T Multiply by T on both sides, T dSsys − dHsys ≥ 0 Move things to the other side (which means flip the ≥ and change the signs), dHsys − T dSsys ≤ 0 INTERESTING, we have a criterion for spontaneity in terms of the system alone. Define a new state function, G = H − TS 3 MELTING AND BOILING OF WATER 3 Note that this is a state function since it is a combination of other state functions. For our system at constant temperature, dG = dH − T dS − SdT = dH − T dS (constant temperature) And this is exactly what we have above (dHsys − T dSsys ≤ 0), dG ≤ 0 (<)spontaneous; (=)equilibrium Consider ∆G for a general isothermal process, A ! B, ∆G = ∆H − T ∆S ∆G ∆H = − ∆S T T where the first term on the rhs is the entropy from exchange with the surroundings (increase thermal disorder in the surroundings), and the second term on the rhs is the entropy produced in the system. 3 Melting and boiling of water (Taken from McQuarrie and Simon pgs. 305-306 Gold or 885-886 Red) Consider a process where the sign of ∆G changes sign over a very small change in tempera- ture. Consider the vaporization of liquid water, H2O(l) −! H2O(g) The molar Gibbs energy of vaporization is expressed, ¯ ¯ ¯ ∆vapG = G[H2O(g)] − G[H2O(l)] ¯ ¯ = ∆vapH − T ∆vapS Near 100◦C and at one atm, ¯ −1 ∆vapH = 40:65 kJ · mol ¯ −1 −1 ∆vapS = 108:9 J · K · mol So we have, ¯ −1 −1 −1 ∆vapG = 40:65 kJ · mol − T (108:9 J · K · mol ) 4 NON-PV WORK: ELECTROLYSIS OF WATER 4 And at 373.15 K we have, ¯ −1 −1 −1 ∆vapG = 40:65 kJ · mol − (373:15 K)(108:9 J · K · mol ) = 0 ¯ Notice that ∆vapG = 0 shows us that liquid and vapor are in equilibrium, as expected at 373.15 K. At 373.15 K and 1 atm the Gibbs energy of the liquid and vapor are equal and the transfer from liquid to vapor is a reversible process. Now consider a temperature just below the boiling point. At 363.15 K, ¯ −1 −1 −1 ∆vapG = 40:65 kJ · mol − (363:15 K)(108:9 J · K · mol ) = 1:10 kJ · mol−1 ¯ The positive value of ∆vapG indicates that the formation of water vapor from liquid water is not spontaneous at this temperature and 1 atm. Now consider a temperature just above the boiling point. At 383.15 K, ¯ −1 −1 −1 ∆vapG = 40:65 kJ · mol − (383:15 K)(108:9 J · K · mol ) = −1:10 kJ · mol−1 ¯ The negative value of ∆vapG indicates that the formation of water vapor from liquid water is a spontaneous process at this temperature and 1 atm. *You can check to see that the same thing happens for the melting of water given that at 273.15 K (example 8-1 on pg 306 Gold or 22-1 on pg 886 Red in McQuarrie and Simon), ¯ −1 ∆fusH = 6:01 kJ · mol ¯ −1 −1 ∆fusS = 22:0 J · K · mol 4 Non-PV work: Electrolysis of water (This is from example 8-2 on pg 307 Gold or 22-2 on pg 887 Red of McQuarrie and Simon.) You may have heard a lot of talk about the hydrogen economy. The idea is to burn hydrogen for energy, which is nice and clean since it leaves you with water, 1 H (g) + O (g) −! H O(l): 2 2 2 2 4 NON-PV WORK: ELECTROLYSIS OF WATER 5 The next question should be, \Where do I get the hydrogen?" The answer is usually that you can use electrical power (hopefully produced in a clean way, waves, sun, :::) to split water. This means that we turn the equation around and provide electrical work to run the reaction, electrolysis. 1 H O(l) −! H (g) + O (g) 2 2 2 2 So, your boss at the water splitting plant asks you, \What is the minimum voltage required to split water into hydrogen and oxygen?" (Note at constant T and P , such as 298.15 K and 1 bar) First, you remember the definition of electrical work, work = charge × voltage and then you remember that you already know how to calculate the electrical (non-PV ) work for a reaction, it's just the Gibbs energy by definition. So you use the tables to calculate ∆rH and ∆rS, and using these you calculate, ¯ ∆G = wnonPV (reversible, constant T and P ) ∆G¯ = 237:1 kJ · mol−1 (reversible, T = 298:15 K and P = 1 bar) Now to get the voltage all you need is the charge. The oxidation state of hydrogen goes from +1 to 0 and that of oxygen goes from −2 to 0. This means two electrons are transferred for every H2O(l) molecule. That is two time Avagadro's number of electrons per mole. The total charge of two moles of electrons is, total charge per mol = 2(1:602 × 10−19 C · e−1)(6:022 × 1023 e) = 1:929 × 105 C So we can now calculate the minimum (reversible) voltage required, ∆G¯ ∆G¯ = w = charge × volts −! volts = non-PV charge Use the fact that 1 J = 1 CV And we have, 237:1 × 103 J · mol−1 volts = = 1:23 volts 1:929 × 105 C · mol−1 5 ENTROPY FROM PVT EQUATION OF STATE 6 5 Entropy from PVT equation of state (This is example 8-3 on pgs 309-310 Gold or 22-3 on 889-890 Red of McQuarrie and Simon) ¯ ¯ ¯ Calculate ∆S for an isothermal expansion from V1 to V2 for a gas that obeys the equation of state, P (V¯ − b) = RT Using our derived expression we have, ¯ Z V2 @P ∆S¯ = dV¯ V¯1 @T V¯ ¯ Z V2 dV¯ = R ¯ V¯1 V − b ¯ V2 − b = R ln ¯ V1 − b Notice that we got this result back in chapter 6, but we had to know that dU = 0 for an isothermal process that followed the given equation of state. That information was not required here. 6 Correction for non-ideality This is adapted from pgs 899-910 in Red McQuarrie and Simon Remember when we calculated S for diatomic nitrogen there was a \correction for non- ideality" added to the total.
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