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Chem 141 Lectures 33- Lecture 33 So far when we have talked about equilibrium, we have learned how to figure out equlibrium compositions from equilibrium constants, but at this point we haven’t even speculated about why in one reaction reactants are favored, but in another products are favored. In other words, when will any chemical reaction strongly favor products, i.e., have a large equilibrium constant, and when will it favor reactants , i.e., when will the equilibrium constant be small. In addition, at various points we've made qualitative statements, mostly through the use of LeChatelier's principle, about the effect of changing reaction conditions, such as pressure or temperature, on a chemical equilibrium. Is there any way to calculate the change in the value of an equilibrium constant when the temperature, for example, is changed? There is a subject called chemical thermodynamics that can answer these questions quantitatively. Thermodynamics can tell us two things that are extremely important in chemistry. The first of these is which processes will occur spontaneously , i.e., without doing work on the system. Whether a solution will form when two substances are mixed is one example of a question which thermodynamics can answer. The second thing which thermodynamics tells us is what the conditions are for a system to be in equilibrium. In this chapter, we will concentrate on understanding when a process will be spontaneous, i.e, will occur without doing work on the system, and practice calculations with some special functions of thermodynamics. In order to predict the things that we want to know, it is necessary to describe
our system precisely. In general, this means knowing the temperature, pressure, and mole number of each of the components of the system. These variables taken together are called the state of the system. For example, suppose I have a 5.00 mol sample of Ar(g) at 298K and 1.00 atmosphere pressure. This sample is in a different state than a 2. mol sample of Ar( l ) at 30K and 4.00 atm pressure. All the properties of the system will depend on the state of the system. One of the reasons that states are so important in thermodynamics is that the functions that we will use to predict spontaneity are a special type called state functions. What makes state functions special is that a change in a state function depends only on the initial state and the final state, and not on the path you take from one state to another. For example, last week when we talked about the enthalpy change for a reaction it was given by Hrxn = Hproducts - Hreactants. An example of this is that for the
reaction C(gr) + 2 H 2 (g) CH 4 (g), Hrxn = HCH4(g) - 2 HH2(g) -HC(gr) no matter how many
steps we take to get from our reactants to our products. We also have some names for the conditions under which we run our experiments. An isothermal process is one in which we keep the temperature constant. An isobaric process is one in which the pressure is kept constant. Finally, if we run our experiment so that no heat can flow into or out of our system, then the processes that occur are called adiabatic processes. All of the conclusions that we draw from thermodynamics are based on three laws. We dealt in some detail with the first law of thermodynamics last week. The first law states simply that “The energy of the universe is constant”. To reiterate, the importance of this law is that no process is possible if energy must be created or
natural tendency toward disorder in the way that an abandoned building gradually decays. Since entropy is a measure of disorder, the greater the entropy of a system, the greater the disorder of the system. Let’s look at some systems of chemical interest to see which has greater disorder and therefore greater entropy. As the first case let’s consider, a solution vs. the solute and solvent that make up the solution. At first we have only solute next to solute and solvent next to solvent. When the solution is formed, the solute is dispersed throughout the solvent. WHICH IS MORE DISORDERED? Therefore the entropy of a solution is greater than the entropy of the separated solute and solvent. We symbolize the entropy as S. Entropy is a state function, and therefore we can calculate the entropy change for any process by taking the difference between the initial and final entropies , i.e., S = Sf - Si. For the process of solution, the change in entropy would be
S = Ssolution - Ssolvent - Ssolute. Since the entropy of the solution is higher, we can conclude that Ssolution > 0. Another example to consider would be the entropies of the solid, liquid and vapor phases of the same chemical. In the solid, each particle is assigned a fixed position, and everything is regularly spaced. In the liquid, there is a little more space between particles, and the particles can move around each other. WHICH IS MORE DISORDERED, THE SOLID OR THE LIQUID? THEREFORE, WHICH HAS THE HIGHER ENTROPY, THE SOLID OR THE LIQUID? Now let’s consider the gas. In the gas, the particles have a great deal of space between them and can move around with almost complete freedom. WHICH HAS THE HIGHEST DEGREE OF DISORDER, A SOLID, LIQUID, OR GAS? WHICH HAS THE HIGHEST ENTROPY? Suppose we freeze a gas. The process can be written gas solid. The
entropy of freezing will be given by Sfreezing = Ssolid - Sgas. Since the entropy of the solid is less than the entropy of the gas we conclude that Sfreezing < 0. WOULD SVAPORIZATION BE POSITIVE OR NEGATIVE? Heating an object can affect the amount of disorder in the system. Remember that temperature is a measure of kinetic energy, the energy of motion. In other words, the higher the temperature of something is, the more its particles are moving. Let’s figure out the effect of motion on disorder by considering the two extreme cases. In one case we'll consider a group of 9 particles which are not moving at all. Now consider a group of particles that are the same distance apart as the first group, but are whizzing around. Clearly, the group which is stationary is more ordered. Since the only difference between the two groups of particles is motion, we conclude that increasing motion increases entropy. Since heating a system increases the amount of motion, heating the system also increases the entropy. We symbolize this by saying Sheating > 0. Let me ask you a question. WHICH HAS GREATER ENTROPY, WATER AT 350K OR WATER AT 300K? Finally let’s consider a couple of chemical reactions. First let’s look at the reaction H 2 (g) 2 H(g). For simplicity let’s look at just four H 2 molecules. At first all the hydrogens are combined and can only move in pairs. If we break the bond of all four hydrogens, we now have eight hydrogen atoms, all of which can move by themselves. WHICH IS MORE DISORDERED? WHICH HAS GREATER ENTROPY? The entropy change for the reaction is Srxn = 2 SH - SH2. Since two H atoms have a greater entropy than one H 2 , Srxn > 0 for
system gets more and more ordered as we move toward the solid, the entropy keeps decreasing. Now consider a solid at a given temperature. Remember that the lower the temperature is the lower the entropy is. Now if we have a solid at 0K, we have the lowest possible temperature and therefore the lowest possible entropy. It turns out that if we have a perfect crystal of a pure compound, then this lowest possible entropy is 0. The result of this is that by starting with pure crystals at 0K, we can figure out the entropy of any substance at any temperature. These entropies depend on external pressure, so to simplify things, they are tabulated at one standard pressure, one atm. These one atmosphere entropies are called Standard Entropies , and are symbolized by S. As we've said before, entropies also depend on temperature, so when we report a standard entropy, we must specify the temperature. So for example, the standard entropy for hydrogen gas at 298K would be written as S 2980 (H 2 (g)).
Just as we used standard heats of formation to calculate standard heats of reaction
with the equation Hrxn^0 = Hf(products) - Hf(reactants), we can use our absolute
entropies to calculate entropies of reaction with Srxn = S (products) - S (reactants). Lets practice on the two reactions we've already mentioned. For the reaction
H 2 (g) 2 H(g), our standard entropy of reaction at 298K will be Srxn,298 = 2 * S 298 (H(g)) - S 298 (H 2 (g)). If we look in a table of standard entropies, we find that S 298 for H 2 (g) is 130.
J/mol K, and S 298 for H(g) is 114.6 J/mol K. Therefore our entropy of reaction is
Srxn, 298 = 2 * 114.6 - 130.6 = 98.6 J/mol K. The reason we use 2 times the entropy of the H atoms is that we are taking the difference of the total entropy of our products from the total entropy of our reactants. If we write our reaction as H 2 (g) H(g) + H(g) we see that the total entropy of the products is the entropy of 2 H atoms. As a second example we'll calculate the entropy of the reaction 2 H 2 (g) + N 2 (g) N 2 H 4 ( l )
at 298 K. Our entropy will be given by S 298 = S 298 (N 2 H 4 ( l )) - S 298 (N 2 (g)) - 2 * S 298 (H 2 (g)).
We already know the standard entropy of H 2 (g), and our table tells us that S 298 (N 2 H 4 (l)) = 121.2 J/ mol K and S 298 (N 2 ) = 191.5 J/ mol K. Therefore the reaction entropy is
Srxn,298 = 121.2 - 2*130.6 -191.5 = -331.5 J/ mol K. Do these entropy changes we've talked about correctly predict reaction spontaneity? Let’s make a list of some of our processes and the signs of their entropy changes. For the reaction H 2 (g) 2 H(g), S was positive. For Svaporization, S was
positive, as was Ssolution. We know that H 2 will form spontaneously from H atoms, that at low enough temperature condensation will be spontaneous, and that for some substances solutions will not form. Yet the second law says that the entropy of the universe increases for a spontaneous process. The answer to our problem lies in the word universe. Remember that the universe is system + surroundings, so that when we say the entropy of the universe is increasing we write Ssystem + Ssurroundings > 0 for a spontaneous process. All of the entropies that we have calculated so far have been entropies of the system alone. In order to make a prediction about spontaneity we need to
equilibrium. When we make a reaction mixture, G depends on the concentrations of all of our
reactants and products, i.e., G depends on concentration. Since as the reaction proceeds the concentrations change, G changes as the reaction proceeds. In general
as a reaction proceeds, it approaches a state of dynamic equilibrium, so G tends to approach 0, as the reaction goes on. We define the standard free energy, G , as the free energy difference which accompanies the conversion of reactants in their standard states to products in their standard states. We can calculate these using standard free energies of formation, G , which are defined as the free energy change when a mol of a substance is formed from its reactants in their reference states. We usually can look up free energies of formation in tables. In addition, just as the heat of formation of elements in their stablest states is defined to be 0, Gf of elements in their stablest states is defined to be 0. For example, since hydrogen's stablest form at 298K is H 2 (g), Gf,298(H 2 (g)) = 0. WHAT IS THE FREE ENERGY OF FORMATION OF HG(L) AT 298K? We calculate G of reaction with the rule Grxn = Gf (products) - Gf (reactants). For example, for the reaction
2NH 3 (g) + 7/2 O 2 (g) 2NO 2 (g) + 3 H 2 O(g),
Gf(NH 3 (g)) = -16.64 kJ/mol, Gf(NO 2 (g)) = 51.84 kJ/mol, and Gf(H 2 O(g)) = - 228.60 kJ/mol. WHAT ISGF(O 2 (G))? For this reaction,
Grxn = 2 x Gf(NO 2 (g)) + 3 x Gf(H 2 O(g)) -2 x Gf(NH 3 (g))
= 2 x (51.84) + 3 x (-228.60) - 2 x (-16.64) = -548.84 kJ/mol. When we introduced the Gibbs Free Energy, we defined it by G = H - TS. This equation is important because it tells us about reaction spontaneity and about
chemical equilibrium. To reiterate, if we have a reaction like N 2 + 3H 2 2NH 3 , and G
is < 0, the forward reaction, formation of ammonia, will occur spontaneously. If G > 0, the reverse reaction, decomposition of ammonia, will occur spontaneously. Finally, when G = 0, we will have an equilibrium between reactants and products. Our definition of G shows that we can calculate standard free energies from standard enthalpies, and standard entropies, i.e., at 298K, G 298 = H 298 - TS 298 . For example, for the reaction C 2 H 4 C 2 H 2 + H 2 , Hrxn,298 = Hf,298(C 2 H 2 ) + Hf,298(H 2 ) - Hf,298(C 2 H 4 ). Looking in any table of thermodynamic data, we find that the standard heats of formation at 298 of C 2 H 4 and C 2 H 2 are 52.26 kJ/mol and 226.73 kJ/mol respectively. DOES ANYONE REMEMBER WHAT THE STANDARD HEAT OF FORMATION OF H 2 IS? So Hrxn,298 = 226.73 - 52.26 kJ/mol = 174.47 kJ/mol. For this reaction Srxn,298 = S 298 (C 2 H 2 ) + S 298 (H 2 ) - S 298 (C 2 H 4 ). HOW DO WE FIND ABSOLUTE ENTROPIES? When we look them up we find that they are 200.94, 130.684 and 219.56 J / mol K respectively and therefore Srxn,298 = 200.94 + 130.684 - 219.56 = 112.06 J / mol K.
important equations involving G. This equation is G = G + RT ln Qact.
In this equation G is the standard free energy of reaction, R is the gas constant, T is the temperature in K, ln stands for the natural log function, and Q is the reaction quotient calculated with activities. It is important to reiterate that in this equation we must use Qact, and not for example Qp. Another thing we need to note about this equation is the units of the gas constant. Notice that since our G's are in units of kJ/mol, it is inconvenient to use the usual gas constant units of L atm/mol K. Fortunately, L-atm, which is a unit of energy, can be expressed in J. If we do this the gas constant becomes 8.314 J/mol K. Let’s use this equation to calculate G for a reaction where we start with two different sets of concentrations of our reactants and products, and see how these different concentrations affect G and therefore the spontaneity of the reaction. For the reaction
H 2 (g) + I 2 (g) 2HI(g)
at 298K, pH2 = .2 atm, pI2 = .2 atm and pHI = .5 atm. Remember that G = G + RT ln Q,
so in order to calculate G we need to know G and Q. G = 2Gf(HI) - Gf(H 2 (g)) - Gf(I 2 (g)). If we look these up we find that Gf(HI) = 1.3 kJ/mol, and Gf(I 2 (g)) = 19. kJ/mol, and since H 2 (g) is an element in its stablest form, Gf(H 2 (g)) = 0. Combining these yields Grxn = 2*1.3 - 19.3 = -16.7 kJ/mol.
Q (^) aa (^) H^ HI aI (^) p (^) H patm^ HI atmp (^) I atm (^) (^2 2 ) (^2 2 2 )
b g
d id i
Combining these yields G = -16.7 kJ/mol + 8.314 J/K mol * 298K * ln 6.25 * 1 kJ/1000J = -12. kJ/mol. Since G < 0, the forward reaction, formation of HI, will be spontaneous. What if our initial partial pressures were pH2 = .05 atm, pI2 = .025 atm and pHI =
2.0 atm? Now Q (^) 0 5. ^2 2 0 25. 3200 , and G = -16.7 kJ + 8.314 J/K mol * 298K * ln
3200 * 1 kJ/1000J = 3.2 kJ/mol. SINCE G > 0, WHICH WILL SPONTANEOUSLY OCCUR, FORMATION OF HI, THE FORWARD REACTION, OR DECOMPOSITION OF HI, THE BACKWARD REACTION? So we see that by using standard free energies and the reaction quotient, we can predict which direction of a reaction will be spontaneous, the forward reaction or the reverse reaction. Lecture 35 Lets continue looking at the equation G = G + RT ln Qact, and see what else we can learn from it. Lets consider a special case that is of tremendous importance in chemistry. Remember that when G = 0, our system is in equilibrium. If we plug a
value of 0 for G into our equation we get the equation 0 = G + RT ln Kact, where Kact, the equilibrium constant, is the value of Qact at equilibrium, i.e., Kact = (Qact)eq. We can rewrite this as G = -RT ln Kact,
Why don't you all run this through your calculators and tell me what this comes out to? [846]. WHAT DOES THIS TELL US? [Products are favored.] In this case G < 0 and products dominate our equilibrium mixture. Now lets just try other values of G and see what values of K we get. Why don't
you suggest some values of G and we'll figure out what K is. Lets plug this into K = exp(-G/RT). (Get 3 or 4 examples. Prompt for both positive and negative G, interpret using Q to show reactants or products favored.) Our conclusion is that a positive G yields a K < 1 and therefore favors reactants, while a negative G yields K
1 and therefore favors products. Now we can understand the difference between G and G. G depends on concentration. It changes its value as the reaction proceeds. We can tell whether a given reaction mixture will change spontaneously by calculating the value of G. If G is
positive, the forward reaction will be spontaneous, while if G is negative, the reverse reaction will be spontaneous. G, which is defined only at one pressure, 1 atm, or one concentration, 1 M, is a constant at a given temperature for each reaction. It does not depend on composition, and does not change as the reaction proceeds. Therefore it does not predict spontaneity. What G does tell us is whether the reactants or products will be favored when the reaction reaches equilibrium. The rule is if G > 0, then reactants will dominate. If G < 0 then products will dominate. Finally if G = 0 neither reactants nor products will be favored. We've just shown how to calculate Kact from G by using Kact = e-G/RT. Let’s
see how to calculate G from K. Let’s use the reaction
H 2 (g) + I 2 (g) 2HI(g)
as an example. Suppose we begin with a mixture at 699K in which pH2 = .640 atm, pI2 = .571 atm, and PHI = 0. When the reaction reaches equilibrium, the pressures are pH2 = .167 atm, pI2 = .0980 atm and pHI = .946 atm. The activities will be .167, .0980 and. respectively, and Kact = .946^2 /.167 * .0980 = 54.7. Now that we know Kact, we use G = -RT ln K = -8.314 J/K mol * 298 K * ln 54.7 = -9,910 J/mol = -9.9 kJ/mol. Lets try the same game we did for G, and suggest values of Kact, and figure out what G is. [Prompt for values of Kact greater than and less than 1]. There are two ways that temperature effects the equilibrium constant. The first way in which temperature affects the equilibrium constant is shown by the equation, K = exp(- G /RT). If G > 0, K will increase when the temperature increases. For
example consider a reaction with G 298 = 15 kJ/mol. At 298 K the equilibrium constant is given by Kact = exp(-G/RT) = exp(-15000J/mol/(8.314 J/K mol)* (298K) = 2.35 x 10-3. Now consider the same reaction at 398 K. At this temperature Kact = exp(-15000J/mol)/(8.314J/K mol)(398K) = 1.07 x 10-2. Even though the reaction still favors reactants, shifting the temperature has moved the equilibrium more toward the product side. So we see that temperature effects equilibria. What happens when G is negative? Suppose G is -12 kJ/mol at both 298K and 398K. HOW DO WE CALCULATE KACT AT 298? WHAT IS KACT? (126.9) WHAT IS KACT AT 398K? (37.6) Notice that when G is negative and we increase the temperature, that the equilibrium shifts toward the reactant side. The other reason that the equilibrium constant changes with temperature is
H is negative and S is negative, products will be favored at low temperatures and reactants will be favored at high temperatures. Finally, if H is positive and S is
negative, reactants will always be favored, while if H is negative and S is positive, products will always be favored. So we see that we can use thermodynamics not only to predict whether reactants or products will be favored in a chemical reaction, but in addition and powerfully, whether we can use temperature to change that. Notice that we have two distinct sources for the way temperature affects equilibrium, the change in G given by G H 298 -TS 298 , and the change due to
the equation K = exp(-G/RT). It would be useful if we could develop one equation that combines both of these effects. We can do this by first taking the ln of both sides of our second equation to give, ln K = -G/RT. Now we substitute H - TS for G to give ln K = - (^^ H -T S^ RT^ ^^ ^ )= - RT H^ ^ + RS .
This is an equation for a straight line, y = mx + b, with y = ln K, m = H/R, x = 1/T, and b = S/R. If we plot ln K vs. 1/T we get a straight line. The slope of a straight line is the change in y over the change in x so we can write,
2 1 2 1
ln ln 1 1
K K H
R
T T
We can rewrite this as
ln ( KK^21 )= -^ RH^ ( T^1 2 - T^1 1 ).
We can get the other useful form of this equation by taking the exponential of both sides and multiplying it out to get
K 2 (^) K e 1 ^ ^ HR^^ ^ T^12^^ T^11
This gives us the temperature dependence of the equilibrium constant. We only need to know the original equilibrium constant and H. This equation is particularly useful,
because H changes very slowly with temperature, so as long as the temperature change isn’t too large, we can assume that H doesn't change when the temperature changes.
Let’s try using this a couple of times. [Someone give me K 1 , H > 0, T 1 and T 2 ] Let’s plug these into our equation. [What is the new K?] Let’s try this for another K 1 , T 1 , and T 2 , but this time with H < 0. So we see that for H > 0, as T increases the equilibrium constant increases, i.e., favors products more. In contrast, if H < 0, as T increases the equilibrium constant decreases.
