# Thermodynamics. When a gas expands, the energy of the gas molecules is spread out more widely.

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1 Thermodynamics 7a Thermodynamics is a funny subject. The first time you go through it, you don t understand it at all. The second time you go through it, you think you understand it except for one or two small points. The third time you go through it, you know you don t understand it, but by that time you are so accustomed to it that it doesn t bother you anymore. -Arnold Sommerfeld 1. Enthalpy Chemical measurements are usually carried out in open containers, where the pressure is constant. If the reaction pulls air into the flask, work is being done on the system by the surroundings. If gas goes from the flask to the surroundings, work is being done by the system. We can still measure the amount of heat given off or taken in by the reaction, but it is no longer equal to the change in the internal energy of the system. Some of the heat is converted to work. We get around this problem by introducing a term called enthalpy (H). Enthalpy is the sum of the system's internal energy plus the product of pressure and volume. Thus: H = E + PV The change in the enthalpy of a system is equal to the change in its internal energy plus the change in the product of the pressure times the volume of the system. ΔH = ΔE + P ΔV Combining this equation with the first law of thermodynamics, ΔE = q - P ΔV, we end with: ΔH = q That is, the change in enthalpy will be equal to the heat flow. This will prove useful later on. 2. Entropy and the Second Law If the energy in the Universe remains constant, if energy cannot be consumed, then what drives chemical reactions? It is the dispersal of energy. Reactions proceed when they allow energy to spread out. This tendency of energy to disperse is measured by something called "entropy." When a hot object cools, the heat energy which was concentrated in the object spreads to the molecules air and table molecules, for example which touch the hot object. When you drop an object its potential energy is converted to kinetic energy which then changes very quickly to thermal energy when it hits the ground. When a gas expands, the energy of the gas molecules is spread out more widely. When ice melts, thermal energy from a (warm) external source spreads into the ice and breaks the bonds holding the water molecules in their crystal lattice. Even a spontaneous, exothermic reaction, a case where entropy doesn't seem necessary, is still a case where energy is spread out. Heat is transferred from the bonds of reactant to thermal energy in the system and its surroundings. -1-

2 The Second Law of Thermodynamics states that in any spontaneous process, the entropy of the Universe increases. All of the above processes are spontaneous (or reversible) since energy is being dispersed. An example of a non-spontaneous (reversible) process would be an ice cube at 0 C in a glass of water at C. Energy is being (slowly) transferred; but the temperature difference is so small that the system is virtually at equilibrium. The change in entropy caused by one of these processes is calculated by taking the heat which is transferred and dividing it by the temperature at which the transfer occurs. Thus: ΔS = q T This equation is Clausius' original, thermodynamic definition of entropy, proposed around You will often see it written with a negative sign! This is because in physics and engineering a positive heat flow is from the system to the surroundings. What is the entropy change associated with melting 1 gram of ice? Melting ice requires that 6.01 kjoules/mol flow into the system. (This is known as the enthalpy or the heat of fusion.) Thus: J 1mol 6010 mol 18.0g 1g S 273 K ΔS = 1.22 J K Notice the entropy is measured in joules per degree, not in kilojoules. This is not an obvious point. 3. The Statistical Definition of Entropy A more recent (around 1900) definition of entropy is the statistical definition. This definition, which was put forth by Ludwig Boltzmann, gives the same results but in a different way. The statistical definition says that entropy is determined by the number of microstates the number of different arrangements in which a system can exist. The relevant equation is: S = k ln W Here k is a constant, known as the Boltzmann constant, which is equal to R divided by Avogadro's Number. W is the number of microstates the ways in which a system can be arranged. The importance of the number of arrangements can be seen in a system containing four molecules (molecules A, B, C, and D) in two connected bulbs (bulbs 1 and 2). In the first column, all four molecules are in bulb #1 and there is only one possible arrangement (one possible microstate). In the second column, where one molecule is in bulb #2 and the remainder is in bulb #1, there are four ABCD ABCD ABCD ABCD ABCD Possible Arrangement of 4 molecules in 2 bulbs possible states. The column with the most possible arrangements, and hence the most likely configuration, is column 3, where the number of molecules in each bulb is equal. This column has the greatest entropy. Consider a crystal of CO at 0 K. Because carbon monoxide is symmetrical, each molecule of CO can -2-

3 be oriented either as C O or as O C. Assuming that each configuration has the same energy; we can calculate the entropy of one mole of CO at 0 K as follows: S = k ln W Since k = R/N, where N is Avogradro's Number S = (R/N) ln W Since the number of ways in which one mole of CO molecules can be arranged is 2 N S = (R/N) ln 2 N = (R/N) (N ln 2) S = R ln 2 S = = 5.76 joules This is fairly close to the experimental value of 4.0 joules/mole, and even this small difference can be explained. Since carbon monoxide has a slight dipole, adjacent molecules which are oriented CO to CO are energetically favored over molecules which are CO to OC. Therefore states with an excess of CO to CO orientations are favored and the crystal is not truly random, as our calculation assumes. 4. Entropy Summarized Often it is possible to predict the sign of the entropy change in a reaction. Consider the reaction: 2 H 2 S(g) + SO 2 (g) 3 S(s) + 2 H 2 O(g) Here the number of gas molecules decreases. Therefore the entropy of the system decreases. This is in fact quite general. The translational entropy of a gas is so large that a reaction in which the number of gas molecules increases will necessarily be associated with an entropy increase. Similarly a decrease in the number of gas molecules will be associated with decreasing entropy. You can usually predict whether entropy will increase or decrease by knowing the following: Entropy of a gas > entropy of a liquid > entropy of a solid Dissolving a solute increases its entropy greatly. Heating something, a solid, liquid or gas increases its entropy. Allowing a gas to expand increases its entropy. Mixing two substances, whether they are solids, liquids or gases, increases their entropy. Although these can all be explained in terms of the dispersal of energy, explanations are often made by equating entropy with disorder. Strictly speaking entropy is associated with disorder, but the analogy is often overstated. Consider the use of entropy to describe a macroscopic system such as a messy desk. Messing up a desk does not occur spontaneously and does not involve a heat flow. It is not truly an entropy increase. 5. The Third Law and the Zero of Entropy Enthalpies are defined in terms of an arbitrary zero an element in its standard state at 25 C. Thus there is no such thing as an absolute enthalpy, only an enthalpy change. Entropy, however, is different. Since entropy is the dispersal of energy, the less energy there is to disperse, the lower the entropy. In other words as temperature approaches zero, so does entropy. This is so fundamental that it makes up the third law of thermodynamics. "The zero of entropy is a perfect crystal of a pure substance at zero degrees Kelvin." (This makes sense since a perfect crystal at absolute zero has only one possible configuration.) So enthalpy is given as a standard enthalpy of formation, ΔH f. However entropy is not given as a change but as standard entropy, S. -3-

4 6. Free Energy A quantity called the Gibbs Free Energy (G) incorporates both entropy and enthalpy. The sign of ΔG determines the spontaneity of a reaction and is therefore of great interest to chemists. The Gibbs free energy is calculated from the equation: ΔG = ΔH - TΔS A reaction with a negative ΔG is spontaneous; a reaction with a positive ΔG is not spontaneous; and a reaction with zero ΔG is at equilibrium. Looking at the above equation and the affect of the sign of ΔG on spontaneity, we can make the following table. Make sure that you agree. ΔH ΔS Is it spontaneous? 1) + + At high temperature 2) + - Never 3) - - At low temperature 4) - + Always Case 1, where high temperature is needed to make the reaction spontaneous, is of greatest concern to the chemist. An example would be the boiling of water. Steam has a both higher entropy and a higher enthalpy than water, and the boiling process requires a high temperature. Let us calculate the temperature at which it happens. From a table of thermodynamic data we find that for boiling water ΔH = -242 kj -(-286 kj) = 44 kj/mol. Similarly ΔS = 189 J - 70 J = 119 J/mol. The boiling point of water is the temperature at which the liquid and gaseous forms of water are in equilibrium. At higher temperature the gas phase is favored and at lower temperature the liquid phase is favored. But what we are looking for is the equilibrium temperature, the temperature at which ΔG = 0. ΔH f S H 2 O(l) -286 kj 70 J H 2 O(g) -242 kj 189 J Thermodynamic properties of water We can calculate the transition temperature, the temperature at which equilibrium occurs and at which ΔG = 0. But in order to do so, we need values for ΔH and ΔS. We can take them from a thermodynamic table, but these data are correct only at 273 K. So we must assume that both ΔH and ΔS are independent of temperature. This is a good assumption for the enthalpy but is less good for entropy. Nevertheless, we will assume it to be true for both. We can calculate: ΔG = ΔH - TΔS 0 = 44 kj - T (.119 kj) T = 369 K = 96 C This is clearly wrong, since the normal boiling point of water is 100 C. The error comes from using S values at temperatures other than 25 C. But the result is close and this is how the calculation is done. You should know the method. -4-

5 The thermodynamic tables found in chemistry textbooks not only contain enthalpy values (ΔH f ) and entropy values (S ), they also contain free energy values (ΔG f ). Like energy and enthalpy, free energy is a state function. These tables are used to calculate the ΔG associated with a reaction, but only at 25 C. If you need to calculate free energy change at any other temperature, calculate ΔH and ΔS from the tables and then use them to calculate ΔG. Gibbs Free Energy is not conserved and, despite its units, is not really energy. It is sometimes called the "Gibbs Function." However, the Gibbs Free Energy does have a relationship to work. Work is not a state function and the amount of work obtained from a chemical process depends on how it is carried out. What we can calculate is the maximum amount of work obtainable from a chemical process. THIS is equal to the free energy change. 7. Equilibrium Constants The free energy change determines whether a reaction is spontaneous. The equilibrium constant, however, is more generally useful and can be calculated from the free energy change. Thus the following equation, where Q is the equilibrium quotient, gives a reaction's free energy change. ΔG = ΔG + RT ln Q For a system at equilibrium, K = Q, ΔG = 0, and the following equation is used: ΔG = -RT ln K You use 8.31 J/mol K as the gas constant and get a free energy in units of joules. Do NOT use as the gas constant or expect the energy to be in kj! Notice, also, that K depends on ΔG and not ΔG. This is because the free energy change at equilibrium is always zero, which would lead to the equilibrium constant being one. Let's calculate the equilibrium constant for: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) From the table on the right we know that the free energy change is -34 kj/mol at 25 C. So -34,000 = (ln K) ln K = = 13. Since this number has two significant digits, neither of which come after the decimal we can calculate K = ΔH f ΔG f S kj/mol kj/mol J/K mol N 2 (g) H 2 (g) NH 3 (g) Of course if we had mistakenly used -34 kj instead of -34,000 J, we would calculate ln K = and K = Mistakes in this type of calculation often give you a K value of approximately one. So if you get an answer of K = 1, CHECK IT! Beware of equilibrium constants close to 1! -5-

6 Now let us do the same calculation at 100 C. Since this is not at standard temperature we can no longer use "G " but will instead use "G'" to represent the free energy at standard concentrations but at a non-standard temperature. Also because of the non-standard temperature, we must calculate the free energy change from the changes in enthalpy and entropy. Thus: ΔG' = ΔH - T ΔS ΔG' = (2-46,000) ( ) ΔG' = -92, ,227 ΔG' = -17,773 J/mol Convince yourself that his is known to two significant digits. Now we calculate K ΔG' = - RT lnk -17,773 = ln K ln K = 5.73 The ln K is known to two significant digits, one of which is after the decimal. So K is only known to one significant digit. K = = Other Statements of the Three Laws In this house we obey the laws of thermodynamics! - Homer Simpson The Second Law can be stated in a number of ways. These include: A) Heat will not flow spontaneously from a cold object to a hot object. B) A system which is free of external influences becomes more disordered with time. This disorder can be expressed in terms of a quantity called entropy. C) In any irreversible process (which includes almost everything) the entropy of the Universe increases. D) You cannot create a heat engine which extracts heat and converts it to useful work. If you could, an air conditioner would produce energy instead of consuming it. A device which violates (D) is referred to as a "perpetual motion machine of the second kind." In order to convert heat to work, you must take heat from a hot object and move it to a colder object. The Third Law can also be stated as: It is impossible to reach absolute zero in a finite number of steps. Now let us conclude with another version of the three laws, attributed to the author C. P. Snow 1) You can't win. 2) You can't break even. 3) You can't get out of the game (because you can t reach absolute zero). -6-

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