1. Thermodynamics1 Chemical Thermodynamics (Chapter 19)

2. Thermodynamics2 General Review Thermodynamics is the study of energy transformations We often study transformations that involve the exchange of energy between a system and its surroundings –In chemical thermodynamics the system is generally defined as the chemicals under study –The surroundings are the rest of the universe –We ordinarily try to set boundaries to study a limited portion of the the universe, such as the contents of a calorimeter One key quantity that is frequently studied is the Internal Energy (E) of a system The Internal Energy of a system includes all forms of energy, and is much too complex to measure directly We can, however, study the transfer of such energy between a system and its surroundings

3. Thermodynamics3 General Review We can say: E f - E i  E = q + w Where q and w are the heat and work exchanged with the surroundings, respectively q can be measured experimentally, generally by studying the heat transferred to / from the surroundings In chemical systems, w is generally observed as pressure- volume work (P  V) as the system expands to do work on its surroundings, or contracts as the surroundings do work on the system Not that, though we don’t know the actual values of E f and E i, it is still possible to experimentally determine the exchange between the system and its surroundings

4. Thermodynamics4 General Review To simplify our study of energy transformations, we generally study quantities called state functions State functions are defined as thermodynamic quantities whose values do not depend on their history Examples of state functions include temperature (T), potential energy (PE), and one studied previously, change in enthalpy (  H) State functions, as opposed to process functions, like work, are given capital-letter symbols

5. Thermodynamics5 General Review Normally q, the heat transfer, is not a state function. However, when q is measured under constant pressure, designated a q p, it becomes a state function: q p =  H Thus enthalpy is defined as the heat change in a system measured under constant pressure The constant pressure requirement is not particularly stringent since most reactions are studied in open containers under relatively constant atmospheric pressure

6. Thermodynamics6 General Review Hess’s Law, which states the change in enthalpy observed for two states in a system is independent of the pathway taken during the change This law based on the First Law of Thermodynamics and on the fact that enthalpy changes are state functions. The First Law of Thermodynamics can be stated in many ways, but it is an expression of the principle of conservation of energy - the total amount of energy in the universe is constant - energy can neither be created nor destroyed.

7. Thermodynamics7 General Review Some other considerations when using Hess’s Law: –Enthalpic quantities are extensive quantities, meaning that the amount of heat exchanged depends on the amount of material undergoing change –The physical state of substances in a reaction, gas, liquid, or solid, affects the enthalpy calculation –When a chemical equation is written backwards, the sign of the enthalpy change for the reaction is reversed The methods used in solving enthalpy problems using Hess’s law will, with a few modifications, also be used for other quantities that are also state functions.

8. Thermodynamics8 General Review Lets review the application of Hess’s Law by determining the enthalpy change for the combustion of propane, as shown below: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) To determine the enthalpy of combustion, we will need to use enthalpy of formation values from tables in the back of the text. It is important to remember that the values in the tables represent the enthalpies of formation for a substance from its elements, under standard conditions (1 M concentrations, 1 atm pressure and 25 o C)

9. Thermodynamics9 General Review We remember that elements, in their normal elemental state, have no formation values For other substances in the equation, we write formation equations, including their enthalpy values from the table: 3 C(s) + 4 H 2 (g)C 3 H 8 (g)  H = -103.85 kJ/mol C(s) + O 2 (g) CO 2 (g)  H = -393.5 kJ/mol H 2 (g) + O 2 (g)H 2 O(g)  H = -241.82 kJ/mol These equations, and their accompanying enthalpies, can now be adjusted and combined to give the target equation, and its enthalpy

10. Thermodynamics10 General Review We will make the following adjustments to the equations and enthalpies –The first equation will be written backward because C 3 H 8 is a reactant, not a product. The sign of the enthalpy will also be reversed. –The second equation, and its accompanying enthalpy, will be tripled because our target equation contains three CO 2 –The third equation, and its accompanying enthalpy, will be quadrupled because our target equation contains four H 2 O Now, if the contributing equations can be combined to give the target equation, the combination of their adjusted enthalpies will represent the target equation

11. Thermodynamics11 General Review C 3 H 8 (g) 3 C(s) + 4 H 2 (g)  H = -103.85 kJ/mol 3[C(s) + O 2 (g) CO 2 (g)]  H = 3(-393.5 kJ/mol) 4[H 2 (g) + O 2 (g)H 2 O(g)]  H = 4(-241.82 kJ/mol) or C 3 H 8 (g) 3 C(s) + 4 H 2 (g)  H = 103.85 kJ/mol 3C(s) + 3 O 2 (g) 3 CO 2 (g)  H = -1180.5 kJ/mol) 4H 2 (g) + 2 O 2 (g)4H 2 O(g)  H = -967.28 kJ/mol) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g)  H = -2043.93 kJ/mol Since the adjusted contributing equations can be combined to give the correct target equation, Hess’s Law states that the combination of their adjusted enthalpies will give the correct overall enthalpy

12. Thermodynamics12 General Review Having demonstrated, by combining equations, how Hess’s Law works, be aware that a simpler method, defined below, is usually employed by chemists That is, taking into consideration the coefficients in the balanced equation, the summation of the enthalpies of formation of the products minus the summation of the enthalpies of formation of the reactants can be combined to give the enthalpy of the reaction Verify this shortcut approach with the problem just done

13. Thermodynamics13 Beyond Enthalpy Thermodynamics is often concerned with the question: can a reaction occur? First Law of Thermodynamics: energy is conserved. Any process that occurs without outside intervention is spontaneous. When two eggs are dropped they spontaneously break. The reverse reaction (two eggs leaping into your hand with their shells back intact) is not spontaneous. We can conclude that a spontaneous process has a direction.

14. Thermodynamics14 Spontaneous Processes A process that is spontaneous in one direction is not spontaneous in the opposite direction. The direction of a spontaneous process can depend on temperature: Ice turning to water is spontaneous at T > 0  C. Water turning to ice is spontaneous at T < 0  C.

15. Thermodynamics15 Spontaneous Processes Reversible and Irreversible Processes A reversible process is one that can go back and forth between states along the same path. –When 1 mol of water is frozen at 1 atm at 0  C to form 1 mol of ice, q =  H fus of heat is removed. –To reverse the process, q =  H fus must be added to the 1 mol of ice at 0  C and 1 atm to form 1 mol of water at 0  C. –Therefore, converting between 1 mol of ice and 1 mol of water at 0  C is a reversible process. Allowing 1 mol of ice to warm is an irreversible process. To get the reverse process to occur, the water temperature must be lowered to 0  C.

16. Thermodynamics16 Spontaneous Processes Chemical systems in equilibrium are reversible. In any spontaneous process, the path between reactants and products is irreversible. Thermodynamics gives us the direction of a process. It cannot, however, predict the speed at which the process will occur. Generally, nature favors lower energy states, that is, spontaneous processes are usually exothermic. Some processes, such as the spontaneous melting of ice at room temperature, are endothermic. Why are some endothermic reactions spontaneous?

17. Thermodynamics17 Entropy and the Second Law of Thermodynamics Spontaneous Expansion of a Gas Why do spontaneous processes occur? Consider an initial state: two flasks connected by a closed stopcock. One flask is evacuated and the other contains 1 atm of gas. The final state: two flasks connected by an open stopcock. Each flask contains gas at 0.5 atm. The expansion of the gas is isothermal (i.e. constant temperature). Therefore the gas does no work and heat is not transferred. Why does the gas expand?

18. Thermodynamics18 Entropy and the Second Law of Thermodynamics Spontaneous Expansion of an Ideal Gas

19. Thermodynamics19 Entropy and the Second Law of Thermodynamics Entropy Entropy, S, is a measure of the disorder of a system. Spontaneous reactions proceed to lower energy or higher entropy (nature favors disorder). In ice, the molecules are very well ordered because of the H-bonds. Therefore, ice has a low entropy. As ice melts, the intermolecular forces are broken (requires energy), but the order is interrupted (so entropy increases). Water is more random than ice, so ice spontaneously melts at room temperature.

20. Thermodynamics20 Entropy and the Second Law of Thermodynamics Order in ice crystals

21. Thermodynamics21 Entropy and the Second Law of Thermodynamics Entropy There are often multiple entropic effects occurring simultaneously. For example, when an ionic solid is placed in water two things happen: –the water organizes into hydrates about the ions (so the entropy decreases), and –the ions in the crystal dissociate (the hydrated ions are less ordered than the crystal, so the entropy increases).

22. Thermodynamics22 Entropy and the Second Law of Thermodynamics Entropy

23. Thermodynamics23 Entropy and the Second Law of Thermodynamics Entropy Generally, when an increase in entropy in one process is associated with a decrease in entropy in another, the increase in entropy dominates. Entropy is a state function. For a system,  S = S f - S i If  S > 0 the randomness increases, if  S < 0 the order increases.

24. Thermodynamics24 Entropy and the Second Law of Thermodynamics Disorder can also be considered from a probability perspective. For example, consider a simple system containing two chambers, separated by a stopcock, where two molecules are located in one of the chambers, and the other is evacuated. –When the stopcock is opened and the gas molecules are free to move independently, several new states become possible –The number of new states possible is given as: Number of states = 2 n, where n represents the number of gas molecules in the system in this case n = 2.

25. Thermodynamics25 Entropy and the Second Law of Thermodynamics The experiment described previously is shown below. Note that the initial condition (a) with two molecules gives rise to 2 2 = 4 equally possible states (b). Now consider a more complex system containing a much larger number of molecules, such as a mole (6.02 x 10 23 ) molecules. –The number of possible states would be –The vast majority of these states would put equal numbers of molecules randomly in each chamber. Only two states would put all molecules in one chamber or the other.

26. Thermodynamics26 Entropy and the Second Law of Thermodynamics Ludwig Boltzmann formalized the statistical approach to entropy with his famous equation (engraved on his headstone!) S = k ln W Where k = Boltzmann’s constant: k = 1.381 x 10 -23 J K -1 ln W represents the natural log of the number of ways a system can be arranged

27. Thermodynamics27 Entropy and the Second Law of Thermodynamics While the statistical approach is interesting, it will be more useful for us to consider entropy in terms of disorder and other thermodynamic quantities. Specifically, it can be shown that: If a system changes reversibly between state 1 and state 2. Then, the change in entropy is given by: –at constant T where q rev is the amount of heat added reversibly to the system. (Example: a phase change occurs at constant T with the reversible addition of heat.) –Note that reversible processes are generally associated with equilibrium systems

28. Thermodynamics28 Entropy and the Second Law of Thermodynamics The relationship suggests that changes in entropy are related to changes in heat. Does this seem reasonable? Remember, in our modern understanding, heat is defined in terms of molecular motion. Intuitively we understand that a system containing more heat will certainly exhibit greater disorder, especially in terms of degrees of freedom of translational, rotational and vibrational motion.

29. Thermodynamics29 Entropy and the Second Law of Thermodynamics Additional Entropy Considerations Since motion is associated with disorder, entropy values for the physical states of a substance are not the same. In general: S gases > S liquids > S solids Entropy generally increases with increasing molar mass Entropy generally increases with increasing numbers of atoms in molecules Entropy values for elements are NOT zero As with enthalpy, the entropy change associated with a process can be calculated by subtracting the total entropy table values of the reactants from the total entropy table values of the products Also note that entropy values are reported in joules, not kilojoules

30. Thermodynamics30 Entropy and the Second Law of Thermodynamics The Second Law of Thermodynamics Let’s now take our understanding of entropy and derive, in an informal manner, the Second Law of Thermodynamics. Let’s begin by looking a little more closely and the formula we just discussed. For a reversible process at constant temperature: Now, since q p =  H at constant pressure, we can say at constant temperature and pressure for a reversible process:

31. Thermodynamics31 Entropy and the Second Law of Thermodynamics The Second Law of Thermodynamics Now, if we multiply through by T and gather terms we get:Now, if we multiply through by T and gather terms we get:  H - T  S = 0 The quantity,  H - T  S, is symbolized as  G and is called the Free Energy (formerly Gibb’s Free Energy) of the process.The quantity,  H - T  S, is symbolized as  G and is called the Free Energy (formerly Gibb’s Free Energy) of the process. This quantity for a reversible process, one at equilibrium, has a value of  G = 0.This quantity for a reversible process, one at equilibrium, has a value of  G = 0. For processes not at equilibrium,  G has a nonzero value.For processes not at equilibrium,  G has a nonzero value.

32. Thermodynamics32 Entropy and the Second Law of Thermodynamics The Second Law of Thermodynamics When all of the thermodynamic quantities are combined we get an important master thermodynamic equation:When all of the thermodynamic quantities are combined we get an important master thermodynamic equation:  G =  H - T  S Spontaneous processes involve either exothermic reactions (  H is negative) or increases in entropy (  S is positive). When those conditions occur,  G will have a negative sign. In reality, when the effects of  H and  S combine so that  G has a negative sign, we can state that the process will be spontaneous. Indeed,  G is the key indicator of reaction spontaneity.Spontaneous processes involve either exothermic reactions (  H is negative) or increases in entropy (  S is positive). When those conditions occur,  G will have a negative sign. In reality, when the effects of  H and  S combine so that  G has a negative sign, we can state that the process will be spontaneous. Indeed,  G is the key indicator of reaction spontaneity. We can summarize with three important generalizations:We can summarize with three important generalizations: –For reversible, equilibrium processes:  G = 0 –For spontaneous processes  G < 0 –For nonspontaneous processes  G > 0

33. Thermodynamics33 Entropy and the Second Law of Thermodynamics When we consider the thermodynamic equation just discussed, we can make four important generalizations, based on  H and  S values: Enthalpy sign Entropy sign Spontaneity 1 -+Spontaneous at all temperatures 2 --Spontaneous only at lower temperatures 3 ++Spontaneous only at higher temperatures 4 +-Nonspontaneous at all temperatures

34. Thermodynamics34 Entropy and the Second Law of Thermodynamics Useful as the generalizations in the previous slide are, even more important conclusions can be drawn from the Second Law of Thermodynamics, which we will derive and define now. We will not do a rigorous derivation, but the one we use should be complete enough to justify the validity of the law. First, lets make some necessary assumptions: For a process, the entropy for the universe is:  S univ =  S sys +  S surr For a reversible process at equilibrium the entropy change for the universe is zero, as disorder is exchanged only between the system and its surroundings, and we can say:  S sys = -  S surr

35. Thermodynamics35 Entropy and the Second Law of Thermodynamics  G sys for a reversible process can be defined as:  G sys =  H sys - T  S sys Dividing through by T we get: Now, since and since  S sys = -  S surr with substitution we get: Multiplying through by -1, we get:

36. Thermodynamics36 Entropy and the Second Law of Thermodynamics Finally, since we know:  S univ =  S sys +  S surr Substituting we get: This equations states mathematically that a negative value for  G will be accompanied by an increase in entropy for the universe. This is a statement of the Second Law of Thermodynamics: A spontaneous process will lead to an increase in entropy in the universe.

37. Thermodynamics37 Entropy and the Second Law of Thermodynamics The consequences of the Second Law of Thermodynamics are profound. –All spontaneous processes in the universe, and there are many, will lead to greater disorder in the universe –Energy transformations will always be less than 100% efficient, with entropy changes accompanying all of them –The Second Law states that energy sources will eventually run down as energy becomes more evenly dispersed

38. Thermodynamics38 Third Law of Thermodynamics This would be an appropriate time to define the Third Law of Thermodynamics. It states: A state of zero entropy can only be found in a perfectly pure crystalline solid at absolute zero. Since perfect purity is impossible, and since achieving absolute zero is not possible, the Third Law refers to a theoretical state - approachable but not actually attainable.

39. Thermodynamics39 Three Laws of Thermodynamics First Law: The total amount of energy in the universe is constant - energy can be neither created nor destroyed. Second Law: Spontaneous processes will lead to and increase in entropy in the universe. Third Law: A state of zero entropy can only exist in a perfectly pure crystalline solid at absolute zero Someone once said whimsically that the three laws can be summarized as: You can’t win; you can’t break even; and you gotta play.

40. Thermodynamics40 Thermodynamics Problems There are many useful types of problems that can be solved using thermodynamic relationships. For example, if the heat of fusion (melting) is 6.00 kJ/mol, and the heat of vaporization is 40.65 kJ/mol, compare the entropy change observed when a mole of ice melts at 0 o C to the entropy change observed when a mole of liquid water vaporizes at 100 o C. Remember, for reversible processes such as these. Remember also that Kelvin temperatures must be used in these calculations. Ans:  S fusion = 22.0 J/mol-K  S vaporization = 109 J/mol-K

41. Thermodynamics41 Entropy and the Second Law of Thermodynamics The Second Law of Thermodynamics The second law of thermodynamics explains why spontaneous processes have a direction. In any spontaneous process, the entropy of the universe increases.  S univ =  S sys +  S surr : the change in entropy of the universe is the sum of the change in entropy of the system and the change in entropy of the surroundings. Entropy is not conserved:  S univ is increasing.

42. Thermodynamics42 Entropy and the Second Law of Thermodynamics The Second Law of Thermodynamics For a reversible process:  S univ = 0. For a spontaneous process (i.e. irreversible):  S univ > 0. Note: the second law states that the entropy of the universe must increase in a spontaneous process. It is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases. For an isolated system,  S sys = 0 for a reversible process and  S sys > 0 for a spontaneous process.

43. Thermodynamics43 The Molecular Interpretation of Entropy A gas is less ordered than a liquid that is less ordered than a solid. Any process that increases the number of gas molecules leads to an increase in entropy. When NO(g) reacts with O 2 (g) to form NO 2 (g), the total number of gas molecules decreases, and the entropy decreases.

44. Thermodynamics44 The Molecular Interpretation of Entropy

45. Thermodynamics45 The Molecular Interpretation of Entropy There are three atomic modes of motion: –translation (the moving of a molecule from one point in space to another), –vibration (the shortening and lengthening of bonds, including the change in bond angles), –rotation (the spinning of a molecule about some axis).

46. Thermodynamics46 The Molecular Interpretation of Entropy Energy is required to get a molecule to translate, vibrate or rotate. The more energy stored in translation, vibration and rotation, the greater the degrees of freedom and the higher the entropy. In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, this is a state of perfect order. Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero. Entropy changes dramatically at a phase change.

47. Thermodynamics47 The Molecular Interpretation of Entropy

48. Thermodynamics48 The Molecular Interpretation of Entropy As we heat a substance from absolute zero, the entropy must increase. If there are two different solid state forms of a substance, then the entropy increases at the solid state phase change. Boiling corresponds to a much greater change in entropy than melting. Entropy will increase when –liquids or solutions are formed from solids, –gases are formed from solids or liquids, –the number of gas molecules increase, –the temperature is increased.

49. Thermodynamics49 Calculations of Entropy Changes Absolute entropy can be determined from complicated measurements. Standard molar entropy, S  : entropy of a substance in its standard state. Similar in concept to  H . Units: J/mol-K. Note units of  H: kJ/mol. Standard molar entropies of elements are not zero. For a chemical reaction which produces n products from m reactants:

50. Thermodynamics50 Gibbs Free Energy For a spontaneous reaction the entropy of the universe must increase. Reactions with large negative  H values are spontaneous. How to we balance  S and  H to predict whether a reaction is spontaneous? Gibbs free energy, G, of a state is G = H - TS For a process occurring at constant temperature  G =  H - T  S.

51. Thermodynamics51 Gibbs Free Energy There are three important conditions: –If  G < 0 then the forward reaction is spontaneous. –If  G = 0 then reaction is at equilibrium and no net reaction will occur. –If  G > 0 then the forward reaction is not spontaneous. (However, the reverse reaction is spontaneous.) If  G > 0, work must be supplied from the surroundings to drive the reaction. For a reaction the free energy of the reactants decreases to a minimum (equilibrium) and then increases to the free energy of the products.

52. Thermodynamics52 Gibbs Free Energy

53. Thermodynamics53 Gibbs Free Energy Standard Free-Energy Changes We can tabulate standard free-energies of formation,  G  f (c.f. standard enthalpies of formation). Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M concentration (solution), and  G  = 0 for elements.  G  for a process is given by The quantity  G  for a reaction tells us whether a mixture of substances will spontaneously react to produce more reactants (  G  > 0) or products (  G  < 0).

54. Thermodynamics54 Free Energy and Temperature Focus on  G =  H - T  S: –If  H 0, then  G is always negative. –If  H > 0 and  S < 0, then  G is always positive. (That is, the reverse of 1.) –If  H < 0 and  S < 0, then  G is negative at low temperatures. –If  H > 0 and  S > 0, then  G is negative at high temperatures. Even though a reaction has a negative  G it may occur too slowly to be observed. Thermodynamics gives us the direction of a spontaneous process, it does not give us the rate of the process.

55. Example 1 Thermodynamics55 Let’s consider a real free energy calculation for a real reaction. Using standard free energies of formation from the tables at the back of the book, calculate the free energy change for the following reaction under standard conditions: 3 O 2 (g) 2O 3 (g) From the tables: For O 2 (g): For O 3 (g): Ans: 326.8 kJ/mol

56. Free Energy Calculations: Considerations Thermodynamics Under standard temperature conditions (T = 25 o C), the calculation of  G is simple and direct, and is done as has been shown previously where the  G values from the table can be used directly, by subtracting the summation of reactant values from the summation of product values. The example just done for ozone is typical. Note that when temperatures depart from the standard value,  G, which is dramatically affected by temperature, must be calculated by using  H and  S, which are relatively unaffected by temperature using the standard equation:  G =  H - T  S

57. Example 2 Thermodynamics57 Substance (kJ/mol)(J/mol  K)(kJ/mol) H 2 (g)0130.580 N 2 (g)0191.500 NH 3 (g)-46.19192.5-16.66 Calculate  G for the Haber Process, shown below, at 25 o C and at 500 o C. Note that  G (like  H) for elements, in their most stable elemental state, have a value of zero. Note also that the table values for S have different units (J/mol  K) that must be made compatible with the other quantities. 3H 2 (g) + N 2 (g) 2NH 3 (g) From the table: Ans: (at 25 o C)  G = -33.32 kJ/mol(at 500 o C)  G = 60.86 kJ/mol Would it be possible to determine at what temperature the process ceases to be spontaneous?

58. Thermodynamics58 Free Energy and Temperature

59. Thermodynamics59 Free Energy and the Equilibrium Constant Recall that  G  and K (equilibrium constant) apply to standard conditions. Recall that  G and Q (equilibrium quotient) apply to any conditions. It is useful to determine whether substances under any conditions will react:

60. Thermodynamics60 Free Energy and the Equilibrium Constant At equilibrium, Q = K and  G = 0, so From the above we can conclude: –If  G  1. –If  G  = 0, then K = 1. –If  G  > 0, then K < 1.

61. Example 3 Thermodynamics61 Using the equation that relates the equilibrium constant to the standard free energy change, estimate DG for the ionization of acetic acid, which has an equilibrium constant of 1.76 x 10 -5 M at 25 o C. Ans: 27.1 kJ/mol

62. Example 4 Thermodynamics62  G for the hydrolysis (break down) of a peptide bond, binding two amino acids, represented below, is estimated to be -10 kJ/mol. Estimate the equilibrium constant for the hydrolysis reaction. AA 1 -AA 2 + H 2 O AA 1 + AA 2 Ans: 0.018