1. Thermodynamics The universe is in a state of constant change, the only invariant is Energy

2. Consider …. 2 Gravity causes molecules of water move turbine blades turbines move coils of wire in magnetic fields moving magnetic fields move electrons moving electrons drive chemical reactions in a battery chemical reactions power your phone creating light and sound

3. Consider ….  In this example gravity is responsible for a working smartphone  Gravity does work on the blades  The turning blades do work on electrons  Electrons do work in chemical reactions  Chemical reactions do work on a speaker and power LEDs in the screen, power radios etc  The ability of something to do work on something else is transferred from gravity to water to electrons to chemical reactions to moving magnetics and moving air and light from the screen  In this example gravity is responsible for a working smartphone  Gravity does work on the blades  The turning blades do work on electrons  Electrons do work in chemical reactions  Chemical reactions do work on a speaker and power LEDs in the screen, power radios etc  The ability of something to do work on something else is transferred from gravity to water to electrons to chemical reactions to moving magnetics and moving air and light from the screen 3

4. Consider …. 4 Gravity causes hydrogen atoms fuse to make Helium and a little bit of mass is converted into light and heat etc Photons are absorbed by chlorophyll and used to power photosynthesis We extract the oil make biofuel ignite the biofuel and excess energy released gives the molecules to power to move pistons

5. Consider ….  In this example Gravity causes fusion reactions in the sun are responsible for a working car  Mass is converted to light  light moves electrons  Electrons do work in chemical reactions to create sugars oils etc  Oils react with oxygen to create fast moving CO 2 and H 2 O  Molecules push pistons and drive the car  The ability of something to do work on something else is transferred from the sun to chlorophyll to electrons to chemical reactions to moving molecules and moving pistons and moving wheels  In this example Gravity causes fusion reactions in the sun are responsible for a working car  Mass is converted to light  light moves electrons  Electrons do work in chemical reactions to create sugars oils etc  Oils react with oxygen to create fast moving CO 2 and H 2 O  Molecules push pistons and drive the car  The ability of something to do work on something else is transferred from the sun to chlorophyll to electrons to chemical reactions to moving molecules and moving pistons and moving wheels 5

6. Energy: the capacity to do work  In each step in the previous examples a capacity to do work is transferred from one thing to another, this is called energy  Gravitational energy is transferred into kinetic energy, into electrical energy, chemical energy, sound and light energy etc  All dynamic processes in the universe are due to the flow of energy  Thermodynamics is the study of energy (heat) flow and the laws that govern it  Since we want to understand chemical transformation we need to understand energy transformation  In each step in the previous examples a capacity to do work is transferred from one thing to another, this is called energy  Gravitational energy is transferred into kinetic energy, into electrical energy, chemical energy, sound and light energy etc  All dynamic processes in the universe are due to the flow of energy  Thermodynamics is the study of energy (heat) flow and the laws that govern it  Since we want to understand chemical transformation we need to understand energy transformation 6

7. Energy  Energy is a universal invariant  It can change from one form to another but cannot be created or destroyed  It is measured in Joules (J)  There is potential energy (energy that something has because of where and what it is) and kinetic energy (the energy is has because of how fast it is moving)  The lower the potential energy the more stable something is. Potential energy can be negative  When some process happens, generally it is to lower the potential energy  The study of energy helps us to predict whether a process is spontaneous or not  Energy is a universal invariant  It can change from one form to another but cannot be created or destroyed  It is measured in Joules (J)  There is potential energy (energy that something has because of where and what it is) and kinetic energy (the energy is has because of how fast it is moving)  The lower the potential energy the more stable something is. Potential energy can be negative  When some process happens, generally it is to lower the potential energy  The study of energy helps us to predict whether a process is spontaneous or not 7

8. molecules and energy  molecules are in a constant state of motion  electrons are orbiting nuclei, so they experience a potential energy due to the Coulomb attraction between charges which they share with the nuclei  They have kinetic energy from ‘orbiting’  The nuclei also are in motion, either by vibrating against other nuclei, or else the whole molecule is moving through space, as translation (related to temperature) or rotating  When molecules react, some bonds break and new bonds form. In an exothermic reaction the new bonds are more stable (they have a smaller or more negative potential energy)  molecules are in a constant state of motion  electrons are orbiting nuclei, so they experience a potential energy due to the Coulomb attraction between charges which they share with the nuclei  They have kinetic energy from ‘orbiting’  The nuclei also are in motion, either by vibrating against other nuclei, or else the whole molecule is moving through space, as translation (related to temperature) or rotating  When molecules react, some bonds break and new bonds form. In an exothermic reaction the new bonds are more stable (they have a smaller or more negative potential energy) 8

9. What is Thermodynamics?  Thermodynamics is a branch of physics concerned with energy flow. Historically it had an emphasis on heat, temperature and their relation to energy and work.  Study of energy changes accompanying chemical and physical changes to a system  Defines systems using a few macroscopic (measurable) variables, such as internal energy, entropy, temperature and pressure  In chemistry, thermodynamics predicts if reactions occur and how the equilibrium constant changes with temperature  Thermodynamics is a branch of physics concerned with energy flow. Historically it had an emphasis on heat, temperature and their relation to energy and work.  Study of energy changes accompanying chemical and physical changes to a system  Defines systems using a few macroscopic (measurable) variables, such as internal energy, entropy, temperature and pressure  In chemistry, thermodynamics predicts if reactions occur and how the equilibrium constant changes with temperature 9

10. 10 First Law of Thermodynamics  you can’t get something for nothing  First Law of Thermodynamics: Energy cannot be Created or Destroyed  the total energy of the universe cannot change  though you can transfer it from one place to another and from one form of energy to another  In thermodynamics we will often arbitrarily talk of the system and the surroundings. Then in a process (say a reaction in a test tube)  ΔE univ = 0 = ΔE sys + ΔE surr (1)  you can’t get something for nothing  First Law of Thermodynamics: Energy cannot be Created or Destroyed  the total energy of the universe cannot change  though you can transfer it from one place to another and from one form of energy to another  In thermodynamics we will often arbitrarily talk of the system and the surroundings. Then in a process (say a reaction in a test tube)  ΔE univ = 0 = ΔE sys + ΔE surr (1)

11. 11 First Law of Thermodynamics Conservation of Energy  For an exothermic reaction, products have a lower potential energy than the reactants and the extra “lost” heat ΔE (same as ΔH if P constant) from the system goes into the surroundings  two ways energy “lost” from a system,  converted to heat, q  used to do work, w ΔE = q + w(2) Sign Conventions  Heat q added to a system q > 0  Heat q leaves the system q < 0  Work done W on the system W > 0  Work done W by the system W < 0 Conservation of Energy  For an exothermic reaction, products have a lower potential energy than the reactants and the extra “lost” heat ΔE (same as ΔH if P constant) from the system goes into the surroundings  two ways energy “lost” from a system,  converted to heat, q  used to do work, w ΔE = q + w(2) Sign Conventions  Heat q added to a system q > 0  Heat q leaves the system q < 0  Work done W on the system W > 0  Work done W by the system W < 0

12. Example of Work Done W by the system  In physics W is defined as a force F acting to move an object a distance Δx in the direction of the force F  If in a reaction you create moles of gas the gas expands into the atmosphere  The gas has to expand pushing against the pressure P of the atmosphere (using P = F/A pressure is force/Area)  W = FΔx = -(PA.Δx) = -P.ΔV Example of heat q  In an exothermic reaction at constant pressure ΔH < 0 and q = ΔH  In an endothermic reaction at constant pressure ΔH > 0 and q = ΔH Example of Work Done W by the system  In physics W is defined as a force F acting to move an object a distance Δx in the direction of the force F  If in a reaction you create moles of gas the gas expands into the atmosphere  The gas has to expand pushing against the pressure P of the atmosphere (using P = F/A pressure is force/Area)  W = FΔx = -(PA.Δx) = -P.ΔV Example of heat q  In an exothermic reaction at constant pressure ΔH < 0 and q = ΔH  In an endothermic reaction at constant pressure ΔH > 0 and q = ΔH 12 First Law of Thermodynamics

13.  Energy conservation requires that the internal energy E change in the system equal the heat supplied (q) + work done (w) on the system ΔE = q + w(2) ΔE = ΔH + PΔV(3)  E is the total energy of everything in the system (the kinetic and potential energy of the atoms)  ΔH is the change in the Enthalpy which is the heat change q P measured at constant P  ΔH = ΔE + PΔV = (q P + w) + PΔV  If the gas expands w = -PΔV and ΔH=q P  ΔE (ΔU), ΔH are a state functions  internal energy change independent of how this change occurs  Energy conservation requires that the internal energy E change in the system equal the heat supplied (q) + work done (w) on the system ΔE = q + w(2) ΔE = ΔH + PΔV(3)  E is the total energy of everything in the system (the kinetic and potential energy of the atoms)  ΔH is the change in the Enthalpy which is the heat change q P measured at constant P  ΔH = ΔE + PΔV = (q P + w) + PΔV  If the gas expands w = -PΔV and ΔH=q P  ΔE (ΔU), ΔH are a state functions  internal energy change independent of how this change occurs 13 First Law of Thermodynamics

14. The first law and time reversal  The first law tells us that only processes where there is no net change in the total energy are allowed (energy is conserved) 14

15. The first law and spontaneity  In all observed phenomena the total energy is always the same  The energy at t, E(t) is equal to the energy at time time t+dt,  E(t) = E(t+dt)  So if that is the case why do we always see some processes only going one way?  In all observed phenomena the total energy is always the same  The energy at t, E(t) is equal to the energy at time time t+dt,  E(t) = E(t+dt)  So if that is the case why do we always see some processes only going one way? 15 ✓ ✗

16. The first law and spontaneity  Clearly the first law isn’t the end of the story regarding energy and what happens in processes 16 ✓ ✗

17. 17 Q as an Energy Tax  Thermodynamics originally came about with a desire to understand how heat engines worked  If you want to control energy to do work w, you find you can’t break even, because you always create q  to recharge a battery with 100 kJ of useful energy will require more than 100 kJ  every energy transition results in a “loss” of energy due to q  conversion of energy to heat which is “lost” by heating up the surroundings  Thermodynamics originally came about with a desire to understand how heat engines worked  If you want to control energy to do work w, you find you can’t break even, because you always create q  to recharge a battery with 100 kJ of useful energy will require more than 100 kJ  every energy transition results in a “loss” of energy due to q  conversion of energy to heat which is “lost” by heating up the surroundings

18. 18 Heat Tax fewer steps generally results in a lower total heat tax (more efficient)

19. 19 Factors Affecting Whether a Reaction Is Spontaneous  It turns out that there are two factors that determine the thermodynamic favorability are the enthalpy H and the entropy S.  The change in enthalpy is a comparison of the bond energy of the reactants to the products.  bond energy = amount needed to break a bond.  ΔH  The entropy factors relates to the randomness/orderliness of a system  ΔS  The enthalpy factor is generally more important than the entropy factor  Let’s look at these  It turns out that there are two factors that determine the thermodynamic favorability are the enthalpy H and the entropy S.  The change in enthalpy is a comparison of the bond energy of the reactants to the products.  bond energy = amount needed to break a bond.  ΔH  The entropy factors relates to the randomness/orderliness of a system  ΔS  The enthalpy factor is generally more important than the entropy factor  Let’s look at these

20. Enthalpy  related to the internal energy E, the energy change measured at constant P is ΔH = Δ  pΔV  generally kJ/mol)  ΔH rxn is related to the breaking and forming of chemical bonds. Stronger bonds = more stable molecules smaller (more negative) PE  if products more stable than reactants, energy released  exothermic  ΔH < 0 (negative) – the process is generally spontaneous  if reactants more stable than products, energy absorbed  endothermic  ΔH > 0 (positive) – the process may not be spontaneous  The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions.  Hess’ Law  related to the internal energy E, the energy change measured at constant P is ΔH = Δ  pΔV  generally kJ/mol)  ΔH rxn is related to the breaking and forming of chemical bonds. Stronger bonds = more stable molecules smaller (more negative) PE  if products more stable than reactants, energy released  exothermic  ΔH < 0 (negative) – the process is generally spontaneous  if reactants more stable than products, energy absorbed  endothermic  ΔH > 0 (positive) – the process may not be spontaneous  The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions.  Hess’ Law

21. Using Hess’s Law 21

23. Spontaneity: Enthalpy Driven Processes In many cases, the direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end Cellulose and O 2 have a bigger potential energy than the equivalent amount of carbon dioxide and water The transformation lowers the overall potential energy, C-O and H-O bonds are more stable than C-C and C-H bonds exothermic reactions (Δ H < 0) are spontaneous The extra energy leaves as heat Transformations have accompanying energy changes to parts but overall Δ E=0. Can we tell which transformations will occur spontaneously by studying the energy change in the part that is changing?

24. Spontaneity: Entropy Driven Processes But some processes are spontaneous but not exothermic! These are entropy driven processes, processes where Δ S > 0

25. 25 Entropy S  Entropy, S, is a thermodynamic function that increases as the number of equivalent ways of arranging the atoms/molecules (positions and velocities) in a system to give the appropriate V, U and T increases  S generally J/(K.mol) S = k ln W (6)  k = Boltzmann Constant = 1.38 x 10 -23 J/K  W is the number of energetically equivalent ways accessible, unitless (measure of our lack of knowledge about the system)  Entropy is the energy dispersal per unit temperature  Another way to calculate the change in entropy ΔS at constant T is ΔS = q rev /T(6’)  Where T is in Kelvin and q rev is the amount of heat generated in the process if it happened sufficiently slowly to be at equilibrium the whole time  S is a measure of the unavailability of a system to do work, since q rev is the energy lost to the universe as heat, sound etc, that cannot be refocused and used

26. 26 W W Energetically Equivalent States for the Expansion of a Gas

27. 27 Macrostates → Microstates This macrostate can be achieved through several different arrangements of the particles This macrostate can be achieved through several different arrangements of the particles These microstates all have the same macrostate So there are (3)6 different particle arrangements that result in the same macrostate These microstates all have the same macrostate So there are (3)6 different particle arrangements that result in the same macrostate

28. 28 Macrostates and Probability There is only one possible arrangement that gives State A and one that gives State B There are 6 possible arrangements that give State C Therefore State C has higher entropy than either State A or State B The macrostate with the highest entropy also has the greatest dispersal of energy

29. 29 Changes in Entropy, ΔS  As we shall see the condition ΔS>0 is deemed a favorable entropy change.  Since ΔS = S f - S i = k ln(W f ) – k ln(W i ) = k ln(W f /W i )  for ΔS>0, W f >W i  This happens when the energy is more dispersed in the final state  Processes where ΔS>0 include:  reactions whose products are in a more disordered state.  (solid < liquid < gas)  reactions which have larger numbers of product molecules than reactant molecules.  increase in temperature  solids dissociating into ions upon dissolving  As we shall see the condition ΔS>0 is deemed a favorable entropy change.  Since ΔS = S f - S i = k ln(W f ) – k ln(W i ) = k ln(W f /W i )  for ΔS>0, W f >W i  This happens when the energy is more dispersed in the final state  Processes where ΔS>0 include:  reactions whose products are in a more disordered state.  (solid < liquid < gas)  reactions which have larger numbers of product molecules than reactant molecules.  increase in temperature  solids dissociating into ions upon dissolving

30. Calculate the entropy change for expansion of 1 mole of an ideal gas from 1L to 2L 30 ΔS = S f – S i = k ln(W f /W i ) L = Lk ln(W f /W i ) = R ln(W f /W i ) from the last problem the number of available microstates for each atom is W and it will be proportional to the volume available to the gas V j so W j =pV j. For the entire gas it will be (pV j ) L where p is a constant. L = Avogadro’s constant and L x k = R the molar gas constant. For n particles V i = 1L V f = 2L ΔS= R ln(pV f /pV i )=R ln(2L/1L) = 5.76 J/mol/K note that ΔS > 0 so the gas will spontaneously fill the two flasks and halve the pressure

31. 31 Entropy Change in a State Change  when materials change state, the number of macrostates it can have changes as well  for entropy: solid < liquid < gas  because the degrees of freedom of motion increases solid → liquid → gas  when materials change state, the number of macrostates it can have changes as well  for entropy: solid < liquid < gas  because the degrees of freedom of motion increases solid → liquid → gas

32. 32 Entropy Change and State Change As a solid melts the atoms/molecules have more freedom to move, the energy is more disperse so the entropy increases, this is why ice spontaneously melts above 0 o C

33. 33 Increases in Entropy

34. Qualitative Predictions for ΔS 34 Predict if ΔS > 0 or ΔS < 0 for the following processes H 2 O(l)  H 2 O(g) Ag + (aq)+Cl - (aq)  AgCl(s) 4Fe(s)+3O 2 (g)  2Fe 2 O 3 (s) N 2 (g)+O 2 (g)  2NO(g) Evaporation big increase in V available ΔS > 0 Ions free to move in solution and not so free to move in the solid ΔS < 0 Gas  solid a lot less freedom of movement for the O 2 ΔS < 0 Same number of moles of each side, at this point all we know is ΔS ≈ 0

35. 35 The 2 nd Law of Thermodynamics: Spontaneity  "Energy spontaneously disperses from being localized to becoming spread out if it is not hindered from doing so.”  The total entropy change of the universe must be positive for a process to be spontaneous  for reversible process ΔS univ = 0,  for irreversible (spontaneous) process ΔS univ > 0 ΔS univ = ΔS sys + ΔS surr (7)  if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount  when ΔS sys is negative, ΔS surr is positive  the increase in ΔS surr often comes from the heat released in an exothermic reaction  "Energy spontaneously disperses from being localized to becoming spread out if it is not hindered from doing so.”  The total entropy change of the universe must be positive for a process to be spontaneous  for reversible process ΔS univ = 0,  for irreversible (spontaneous) process ΔS univ > 0 ΔS univ = ΔS sys + ΔS surr (7)  if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount  when ΔS sys is negative, ΔS surr is positive  the increase in ΔS surr often comes from the heat released in an exothermic reaction

36. 36 Temperature Dependence of ΔS surr  when a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings  when a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings  the amount the entropy of the surroundings changes depends on the temperature it is at originally  the higher the original temperature, the less effect addition or removal of heat has  when a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings  when a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings  the amount the entropy of the surroundings changes depends on the temperature it is at originally  the higher the original temperature, the less effect addition or removal of heat has

37. The reaction C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g) has ΔH rxn = -2044 kJ at 25°C. Calculate the entropy change of the surroundings. combustion is largely exothermic, so the entropy of the surrounding should increase significantly ΔH system = -2044 kJ, T = 298 K ΔS surroundings, J/K Check: Solution: Concept Plan: Relationships: Given: Find: ΔSΔST, Δ H

38. 38 Gibbs Free Energy, ΔG  For a spontaneous process ΔS univ > 0  maximum amount of energy from the system available to do work on the surroundings at constant temperature T (9)  when ΔG < 0, there is a decrease in free energy of the system that is released into the surroundings; therefore a process will be spontaneous when ΔG is negative  For a spontaneous process ΔS univ > 0  maximum amount of energy from the system available to do work on the surroundings at constant temperature T (9)  when ΔG < 0, there is a decrease in free energy of the system that is released into the surroundings; therefore a process will be spontaneous when ΔG is negative need to know for whole universe just the sys

39. 39 Thermodynamics and Spontaneity Free Energy  spontaneity is determined by comparing the free energy G of the system before the reaction with the free energy of the system after reaction, it includes both the enthalpy and entropy change of a process ΔG = ΔH – T∙ΔS(9)  if the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable  spontaneity ≠ fast or slow  ΔG sys 0  spontaneity is determined by comparing the free energy G of the system before the reaction with the free energy of the system after reaction, it includes both the enthalpy and entropy change of a process ΔG = ΔH – T∙ΔS(9)  if the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable  spontaneity ≠ fast or slow  ΔG sys 0

40. 40 Gibbs Free Energy, ΔG  process will be spontaneous when ΔG is negative  ΔG will be negative when  ΔH is negative and ΔS is positive  exothermic and more random  ΔH is negative and large and ΔS is negative but small  ΔH is positive but small and ΔS is positive and large  or high temperature  ΔG will be positive when ΔH is + and ΔS is −  never spontaneous at any temperature  when ΔG = 0 the reaction is at equilibrium  process will be spontaneous when ΔG is negative  ΔG will be negative when  ΔH is negative and ΔS is positive  exothermic and more random  ΔH is negative and large and ΔS is negative but small  ΔH is positive but small and ΔS is positive and large  or high temperature  ΔG will be positive when ΔH is + and ΔS is −  never spontaneous at any temperature  when ΔG = 0 the reaction is at equilibrium

41. 41 ΔG, ΔH, and ΔS

42. Chemical Potential Energy The chemical potential – is a form of free energy used for chemical reactions, in spontaneous reactions the chemical potential decreases

43. Free Energy Problem 43 For the reaction N 2 (g)+O 2 (g)  2NO(g) ΔH o =180.7kJ and ΔS o = 24.7 J/K at 298K. Is the reaction spontaneous? If spontaneous ΔG o < 0 ΔG o = ΔH o –TΔS o No!

44. The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has ΔH = +95.7 kJ and ΔS = +142.2 J/K at 25°C. Calculate ΔG and determine if it is spontaneous. Since ΔG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature. ΔH = +95.7 kJ, ΔS = 142.2 J/K, T = 298 K ΔG, kJ Answer: Solution: Concept Plan: Relationships: Given: Find: ΔGΔGT, Δ H, Δ S

45. The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has ΔH = +95.7 kJ and ΔS = +142.2 J/K. Calculate the minimum temperature it will be spontaneous. The temperature must be higher than 673K for the reaction to be spontaneous ΔH = +95.7 kJ, ΔS = 142.2 J/K, ΔG < 0  Answer: Solution: Concept Plan: Relationships: Given: Find: T Δ G, Δ H, Δ S

46. The reaction SO 2(g) + ½ O 2(g)  SO 3(g) has ΔH o = -98.9 kJ and ΔS o = -94.0 J/K at 25°C. Calculate ΔG o at 125 o C and determine if it is spontaneous. Since ΔG is -, the reaction is spontaneous at this temperature, though less so than at 25 o C ΔH o = -98.9 kJ, ΔS o = -94.0 J/K, T = 398 K ΔG o, kJ Answer: Solution: Concept Plan: Relationships: Given: Find: ΔGoΔGo T, Δ H o, Δ S o

47. 47 ΔG Relationships  if a reaction can be expressed as the sum of a series of reactions, the sum of the ΔG values of the individual reaction is the ΔG of the total reaction  ΔG is a state function  if a reaction is reversed, the sign of its ΔG value reverses  if the amounts of materials is multiplied by a factor, the value of the ΔG is multiplied by the same factor  the value of ΔG of a reaction is extensive  if a reaction can be expressed as the sum of a series of reactions, the sum of the ΔG values of the individual reaction is the ΔG of the total reaction  ΔG is a state function  if a reaction is reversed, the sign of its ΔG value reverses  if the amounts of materials is multiplied by a factor, the value of the ΔG is multiplied by the same factor  the value of ΔG of a reaction is extensive

48. Standard Free Energy of Formation ΔG o f 48 ΔG o f is the free energy change when 1 mole of a substance is made from its constituent elements in their standard states at 1atm, 298K We can use the equivalent of Hess’s Law to calculate ΔG o Tables exist in Google land tabulating ΔG f o,, ΔH f o, and ΔS o

49. 49 Elements in their standard states More stable than their constituent elements

50. 50 Calculating ΔG o at other temperatures  At 25 o C we can used ΔG o f : ΔG o reaction = ΣnG o f (products) - ΣnG o f (reactants)  But what about at other temperatures?  Assume the change in ΔH o reaction and ΔS o reaction is negligible ΔG o reaction = ΔH o reaction – TΔS o reaction  At 25 o C we can used ΔG o f : ΔG o reaction = ΣnG o f (products) - ΣnG o f (reactants)  But what about at other temperatures?  Assume the change in ΔH o reaction and ΔS o reaction is negligible ΔG o reaction = ΔH o reaction – TΔS o reaction

51. Calculate ΔG o at 25 o C for the reaction CH 4(g) + 8 O 2(g)  CO 2(g) + 2 H 2 O (g) + 4 O 3(g) standard free energies of formation from Appendix of textbook or google ΔG o, kJ Solution: Concept Plan: Relationships: Given: Find: ΔGoΔGo Δ G o f of prod & react SubstanceΔG o f, kJ/mol CH 4 (g)-50.5 O2(g)O2(g)0.0 CO 2 (g)-394.4 H 2 O(g)-228.6 O3(g)O3(g)163.2

52. 52 ΔG under other Nonstandard Conditions ΔG = ΔG o only when the reactants and products are in their standard states there normal state at that temperature partial pressure of gas = 1 atm concentration = 1 M under nonstandard conditions, ΔG = ΔG o + RTlnQ Q is the reaction quotient ΔG = ΔG o only when the reactants and products are in their standard states there normal state at that temperature partial pressure of gas = 1 atm concentration = 1 M under nonstandard conditions, ΔG = ΔG o + RTlnQ Q is the reaction quotient

53. 53 Temperature Dependence of K Measure K at different T, then plot lnK vs 1/T to get DS and DH

54. 54 Example – Calculating K given DH f and S f  Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Δ H f o (NH 3 )=-46.19 kJ/mol Δ S o (NH 3 )=192.5 J/K, Δ S o (N 2 )=191.5 J/K, Δ S o (H 2 )=130.58 J/K  Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Δ H f o (NH 3 )=-46.19 kJ/mol Δ S o (NH 3 )=192.5 J/K, Δ S o (N 2 )=191.5 J/K, Δ S o (H 2 )=130.58 J/K Δ G° = -RT lnK or lnK = - Δ G o /RT lnK = -46400/(8.314 J/K x 700K) = -7.97 K = e -7.97 = 3.45 x 10 -4 since K is << 1, the position of equilibrium favors reactants Δ H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J Δ S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K Δ G° = -92380 J - (700 K)(-198.2 J/K) = +46400 J

55. 55 Example – ΔG under non—standard conditions  Calculate ΔG at 427 o C for the reaction below if the P N2 = 33.0 atm, P H2 = 99.0 atm, and P NH3 = 2.0 atm N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Given Δ H° = -92.380 kJ and Δ S° = -198.2 J/K  Calculate ΔG at 427 o C for the reaction below if the P N2 = 33.0 atm, P H2 = 99.0 atm, and P NH3 = 2.0 atm N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Given Δ H° = -92.380 kJ and Δ S° = -198.2 J/K Q = P NH3 2 P N2 1 x P H2 3 (2.0 atm) 2 (33.0 atm) 1 (99.0) 3 == 1.2 x 10 -7 Δ G = Δ G° + RTlnQ Δ G = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10 -7 ) Δ G = -46300 J = -46 kJ Δ G° = -92380 J - (700 K)(-198.2 J/K) = +46400 J Δ G = Δ G° + RTlnQ

56. 56 At equilibrium 0.03 atm of H 2 O(g) is present but why? If the vapor pressure of water in the atmosphere is less than 0.03 the water will evaporate (entropy driven)! If the vapor pressure is higher than 0.03 atm water vapor will condense (enthalpy driven)

57. 57 Q

58. 58

59. 59 Heat Flow, Entropy, and the 2 nd Law Another consequence of the 2 nd law is that heat flows from a hot body to a cold body until thermal equilibrium is reached In the example on the right, heat must flow from water to ice in order for the entropy of the universe to increase But why that way round? The 1st law is not violated if more ice was formed? Consider an amount of heat q going from the liquid water at T h to the ice at T c Another consequence of the 2 nd law is that heat flows from a hot body to a cold body until thermal equilibrium is reached In the example on the right, heat must flow from water to ice in order for the entropy of the universe to increase But why that way round? The 1st law is not violated if more ice was formed? Consider an amount of heat q going from the liquid water at T h to the ice at T c

60. 60 Thermodynamics vs. Kinetics Kinetics describes how fast things change Thermodynamics is concerned if they will change and if so what changes we will see in internal energy, temperature, pressure etc

61. 61 Example: Diamond → Graphite Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don ’ t worry, it ’ s so slow that your ring won ’ t turn into pencil lead in your lifetime (or through many of your generations).

62. 62 Reversibility of Process  any spontaneous process is irreversible  it will proceed in only one direction  a reversible process will proceed back and forth between the two end conditions  equilibrium  results in no change in free energy  if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction  any spontaneous process is irreversible  it will proceed in only one direction  a reversible process will proceed back and forth between the two end conditions  equilibrium  results in no change in free energy  if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction

63. 63 The 3 rd Law of Thermodynamics: Absolute Entropy  the absolute entropy of a substance is the amount of energy it has due to distribution of energy through its particles  the 3 rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K  S = k.log(1) = 0  therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy  therefore, the absolute entropy of substances is always S > 0  the absolute entropy of a substance is the amount of energy it has due to distribution of energy through its particles  the 3 rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K  S = k.log(1) = 0  therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy  therefore, the absolute entropy of substances is always S > 0

64. 64 Standard Entropies  Entropies can be calculated using the 3 rd law and the equation  The entropy at 1 atm and 298 K and for 1M solutions is labeled S°  Extensive (depends on the system size)  entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution  Entropies can be calculated using the 3 rd law and the equation  The entropy at 1 atm and 298 K and for 1M solutions is labeled S°  Extensive (depends on the system size)  entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution

66. Calculate Δ S for the reaction 4 NH 3(g) + 5 O 2(g)  4 NO (g) + 6 H 2 O (g) ΔS is +, as you would expect for a reaction with more gas product molecules than reactant molecules standard entropies look up in appendix to textbook or google ΔS, J/K Check: Solution: Concept Plan: Relationships : Given: Find: ΔSΔS S o NH3, S o O2, S o NO, S o H2O, SubstanceS, J/mol/K NH 3 (g)192.8 O2(g)O2(g)205.2 NO(g)210.8 H 2 O(g)188.8

67. 67 Relative Standard Entropies States  the gas state has a larger entropy than the liquid state at a particular temperature  the liquid state has a larger entropy than the solid state at a particular temperature  the gas state has a larger entropy than the liquid state at a particular temperature  the liquid state has a larger entropy than the solid state at a particular temperature Substance S°, (J/mol∙K) H 2 O (l)70.0 H 2 O (g)188.8

68. 68 Relative Standard Entropies Molar Mass  the larger the molar mass, the larger the entropy  available energy states more closely spaced, allowing more dispersal of energy through the states  the larger the molar mass, the larger the entropy  available energy states more closely spaced, allowing more dispersal of energy through the states

69. 69 Relative Standard Entropies Allotropes  the less constrained the structure of an allotrope is, the larger its entropy

70. 70 Relative Standard Entropies Molecular Complexity  larger, more complex molecules generally have larger entropy  more available energy states, allowing more dispersal of energy through the states  larger, more complex molecules generally have larger entropy  more available energy states, allowing more dispersal of energy through the states Substance Molar Mass S°, (J/mol∙K) Ar (g)39.948154.8 NO (g)30.006210.8

71. 71 Relative Standard Entropies Dissolution  dissolved solids generally have larger entropy  distributing particles throughout the mixture  dissolved solids generally have larger entropy  distributing particles throughout the mixture Substance S°, (J/mol∙K) KClO 3 (s)143.1 KClO 3 (aq)265.7

72. Problems  Which of the following processes spontaneous? Answer either, always, sometimes, or never   Exothermic reactions ΔH < 0  Endothermic reactions ΔH > 0  Reactions in which ΔS univ > 0  Reactions in which ΔS sys > 0  Reactions in which ΔS univ < 0  Reactions in which ΔS sys < 0  Reactions in which ΔG sys > 0  Reactions in which ΔG sys < 0   Which of the following processes spontaneous? Answer either, always, sometimes, or never   Exothermic reactions ΔH < 0  Endothermic reactions ΔH > 0  Reactions in which ΔS univ > 0  Reactions in which ΔS sys > 0  Reactions in which ΔS univ < 0  Reactions in which ΔS sys < 0  Reactions in which ΔG sys > 0  Reactions in which ΔG sys < 0  72 Sometimes Always Sometimes Never Sometimes Never

73. Problems 73 Calculate ΔH o, ΔS o and ΔG o for the oxidation of SO 2 (g) at 25 o C and 1 atm 2SO 2 (g) + O 2 (g)  2SO 3 (g) given the following data Estimate the equilibrium constant and the position of the equilibrium for the above reaction at 427 o C

74. Problems 74 2SO 2 (g) + O 2 (g)  2SO 3 (g)

75. Problems 75 2SO 2 (g) + O 2 (g)  2SO 3 (g) Estimate the equilibrium constant and the position of the equilibrium for the above reaction at 427 o C

76. 76 Free Energy and Reversible Reactions  the change in free energy is a theoretical limit as to the amount of work that can be done  if the reaction achieves its theoretical limit, it is a reversible reaction  the change in free energy is a theoretical limit as to the amount of work that can be done  if the reaction achieves its theoretical limit, it is a reversible reaction

77. 77 Real Reactions  in a real reaction, some of the free energy is “lost” as heat  if not most  therefore, real reactions are irreversible  in a real reaction, some of the free energy is “lost” as heat  if not most  therefore, real reactions are irreversible

78. 78 ΔG under nonstandard conditions  from an earlier calculation for expansion