1. Chapter 19: Chemical Thermodynamics Tyler Brown Hailey Messenger Shiv Patel Agil Jose

2. 19.1 Spontaneous Processes

3. Spontaneous Process: A process that proceeds on its own without any outside assistance Occurs in a definite direction “Processes that are spontaneous in one direction are nonspontaneous in the other” (Pg 805 Brown and LeMay)

4. 19.1 Spontaneous Processes Reversible Process: A process in which the system is changed in a way that the system and surroundings can be restored to their original state by exactly reversing the change (can be completely restored to original condition) Reversible processes reverse direction whenever an infinitesimal change is made in some property of the system

5. 19.1 Spontaneous Processes Irreversible Process: Cannot be reversed to return system and surroundings to original state

6. 19.1 Spontaneous Processes Isothermal Process: A process with a constant temperature

7. 19.1 Spontaneous Processes First Law of Thermodynamics: Energy is conserved “Physical and chemical processes have a directional character” (Pg 804) I.e. Sodium and Chlorine come together on their own to make NaCl, but it does not decompose of its own accord.

8. 19.1 Spontaneous Processes Experimental conditions help to determine if a process is spontaneous I.e. When the temperature > 0°C in ordinary atmospheric pressure, ice melting is spontaneous. »In these conditions, liquid water turning in to ice is not spontaneous

9. 19.1 Spontaneous Processes Entropy: The ratio of heat delivered to the temperature at which it is delivered “A reversible change produces the maximum amount of work that can be achieved by the system on the surroundings (w rev = w max )” (Page 807) All spontaneous processes are IRREVERSIBLE

10. 19.1 Spontaneous Processes Tell whether or not the following process is spontaneous Ice melts at -1 degree Celsius

11. 19.1 Spontaneous Processes Tell whether or not the following process is spontaneous Ice melts at -1 degree Celsius This would not be spontaneous because Ice does not melt under 0 degrees Celsius. This reaction would be spontaneous in the reverse reaction

12. 19.1 Spontaneous Processes 19.1 Problems: 7-18

13. 19.2 Entropy and the Second Law of Thermodynamics

14. 19.3 The Molecular Interpretation of Entropy

15. Molecular motion Translational Motion – entire molecule moves in one direction (like throwing a baseball) Vibrational Motion – atoms in the molecule periodically move toward and away from one another Rotational Motion – molecules spin like a top

16. What is Statistical Thermodynamics? It’s the use of tools of statistics and probability to provide the link between macroscopic and microscopic worlds We molecules in bulk in a microstate: a single possible arrangement of the positions and kinetic energies of gas molecules in a specific thermodynamic state

17. So what equation do we use? Boltzmann created this beautiful equation: S = k lnW K is Boltzmann’s constant which is: 1.38 x 10 -23 J/K -This means entropy is the measure of how many microstates are associated with a particular macroscopic state

18. Relationship between microstates and entropy The more microstates, the more entropy Increasing volume, temperature, and number of molecules increases entropy due to larger number of microstates

19. GENERALLY SPEAKING Entropy increases when: -Gases are formed from either solids or liquids -Liquids or solutions are formed from solids -The number of gas molecules increases during a chemical reaction

20. The Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is zero S(0 K) = 0 - Makes sense because when there is no molecular motion (temp is 0 K), there is only ONE microstate

21. Trends Sharp increase of entropy at melting points Sharp increase of entropy at boiling points Entropy of solids < Entropy of liquids < Entropy of gas

22. Example What will the sign of change of entropy be for the following reactions? 2Na (s) + Cl 2 (g) 2NaCl (s) 2H 2 (g) + O 2 (g) 2H 2 O (l)

23. Answers Because of Boltzmann’s equation, we know that the more particles there are in a system, the more microstates there are, so the more entropy there is, so: The reaction to form NaCl has a negative sign for change in entropy, since there are less particles after the reaction The reaction to form water also has a negative sign for change in entropy since there are less particles

24. Example In a chemical reaction two gases combine to form a solid. What do you expect for the sign of change in S? For which of the processes does the entropy of the system increase? A)The melting of ice cubes at -5 degrees Celsius and 1 atm pressure B) dissolution of sugar in a cup of hot coffee C) the reaction of nitrogen atoms to form N 2

25. Answer Entropy decreases when two gases form a solid, so negative sign A) Positive because the ice is melting from a structured solid to unstructured liquid B) Positive because the sugar went from being structured solids to numerous separated particles C) Negative, there is a huge decrease in the number of gas particles (by 1/2 to be exact)

26. Practice, practice, practice! Page 838 #’s 19.27 – 19.40 A closer look on page 810

27. 19.4 Entropy Changes in Chemical Reactions

28. How do we do it? We can’t measure change in entropy for a reaction, but we CAN find the absolute value of the entropy for many substances at any temperature Standard molar entropies are denoted by S ° (pronounced S knot) for any pure substance in their standard state at 1 atm

29. Observations on Standard Entropy Standard molar entropies (SME) of elements at reference temperature 298 K are not zero SME of gases greater than liquids SME generally increase with molar mass SME generally increase with number of atoms in molecular formula

30. So what’s the equation we use? ΔS° = ΣnS(products) – ΣnS(reactants) In english Change in entropy equals sum of entropies of products minus the sum of entropies of reactants

31. Entropy Changes in the Surroundings The change in entropy of the surroundings will depend on how much heat is absorbed or given off by the system Isothermal process equation: ΔS surr =-q sys /T - At constant pressure, q sys is the enthalpy change of the system, ΔH

32. Example Calculate change in standard entropy for the following reaction 2Na (s) + Cl 2 (g) 2NaCl (s)

33. Answer 1. The standard entropy change is equal to the standard entropy of products minus the reactants 2. So we must first find the standard entropy of the products(using a standard entropy table), and multiply them by the number of moles in the reaction Standard entropy of NaCl = 72.1 J/K-mol, so 2(72.1)= 144.2 3. Then we find the standard entropy of reactants Standard entropy of Na = 51.2 J/K-mol, so 2(51.2)= 102.4 J/K-mol Standard entropy of Cl 2 = 223.1 J/K-mol, 223.1 J/K-mol 4. Now we subtract product and reactants 144.2 J/K-mol – (102.4 J/K-mol + 223.1 J/K-mol) = - 181.3 J/K-mol

34. Example Cyclopropane and propylene isomers both have the formula C 3 H 6. Based on the molecular structures shown, which of these isomers would you expect to have the higher standard molar entropy at 25 degrees Celsius? CyclopropanePropylene

35. Answer Propylene because it is much less organized compared to Cyclopropane

36. We love to practice! Page 839 #’s 19.41 – 19.48 Example on page 821, figure 19.5

37. 19.5 Gibbs Free Energy

38. Standard Free Energies of Formation: Used to calculate standard free-energy change for chemical processes

39. 19.5 Gibbs Free Energy Equation for Standard Free-Energy Change: ΔG° = ΣnΔG° f ( products) - ΣmΔG° f (reactants)

40. 19.5 Gibbs Free Energy G = H – TS T is the absolute temperature. The change in free energy of a system (ΔG) is given by: ΔG= ΔH – TΔS

41. 19.5 Gibbs Free Energy If ΔG is negative, the reaction is spontaneous If ΔG is zero, then reaction is at equilibrium If ΔG is positive, the forward reaction is not spontaneous, but the reverse reaction is.

42. 19.5 Gibbs Free Energy For a certain chemical reaction, ΔH = - 35.4 kJ and ΔS = - 85.5 J/K. Calculate ΔG for the reaction at 298 K.

43. 19.5 Gibbs Free Energy For a certain chemical reaction, ΔH = - 35.4 kJ and ΔS = - 85.5 J/K. Calculate ΔG for the reaction at 298 K (19.51 C) ΔG = ΔH –TΔS ΔG = - 35.4 kJ – (298K)(-85.5 J/K) ΔG = - 35.4 kJ + 25500 J ΔG = -35.4 kJ + 25.5 kJ ΔG = -9.9 kJ

44. 19.5 Gibbs Free Energy 19.5 Problems: 49 -70

45. 19.6 Free Energy and Temperature

46. The equation ΔG=ΔH+(–TΔS) in which ΔH is the enthalpy term and –TΔS is the entropy term, can be used to demonstrate how free energy is affected by the change in temperature. The value of –TΔS depends directly on the absolute temperature, T, ΔG will vary with the temperature.

47. 19.6 Free Energy and Temperature T is a positive number at all temperatures other than absolute zero. Both the enthalpy and entropy can be either positive or negative. When ΔS is positive, which means greater randomness than the original state, then the TΔS is negative. When the ΔS is negative, then the TΔS is positive.

48. 19.6 Free Energy and Temperature ΔG will always be negative when the process is spontaneous, and the opposite is true when the process is not spontaneous.

49. 19.6 Free Energy and Temperature H 2 O (s) → H 2 O (l) ΔH>0 ΔS>0 This process is endothermic, meaning the ΔH is positive. Entropy also increases with this process, meaning that ΔS is positive as well, which makes –TΔS a negative value. This means that the –TΔS term dominates, making ΔG negative. A negative ΔG means that this process is spontaneous at T>0 º C

50. 19.6 Free Energy and Temperature Calculate ΔG for 2NO 2(g) → N 2 O 4(g) at 298K using appendix C.

51. 19.6 Free Energy and Temperature ΔH 2NO 2(g) → N 2 O 4(g) = -58.02 kJ/mol ΔS 2NO 2(g) → N 2 O 4(g) = -175.79 J/K

52. 19.6 Free Energy and Temperature ΔG º = -58.02kJ – (298K)(-175.79J/K)(1kJ/1000J) -58.02kJ – (298K)(-0.176kJ/K) -58.02kJ – (-52.45kJ) ΔG º = -5.57kJ This means that at 298K, 2NO 2(g) → N 2 O 4(g) is a spontaneous reaction.

53. 19.6 Free Energy and Temperature Textbook problems: chapter 19, 65,66,79. 80, and 87

54. 19.7 Free Energy and the Equilibrium Constant

55. Most reactions occur in nonstandard conditions To find the free energy in nonstandard reactions, we use the standard free energy change to calculate ΔG = ΔG° + RT ln(Q) R is ideal gas constant = 8.314 J/mol-K T is absolute temperature Q is reaction quotient for that particular mixture of interest

56. Some more info At equilibrium ΔG = 0 Q = equilibrium constant K

57. Trends The more negative ΔG°, is the larger K is The more positive ΔG° is, the smaller K is

58. Example 2CO (g) + O 2 (g) 2CO 2 (g) Calculate the change in free energy for the above reaction at 298K, for a reaction mixture that consists of 1.0 atm O 2, 2.0 atm CO, and 0.75 atm CO 2

59. Answer 1.First we must find Q in order to use the equation provided earlier. So we solve for the reaction quotient: Q=(P CO2 ) 2 / (P CO ) 2 (P O2 )=(.75 atm) 2 / (2.0 atm) 2 (1.0 atm) =.14atm 2. Now calculate standard change in free energy for reaction using tables Products – Reactants, so 2(-394.4kJ/mol) – (2(-137.2kJ/mol) + 0) Change in standard free energy is -514.4 kJ/mol 3. Now we can finally use ΔG = ΔG° + RT ln(Q) ΔG = (-514.4 kJ/mol) + (8.314 J/mol-K)(298 K) (1 kJ/1000J)ln(.14) = (-514.4 kJ/mol) – 4.871 kJ/mol = -519.271 = - 520 kJ/mol

60. Practice Problems Page 841 Problems 19.71 – 19.95 You can find answers on pages A-24 thru A-25

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