1. Question of the Day: 1. Is an endo or exo reaction more likely to be spontaneous? 2. Use the diagram below: If 1 mol of water is produced in the combustion, how much heat is involved? Day 1 2-3

2. How would you diagram the reaction below? CaO(s) + H 2 O(l) → Ca(OH) 2 (s) + 65.2 kJ CaO (s) + H 2 O (l) H + – 65.2 Ca(OH) 2(s)

3. http://wps.prenhall.co m/esm_brown_chemist ry_9/2/660/169060.cw/i ndex.html Homework # 2 Chapter 5 – w/ discussion partner – show me successful screen (100%)

4. 4 Predicting Spontaneity H 2 O ( l ) → H 2 O (g ) ∆H = + 40.7 kJ; 1 atm, 100°C

5. 5 Predicting Spontaneity CaCO 3(s) → CaO (s) + CO 2(g) ∆H = +178.0 kJ; 298 K vs. 1100 K

6. 6 Predicting Spontaneity

7. 7 Entropy Entropy is the measure of the disorder or randomness of a system. Entropy is a state function. Entropy changes follow Hess’ Law. A system has high entropy if it _______________________________________ Is Very dsorigniaezd (Very disorganized) Has freedom of motion

8. 8 Entropy & Spontaneity Reactions will be spontaneous if ________ is increased. Examples ice cubes melting vs. putting water in freezer to make ice cubes room gets messy vs. doing work to put things in order etnopry

9. 9 Standards & Constants Conditions & Units T = 25 o C = 298K P = 1 atm Units: J/K = Joules/Kelvin What is the relationship between entropy and temperature? As T  0 K, S  0 T ↓, S ↓ and T ↑, S ↑

10. 10 Standards & Constants Why are the entropy values for elements and compounds always positive? T always > 0K Therefore everything is always moving How do entropies compare among the phases of matter? S o solid < S o liquid < S o gas

11. 11 Standard Entropy Changes The equation ∆ S o =  S o products –  S o reactants Example: 2HCI (aq) + 2Ag (s) → 2AgCl (s) + H 2(g) 1 atm, 25 o C Predict if this reaction will have + ∆S o or – ∆S o. Calculate ∆S o.

12. Question of the Day: 1. Use the diagram below: If 36 grams of water is produced in the combustion, how much heat is involved? Day 2 2-6

13. Question of the Day: 1. If 32.6 kJ of heat are given off, how many grams of calcium hydroxide are produced? Day 3 2-7 CaO(s) + H 2 O(l) → Ca(OH) 2 (s) + 65.2 kJ 37 grams

14. 14 Standard Entropy Changes The equation  S o =  nS o products –  mS o reactants Example: 2 H 2(g) + O 2(g) → 2 H 2 O (g) 1 atm, 25 o C Predict if this reaction will have + ∆S o or – ∆S o. Calculate ∆S o.

15. 15 Standard Entropy Changes Example: 2HCI (aq) + 2Ag (s) → 2AgCl (s) + H 2(g) 1 atm, 25°C a. Predict if this rxn will have + ∆S° or – ∆S°. b.Calculate ∆S°.

16. 16 Entropy and Molecular Motion Molecules always have positive entropy values because molecules are always moving. How do molecules move? Rotation Translation Vibration -an atom spins around a single bond -entire molecule spins -entire molecule moves in one direction -atoms in molecule get closer together or further apart along a bond

17. ASSIGNMENT: READ section 18.5 and complete #s 46-53

18. C(s, diamond) → C(s, graphite) Hess’s Law Although the enthalpy change for this reaction cannot be measured directly, you can use Hess’s law to find the enthalpy change for the conversion of diamond to graphite by using the following combustion reactions. a. C(s, graphite) + O 2 (g) → CO 2 (g)ΔH = –393.5 kJ b. C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ

19. Hess’s Law Write equation a in reverse to give: c. CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = 393.5 kJ When you reverse a reaction, you must also change the sign of ΔH. a. C(s, graphite) + O 2 (g) → CO 2 (g)ΔH = –393.5 kJ b. C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ C(s, diamond) → C(s, graphite)

20. Hess’s Law If you add equations b and c, you get the equation for the conversion of diamond to graphite. C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = 393.5 kJ C(s, diamond) → C(s, graphite) b. C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ c. CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = 393.5 kJ C(s, diamond) → C(s, graphite)

21. Hess’s Law If you also add the values of ΔH for equations b and c, you get the heat of reaction for this conversion. C(s, diamond) → C(s, graphite) C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = 393.5 kJ C(s, diamond) → C(s, graphite)ΔH = –1.9 kJ

22. Hess’s Law C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = 393.5 kJ C(s, diamond) → C(s, graphite)ΔH = –1.9 kJ

23. How can you determine ΔH for the conversion of diamond to graphite without performing the reaction? CHEMISTRY & YOU

24. How can you determine ΔH for the conversion of diamond to graphite without performing the reaction? CHEMISTRY & YOU You can use Hess’s law by adding thermochemical equations in which the enthalpy changes are known and whose sum will result in an equation for the conversion of diamond to graphite.

25. Suppose you want to determine the enthalpy change for the formation of carbon monoxide from its elements. Carrying out the reaction in the laboratory as written is virtually impossible. Hess’s Law Another case where Hess’s law is useful is when reactions yield products in addition to the product of interest. C(s, graphite)+ O 2 (g) → CO(g)ΔH = ? 1 2

26. Hess’s Law You can calculate the desired enthalpy change by using Hess’s law and the following two reactions that can be carried out in the laboratory: C(s, graphite) + O 2 (g) → CO 2 (g)ΔH = –393.5 kJ CO 2 (g) → CO(g) + O 2 (g)ΔH = 283.0 kJ C(s, graphite)+ O 2 (g) → CO(g)ΔH = –110.5 kJ 1 2 1 2

27. Hess’s Law C(s, graphite) + O 2 (g) → CO 2 (g)ΔH = –393.5 kJ CO 2 (g) → CO(g) + O 2 (g)ΔH = 283.0 kJ C(s, graphite)+ O 2 (g) → CO(g)ΔH = –110.5 kJ 1 2 1 2

28. According to Hess’s law, it is possible to calculate an unknown heat of reaction by using which of the following? A.heats of fusion for each of the compounds in the reaction B.two other reactions with known heats of reaction C.specific heat capacities for each compound in the reaction D.density for each compound in the reaction

29. According to Hess’s law, it is possible to calculate an unknown heat of reaction by using which of the following? A.heats of fusion for each of the compounds in the reaction B.two other reactions with known heats of reaction C.specific heat capacities for each compound in the reaction D.density for each compound in the reaction

30. 30 Gibbs Free Energy Gibbs free energy is the definitive method for determining the spontaneity of a reaction. Gibbs free energy incorporates both enthalpy and entropy. ∆G, the change in free energy, is the driving force of a chemical reaction. Free energy is the energy that a system can release through heat and disorder. The Gibbs-Helmholtz equation  G =  H - T  S

31. 31 Why does the Gibbs- Helmholtz equation work? The role of ∆H If -  H : heat out & spontaneity indicated with - sign If  H: no sign means +  H The role of – T∆S T given in K ! always + “ – ” will change sign of ∆S so spontaneity is indicated in like manner with ∆H

32. 32 How does the value of ∆G determine spontaneity? – ∆G Spontaneous Net release of energy through heat & disorder Becomes more stable + ∆G Not spontaneous Must do work or supply energy to make rxn go Reverse reaction is spontaneous ∆G = 0 Reaction is at equilibrium Forward rate = reverse rate

33. 33 Standardize! Why? Changes in pressure and concentration affect entropy values. P = 1 atm for gases Concentration = 1M for solutions

34. 34 Calculating  G 2 H 2(g) + O 2(g) → 2 H 2 O ( l )

35. 35 Effect of Temperature on Reaction Spontaneity HH SS  G =  H - T  SRemarks –+–Spon. @ all temps +–+Not spon. @ any T ––– OR + Spon. @ low T * Not spon. @ high T ++– OR + Spon. @ high T * Not spon. @ low T * High & low temps are relative to equilibrium temp

36. 36 Finding Equilibrium ∆G = ∆H – T∆S = 0 At equilibrium, the free energy released by the spontaneous reaction is being absorbed by the reverse, nonspontaneous reaction. What is the formula for finding the temperature of equilibrium? T =  H /  S

37. 37 Finding Equilibrium What are the possibilities for the equilibrium temperature? May be an ordinary temp May be extremely high or low May be negative ! Impossible System can NEVER be in equilibrium