1. Question of the Day: 1. A __ enthalpy and __ entropy are good indicators that a reaction is probably spontaneous. Day 3 3-5

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

3. 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 4 3-6

4. 4 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 (NEED STANDARD VALUE HANDOUTS).

5. 5 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.

6. 6 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

7. Question of the Day: 1. If 32.6 kJ of heat are given off, how many grams of calcium hydroxide are produced? Day 5 3-7 CaO(s) + H 2 O(l) → Ca(OH) 2 (s) + 65.2 kJ 37 grams

8. Hess’s Law (a new application) Hess’s law of heat summation states that if you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

9. 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

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

11. 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)

12. 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

13. 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

14. 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

15. 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

16. 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

18. All lab reports due Thursday 3-14. Period 1 Presentations due Wednesday 3-13. Period 3 Presentations due Thursday 3-14.

19. 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 kJ Ca(OH) 2(s)

20. 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%)

21. Question of the Day: Explain Hess’s law to the best of your current ability. Day 5 2-9

22. 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