1. Lecture 294/16/07

2. Changes in temperature - recap Heat = (constant) x (mass) x (change in temp)

3. Changes in state Temperature stays the same during changes of state Gas/Vapor Liquid Solid ENERGY q = mass x constant q = moles x constant

4. Change of state constant??? Depends on two things: Identity of substance Which states are changing

5. Solid/Liquid Heat of fusion Solid  Liquid Endothermic ice  Water (333 J/g or 6 KJ/mol) Heat of crystallization Liquid  Solid Exothermic Water  ice (- 333 J/g or - 6 KJ/mol)

6. Liquid/Gas Heat of vaporization Liquid  Gas Endothermic water  water vapor (40.7 KJ/mol) Heat of condensation Gas  Liquid Exothermic vapor  Water (- 40.7 KJ/mol)

7. Solid/Gas Heat of sublimation Solid  Gas Endothermic Heat of deposition Gas  Solid Exothermic

8. What is the minimum amount of ice at 0 °C that must be added to 340 mL of water to cool it from 20.5°C to 0°C?

9. A rainstorm deposits 2.5 x 10 10 kg of rain. Calculate the quantity of thermal energy in joules transferred when this much rain forms. (∆H cond = - 40.7 KJ/mol) Exothermic or endothermic?

10. 1 st Law of Thermodynamics revisited ∆E = q + w Change in Energy content heat work

11. work (w) = - F x d w = - (P x A) x d w = - P∆V if ∆V = 0, then no work

12. State function property of a system whose value depends on the final and initial states, but not the path driving to Mt Washington route taken vs. altitude change ∆E is a state function q and w are not

13. Change in Enthalpy (∆H or q p ) equals the heat gained or lost at constant pressure ∆E = q p + w ∆E = ∆H + (-P∆V) ∆H = ∆E + P∆V

14. ∆E vs. ∆H Reactions that don’t involve gases 2KOH (aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + 2H 2 O (l) ∆V ≈ 0, so ∆E ≈ ∆H Reactions in which the moles of gas does not change N 2 (g) + O 2 (g)  2NO (g) ∆V = 0, so ∆E = ∆H Reactions in which the moles of gas does change 2H 2 (g) + O 2 (g)  2H 2 O (g) ∆V > 0, but often P∆V << ∆H, thus ∆E ≈ ∆H

15. Enthalpy is an extensive property Magnitude is proportional to amount of reactants consumed H 2 (g) + ½ O 2 (g)  H 2 O (g) ∆H = -241.8 KJ 2H 2 (g) + O 2 (g)  2H 2 O (g) ∆H = -483.6 KJ Enthalpy change for a reaction is equal in magnitude (but opposite in sign) for a reverse reaction H 2 (g) + ½ O 2 (g)  H 2 O (g) ∆H = -241.8 KJ H 2 O (g)  H 2 (g) + ½ O 2 (g) ∆H = 241.8 KJ Enthalpy change for a reaction depends on the state of reactants and products H 2 O (l)  H 2 O (g) ∆H = 88 KJ

16. Constant pressure calorimetry (coffee cup calorimetry) heat lost = heat gained Measure change in temperature of water 10 g of Cu at 188 °C is added to 150 mL of water in a coffee cup calorimeter and the temperature of water changes from 25 °C to 26 °C. Determine the specific heat capacity of copper.