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Lecture 324/19/06. Section 1 Equilibrium Le Chatelier’s Solubility Section 2 Acid/Base equilibria pH Buffers Titration Section 3 Oxidation numbers Balancing.

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Presentation on theme: "Lecture 324/19/06. Section 1 Equilibrium Le Chatelier’s Solubility Section 2 Acid/Base equilibria pH Buffers Titration Section 3 Oxidation numbers Balancing."— Presentation transcript:

1 Lecture 324/19/06

2 Section 1 Equilibrium Le Chatelier’s Solubility Section 2 Acid/Base equilibria pH Buffers Titration Section 3 Oxidation numbers Balancing Redox reactions Galvanic cells Standard reduction potential table

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

5 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 vap = - 44 KJ/mol) Exothermic or endothermic?

6 1 st Law of Thermodynamics revisited ∆E = q + w

7 State function property of a system whose value depends on the final and initial states, but not the path

8 work

9 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

10 ∆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) Reactions in which the moles of gas does not change N 2 (g) + O 2 (g)  2NO (g) Reactions in which the moles of gas does change 2H 2 (g) + O 2 (g)  2H 2 O (g)

11 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 = ? 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 = ? Enthalpy change for a reaction depends on the state of reactants and products H 2 O (l)  H 2 O (g) ∆H = 88 KJ

12 Bond enthalpies

13 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 cofee cup calorimeter and the temperature of water changes from 25 °C to 26 °C. Determine the specific heat capacity of copper.


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