1. THERMOCHEMISTRY

2. Definitions #1 Energy: The capacity to do work or produce heat Potential Energy: Energy due to position or composition Kinetic Energy: Energy due to the motion of the object

3. Definitions #2 Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between forms The First Law of Thermodynamics: The total energy content of the universe is constant

4. State Functions State Functions depend ONLY on the present state of the system ENERGY IS A STATE FUNCTION A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane! WORK IS NOT A STATE FUNCTION WHY NOT???

5.  E = q + w  E = change in internal energy of a system q = heat flowing into or out of the system -q if energy is leaving to the surroundings +q if energy is entering from the surroundings w = work done by, or on, the system -w if work is done by the system on the surroundings +w if work is done on the system by the surroundings

6. Work, Pressure, and Volume Expansion Compression +  V (increase) -  V (decrease) - w results+ w results E system decreases Work has been done by the system on the surroundings E system increases Work has been done on the system by the surroundings 1 L atm = 101.3 J

7. Do problem 27 – Blue book Or Do problem 29 – Purple book

8. Energy Change in Chemical Processes Endothermic: Exothermic: Reactions in which energy flows into the system as the reaction proceeds. Reactions in which energy flows out of the system as the reaction proceeds. + q system - q surroundings - q system + q surroundings

9. Endothermic Reactions

10. Exothermic Reactions

11. Enthalpy – flow of heat abbreviated H, but always represented as  H at constant pressure,  H =  E + P  V The sign of  H indicates if system is gaining or losing energy

12. Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured… …usually by the change in temperature of a known quantity of water 1 calorie is the heat required to raise the temperature of 1 gram of water by 1  C 1 BTU is the heat required to raise the temperature of 1 pound of water by 1  F

13. The Joule The unit of heat used in modern thermochemistry is the Joule 1 calorie = 4.184 joules

14. A Bomb Calorimeter

15. A Cheaper Calorimeter (aka – coffee cup calorimeter)

16. Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. 0.129 Lead (solid) 0.140 Mercury (liquid) 0.385 Copper (solid) 0.709 Carbon (graphite,solid) 0.897 Aluminum (solid) 0.129 Gold (solid) 0.449 Iron (solid) 2.01 Water (vapor) 2.09 Water (solid) 2.44 Ethanol (liquid) 4.18 Water (liquid) Specific Heat (J/g·K) Substance

17. Calculations Involving Specific Heat C or s = Specific Heat Capacity q = Heat lost or gained  T = Temperature change OR

18. Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. 0.129 Lead (solid) 0.140 Mercury (liquid) 0.385 Copper (solid) 0.709 Carbon (graphite,solid) 0.897 Aluminum (solid) 0.129 Gold (solid) 0.449 Iron (solid) 2.01 Water (vapor) 2.09 Water (solid) 2.44 Ethanol (liquid) 4.18 Water (liquid) Specific Heat (J/g·K) Substance

20. Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.” Rules: If the reaction is reversed, the sign of ∆H is changed If the reaction is multiplied, so is ∆ H

21. Hess’s Law

22. Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ Step #1: CH 4 must appear on the reactant side, so we reverse reaction #1 and change the sign on  H. CH 4  C + 2H 2 +74.80 kJ

23. Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ CH 4  C + 2H 2 +74.80 kJ Step #2: Keep reaction #2 unchanged, because CO 2 belongs on the product side C + O 2  CO 2 -393.50 kJ

24. Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ CH 4  C + 2H 2 +74.80 kJ C + O 2  CO 2 -393.50 kJ 2H 2 + O 2  2 H 2 O -571.66 kJ Step #3: Multiply reaction #2 by 2

25. Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ CH 4  C + 2H 2 +74.80 kJ C + O 2  CO 2 -393.50 kJ 2H 2 + O 2  2 H 2 O -571.66 kJ Step #4: Sum up reaction and  H CH 4 + 2O 2  CO 2 + 2H 2 O -890.36 kJ

26. Calculation of Heat of Reaction Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O  H rxn =   H f (products) -   H f (reactants)  H rxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]  H rxn = -890.36 kJ -285.83 kJH2OH2O -74.80 kJCH 4  H f CO 2 -393.50 kJ O2O2 0 kJ Substance