1. THERMOCHEMISTRY

2. Energy The ability to do work or transfer heat.The ability to do work or transfer heat. –Work: Energy used to cause an object that has mass to move. –Heat: Energy used to cause the temperature of an object to rise.

3. 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

4. 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

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 problems Chapter 5 5.255.25 5.27 A and B5.27 A and B 5.31 All5.31 All

7. 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

8. The Joule The unit of heat used in modern thermochemistry is the Joule 1 joule = 4.184 calories

9. A Bomb Calorimeter

10. A Cheaper Calorimeter

11. Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Substance Specific Heat (J/g·K) Water (liquid)4.18 Ethanol (liquid) 2.44 Water (solid) 2.06 Water (vapor) 1.87 Aluminum (solid) 0.897 Carbon (graphite,solid) 0.709 Iron (solid) 0.449 Copper (solid) 0.385 Mercury (liquid) 0.140 Lead (solid) 0.129 Gold (solid) 0.129

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

13. Problems 5.53 a and B5.53 a and B

14. 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???

15. State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

16. State Functions However, we do know that the internal energy of a system is independent of the path by which the system achieved that state.However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. –In the system below, the water could have reached room temperature from either direction.

17. State Functions Therefore, internal energy is a state function.Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state.It depends only on the present state of the system, not on the path by which the system arrived at that state. And so,  E depends only on E initial and E final.And so,  E depends only on E initial and E final.

18. State Functions However, q and w are not state functions.However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its  E is the same.Whether the battery is shorted out or is discharged by running the fan, its  E is the same. –But q and w are different in the two cases.

19. Work When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

20. Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = −P  V

21. 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

22. 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

23. Endothermic Reactions

24. Exothermic Reactions

25. Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume:Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV At constant pressure and volume the change in enthalpy is the heat gained or lost  H = q

26. Enthalpies of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants

27. Enthalpies of Reaction This quantity,  H, is called the enthalpy of reaction, or the heat of reaction.

28. Problems 5.57 A and B

29. 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.”

30. Hess’s Law

31. 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

32. 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

33. 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 Step #3: Multiply reaction #2 by 2 2H 2 + O 2  2 H 2 O -571.66 kJ

34. 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

35. 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) Substance  H f CH 4 -74.80 kJ O2O2 0 kJ CO 2 -393.50 kJ H2OH2O-285.83 kJ  H rxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]  H rxn = -890.36 kJ