1. 1 Chapter 6 EnergyThermodynamics

2. 2 Energy is... n The ability to do work. n Conserved. n made of heat and work. n a state function. n independent of the path, or how you get from point A to B. n Work is a force acting over a distance. n Heat is energy transferred between objects because of temperature difference.

3. 3 The universe n is divided into two halves. n the system and the surroundings. n The system is the part you are concerned with. n The surroundings are the rest. n Exothermic reactions release energy to the surroundings. n Endo thermic reactions absorb energy from the surroundings.

4. 4 Potential energy Heat

5. 5 Potential energy Heat

6. 6 Direction n Every energy measurement has three parts. 1. A unit ( Joules of calories). 2. A number how many. 3. and a sign to tell direction. n negative - exothermic n positive- endothermic

7. 7 System Surroundings Energy E <0

8. 8 System Surroundings Energy E >0

9. 9 Same rules for heat and work n Heat given off is negative. n Heat absorbed is positive. n Work done by system on surroundings is positive. n Work done on system by surroundings is negative. n Thermodynamics- The study of energy and the changes it undergoes.

10. 10 First Law of Thermodynamics n The energy of the universe is constant. n Law of conservation of energy. n q = heat n w = work E = q + w E = q + w n Take the systems point of view to decide signs.

11. 11 What is work? n Work is a force acting over a distance. w= F x d w= F x d n P = F/ area n d = V/area w= (P x area) x (V/area)= P V w= (P x area) x (V/area)= P V n Work can be calculated by multiplying pressure by the change in volume at constant pressure. n units of liter - atm L-atm

12. 12 Work needs a sign n If the volume of a gas increases, the system has done work on the surroundings. n work is negative w = - P V w = - P V n Expanding work is negative. n Contracting, surroundings do work on the system w is positive. n 1 L atm = 101.3 J

13. 13 Examples n What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? n If 2.36 J of heat are absorbed by the gas above. what is the change in energy? n How much heat would it take to change the gas without changing the internal energy of the gas?

14. 14 Enthalpy n abbreviated H n H = E + PV (thats the definition) n at constant pressure. H = E + P V H = E + P V n the heat at constant pressure q p can be calculated from E = q p + w = q p - P V E = q p + w = q p - P V q p = E + P V = H q p = E + P V = H

15. 15 Calorimetry n Measuring heat. n Use a calorimeter. n Two kinds n Constant pressure calorimeter (called a coffee cup calorimeter) n heat capacity for a material, C is calculated C= heat absorbed/ T = H/ T C= heat absorbed/ T = H/ T n specific heat capacity = C/mass

16. 16 Calorimetry n molar heat capacity = C/moles heat = specific heat x m x T heat = specific heat x m x T heat = molar heat x moles x T heat = molar heat x moles x T n Make the units work and youve done the problem right. A coffee cup calorimeter measures H. A coffee cup calorimeter measures H. n An insulated cup, full of water. n The specific heat of water is 1 cal/gºC Heat of reaction= H = sh x mass x T Heat of reaction= H = sh x mass x T

17. 17 Examples n The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K. n A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

18. 18 Calorimetry n Constant volume calorimeter is called a bomb calorimeter. n Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. n The heat capacity of the calorimeter is known and tested. Since V = 0, P V = 0, E = q Since V = 0, P V = 0, E = q

19. 19 Bomb Calorimeter n thermometer n stirrer n full of water n ignition wire n Steel bomb n sample

20. 20 Properties n intensive properties not related to the amount of substance. n density, specific heat, temperature. n Extensive property - does depend on the amount of stuff. n Heat capacity, mass, heat from a reaction.

21. 21 Hesss Law n Enthalpy is a state function. n It is independent of the path. We can add equations to to come up with the desired final product, and add the H We can add equations to to come up with the desired final product, and add the H n Two rules If the reaction is reversed the sign of H is changed If the reaction is reversed the sign of H is changed If the reaction is multiplied, so is H If the reaction is multiplied, so is H

22. 22 N2N2 2O 2 O2O2 NO 2 68 kJ NO 2 180 kJ -112 kJ H (kJ)

23. 23 Standard Enthalpy n The enthalpy change for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions) Symbol Hº Symbol Hº n When using Hesss Law, work by adding the equations up to make it look like the answer. n The other parts will cancel out.

24. 24 Hº= -394 kJ Hº= -286 kJ Example Given calculate Hº for this reaction Given calculate Hº for this reaction Hº= -1300. kJ

25. 25 Example Given Calculate Hº for this reaction Hº= +77.9kJ Hº= +495 kJ Hº= +435.9kJ

26. 26 Standard Enthalpies of Formation n Hesss Law is much more useful if you know lots of reactions. n Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states. n Standard states are 1 atm, 1M and 25ºC n For an element it is 0 n There is a table in Appendix 4 (pg A22)

27. 27 Standard Enthalpies of Formation n Need to be able to write the equations. n What is the equation for the formation of NO 2 ? ½N 2 (g) + O 2 (g) NO 2 (g) ½N 2 (g) + O 2 (g) NO 2 (g) n Have to make one mole to meet the definition. n Write the equation for the formation of methanol CH 3 OH.

28. 28 Since we can manipulate the equations n We can use heats of formation to figure out the heat of reaction. n Lets do it with this equation. C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O n which leads us to this rule.

29. 29 Since we can manipulate the equations n We can use heats of formation to figure out the heat of reaction. n Lets do it with this equation. C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O n which leads us to this rule.