Principles of Reactivity: Energy and Chemical Reactions

1. AP Notes Chapter 5 Principles of Reactivity: Energy and Chemical Reactions

2. Thermodynamics • study of energy transfer or heat flow • Energy • Kinetic Energy • Thermal - Heat • Mechanical • Electrical • Sound • Potential Energy • Chemical • Gravitational • Electrostatic

3. Energy is... • The ability to do work. • Conserved. • made of heat and work. • Work is a force acting over a distance. W=F x d • Power is work done over time. P=W t • Heat is energy transferred between objects because of temperature difference.

4. First Law of Thermodynamics • Law of conservation of energy • Total energy of the universe is constant • Temperature and heat • Heat is not temperature • Thermal energy = particle motion • Total thermal energy is sum of all a materials individual energies

5. The Universe • is divided into two halves. • the system and the surroundings. • The system is the part you are concerned with. • The surroundings are the rest. q into system = -q from surroundings

6. System: region of space where process occurs Surroundings: region of space around system

7. Thermodynamic Properties 1. State functions -depend on nature of process NOT method -path independent -denoted by capital letters E → state function

8. Thermodynamic Properties 2. Path functions -depends on how process is carried out -path dependent -denoted by lower case q & w → path functions

9. Work →w w = -  (PV) w = - P  V - V  P Constant P wP = - P  V Constant V wV = - V  P

10. Work →w w = - P  V assume ideal piston (i.e. no heat loss) w =  E  E = - P  V in isothermics w = -q

11. PV = nRT R=0.0821L atm/mol K In isothermic conditions -q = w = -nRT R=8.314 J/mol K

12. DU = q + w In isothermics DU = 0 so q=-w DU = q - PDV In constant V or constant P w=0 so DU = q

13. Calorie - Amount of heat needed to raise the temperature of 1 gram H20, 1 degree centigrade Nutritional calorie [Calorie]= 1 Kcal SI unit → Joule 1 cal = 4.184 J 1000cal = 1Kcal=4.184KJ

14. Direction of energy flow • Every energy measurement has three parts. • A unit ( Joules or calories). • A number how many. • and a sign to tell direction. • negative - exothermic • positive- endothermic • No change in energy - isothermic

15. Factors Determining Amount of Heat • Amount of material • Temperature change • Heat capacity

16. Specific Heat Amount of heat needed to raise the temperature of 1 gram of material 1degree centigrade Heat = q q = m CpT where Cp = Heat capacity T = T2 – T1

17. CP (J/mol.oC) H2O(s) = 37.11 H2O(l) = 75.35 H2O(g) = 33.60

18. What amount of heat is needed to just boil away 100. mL of water in a typical lab?

19. Surroundings System Energy DE <0

20. Exothermic reactions release energy to the surroundings. q < 0 (-) Exothermic

21. Heat Potential energy

22. Surroundings System Energy DE >0

23. Endothermic reactions absorb energy from the surroundings. q > 0 (+) Endothermic

24. Heat Potential energy

25. 1. How much heat is absorbed when 10.0 g of H20 are heated from 200C to 250C?

26. 2. How much heat is needed to completely vaporize 10.0 g of H2O at its boiling point?

27. Heat of Vaporization Hvap(H2O) = 40.66 kJ/mol Heat of Fusion Hfus(H2O) = 6.01 kJ/mol

28. Energy processes: • Within a phase • q=mHv or mHf • Between phases • q=mCpT

30. 3. How much total heat is used to raise the temperature of 100.0 g H2O from-15o to 125oC?

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

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

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

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

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

36. Calorimetry • Measuring heat. • Use a calorimeter. • Two kinds • Constant pressure calorimeter (called a coffee cup calorimeter) • heat capacity for a material, C is calculated • C= heat absorbed/ DT = DH/ DT • specific heat capacity = C/mass

37. Calorimetry • molar heat capacity = C/moles • heat = specific heat x m x DT • heat = molar heat x moles x DT • Make the units work and you’ve done the problem right. • A coffee cup calorimeter measures DH. • An insulated cup, full of water. • The specific heat of water is 1 cal/gºC • Heat of reaction= DH = sh x mass x DT

39. Bomb Calorimeter

40. Calorimeter Constant heat capacity that is constant over a temperature range where q(cal) = CC x TC (CC = calorimeter constant)

41. 4. Determine the calorimeter constant for a bomb calorimeter by mixing 50.0 mL of water at 25oC with 50.0 mL of water at 60.0 oC. The final temp. attained is 40oC.

42. 5. What is the molar heat of combustion of sulfur, if 2.56 grams of solid flowers of sulfur are burned in excess oxygen in the bomb calorimeter in #4?

43. 6. Determine the molar heat of solution of LiOH. HINT: Find the calorimeter constant first.

44. 7. The heat was absorbed by 815 g of water in calorimeter which had an initial temp of 21.25 C. After equilibrium was reached, the final temperature was 26.72 C.

45. Examples • 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. • 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?

46. Calorimetry • Constant volume calorimeter is called a bomb calorimeter. • 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. • The heat capacity of the calorimeter is known and tested. • Since DV = 0, PDV = 0, DE = q

47. Bomb Calorimeter • thermometer • stirrer • full of water • ignition wire • Steel bomb • sample

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

49. Enthalpy • abbreviated H • H = E + PV (that’s the definition) PV = w • at constant pressure. • DH = DE + PDV • the heat at constant pressure qp can be calculated from • DE = qp + w = qp - PDV • qp = DE + P DV = DH

50. Hess’s Law • Enthalpy is a state function. • It is independent of the path. • We can add equations to come up with the desired final product, and add the DH • Two rules • If the reaction is reversed the sign of DH is changed • If the reaction is multiplied, so is DH