1. Thermodynamics Chapter 15

2. Part I Measuring Energy Changes

3. Thermodynamic Terms System – Surroundings Thermodynamic State of System – noting temperature, pressure, composition, and STATE of the system! (aq) (s) (l) (g) State Function -

4. State Function - Change in energy of the system

5. The First Law of Thermodynamics The First Law of thermodynamics – The amount of energy in the universe is constant Law of Conservation of Energy – Energy can neither be created nor destroyed in ordinary chemical and physical changes Our challenge is to predict which reactions will proceed spontaneously or need a push and which reactions will release energy and which will absorb energy

6. Enthalpy Changes Heat – form of energy that flows between samples of matter due to a temperature difference Enthalpy Change – the quantity of heat (enthalpy) transferred into or out of a system as it undergoes a chemical or physical change at constant pressure

7. Enthalpy Changes (  H) Enthalpy of a system depends on the KE and PE of the system (internal energy) Affected by pressure and temperature and volume Heat cannot be measured directly, only enthalpy changes:  H = H Final - H Initial  H = H substances produced – H Substances Consumed Exothermic  H < 0 Endothermic  H > 0

8. Calorimetry The system is the reactants and products (liquids and solids work the best) The surroundings is the calorimeter and the water Assume pressure is constant, no gaseous products.

9. Calorimetry Energy is neither created nor destroyed…so it must go somewhere when released. Heat released from chemical reaction or cooling object is absorbed by water in the cup. Q = m c  T m c  T = m c  T

10. Bomb Calorimetry Calorimetry – observing temperature change when a system absorbs or releases energy. Bigger devise with a bomb chamber inside a water reservoir, usually measuring combustion reactions. http://www.youtube.com/watch?v=ohyA9a mFfsc http://www.youtube.com/watch?v=ohyA9a mFfsc

11. Calibrating the Calorimeter It is silly to think that no heat is lost to the environment, and to the calorimeter It is necessary to calculate the specific heat capacity of the calorimeter by delivering a known quantity of heat and measuring  T (  H Rxn) = (  H calorimeter) + (  H solution)

12. Calibrating the Calorimeter (EX1) Determining Heat Capacity of Calorimeter 6.850KJ of heat were added to a calorimeter with 100.g H 2 O. The temperature rises from 25.00 o C to 37.54 o C Find Cp in J/ o C

13. Calorimetry after Calibration (EX 2) Determining associated  H of reaction 25.00mL of 0.500M NaOH and 25.00mL of 0.600M CH 3 COOH, both substances start at 23.00oC and the reaction ends at a maximum temperature of 25.947oC. How much heat is lost from the reaction? What is the  H/mole reaction?

14. Part II Thermochemical Equations Writing Equations and using  H to calculate energy

15. Thermochemical Equations A balanced chemical equation with a value for  H Interpret coefficients as # of moles, not # of molecules, so fractional coefficients are ok When stoichiometric amounts react, it is called one mole of reaction (mol rxn)

16. Thermochemical Equations C 2 H 5 OH (l)+ 3O 2 (g)  2CO 2 (g) + 3H 2 O (l) + 1367KJ  H = -1367KJ/mole rxn. The physical states are important to note because physical changes involve heat changes.

17. Thermochemical Equations (EX 3)  H of formation for H 2 SO 4 (l) is -814.0KJ/mole rxn. Write a thermochemical equation for which  H rxn is -814.0kJ/mole

18. Thermochemical Equations (EX 4) Find  Hf/mole for the following reaction at 25 o C and 1atm: P 4 (g)+ 6Cl 2 (g)  4PCl 3 (g)

19. Thermochemical Equations (EX 5) Calculate the change in enthalpy (  H) for the reaction of 23.0g of CaO at 25 o C and 1atm. CaO (s) + H 2 O (l)  Ca(OH) 2 (s)  H =-65.3kJ/mole

20. Part III Using Thermochemical Equations for Predicting Unknown Reactions

21. Hess’s Law The enthalpy change of a reaction is the same whether it occurs by one step or any series of steps (state function) Allows for calculation of  H for difficult to measure reactions

22. Hess’s Law – (EX6) From the following enthalpies of reaction, CaCO 3 (s)  CaO (s) + CO 2 (g)  H = 178.1 CaO(s) + H 2 O (l)  Ca(OH) 2 (s)  H = -65.3 Ca(OH) 2 (s)  Ca 2+ (aq) + 2OH 1- (aq)  H = -16.2 Calculate  Hrxn for: Ca 2+ (aq) + 2OH 1- (aq) + CO 2 (g)  CaCO 3 (s) + H 2 O (l)  H = ?

23. Hess’s Law (EX7)  H o rxn =  H o f products -  H o f reactants Calculate  H o rxn using  H o f data for: SF 6 (g) + 3H 2 O(l)  6HF(g) + SO 3 (g)

24. Hess’s Law (EX8) Can be used to calculate the heat of formation if  H o rxn is known Calculate the heat of formation of Fe 3 O 4 if the thermite reaction has a  H o rxn of - 3350kJ/mole 8Al (s) + 3Fe 3 O 4 (s)  4Al 2 O 3 (s) + 9Fe(s)

25. Hess’s Law – Bond Energies Bond Energy – the energy to break a chemical bond or separate atoms in the gaseous phase HCl (g) + 432kJ  H (g) + Cl (g)

26. Hess’s Law – Bond Energies Greater Bond Energy = Greater Bond Stength (always + values) Bond energy is an approximation of bond enthalpy Can use bond energies to estimate thermodynamic values not readily available, but ONLY FOR GAS PHASE!

27. Hess’s Law – Bond Energies (EX9)  H o rxn =  BE reactants -  BE products Use known bond energy values to calculate the heat of reaction for: CH 4 + Cl 2  CH 3 Cl + HCl See page 610 for posted values

28. Changes in internal energy (  E) Internal energy is the sum of all the energy in the system Kinetic energy of the particles + potential energy of the bonds = total energy Cannot be measured directly, only changes (just like enthalpy)  E = E products – E reactants

29. Changes in internal energy (  E)  E = q + w W = -P  V P  V =  nRT W = -  nRT  if  n = 0, work is not done If  n = +, work is –, Work done by the system If  n = -, work is + Work done by the surroundings

30. Changes in internal energy (  E) CO (g) + H 2 O (g)  H 2 (g) + CO 2 (g) 2 Moles  2 moles  n = 0 Work = 0

31. Changes in internal energy (  E) Zn (s) + 2HCl (aq)  H 2 (g) + ZnCl 2 (aq) 0 Moles  1 moles  n = + Work = -, done by system (system expands)

32. Changes in internal energy (  E) N 2 (g) + 3H2 (g)  2NH 3 (g) 4 Moles  2 moles  n = - Work = +, done by surroundings (system is compressed by surroundings)

33. Changes in internal energy (  E)  E = q + w + q = heat absorbed by system - q = heat released + w = work done on system (compression) - w = work done by system (expansion)

34.  H &  E  H =  E + PV  H = q + w + P  V Useful for physical changes involving changes in volume w = -P  V  at constant pressure and temperature  H = q Works for reactions with no change in moles

35.  H &  E For other reactions where the number of moles of gas changes, the correction factor cannot be ignored  H =  E + P  V P  V =  nRT  H =  E +  nRT Better for chemical processes where there is a change in moles of a gas

36.  H &  E In a bomb calorimeter,  v is not allowed, therefore no work can be done so  H =  E In the standard environment,  V cannot be ignored,  work term cannot be ignored

37.  H &  E (EX10) Calculate the  E and work for the combustion of n-pentane at 25oC.  H = -3523kJ/mole rxn.

38. Part IV Spontaneity

39. Spontaneous – occurs under specified conditions without any continued outside influence Implies that the products are thermodynamically more stable than the products States nothing about the speed of the change

40. Spontaneity vs Enthalpy Not all exothermic changes are spontaneous Not all spontaneous changes are exothermic EX. The melting of water is spontaneous at temperatures over 0 o C and is endothermic

41. 2 nd Law of thermodynamics The universe moves toward a state of greater disorder Spontaneity is favored when heat is released during a change Spontaneity is favored when the change creates an increase in disorder

42. Spontaneity (  S) +  S = Spontaneity is favored -  S = Spontaneity is not favored (  S) universe = (  S) system + (  S) surroundings

43. Entropy Entropy – a measure of the disorder of the system (S) Third Law of Thermodynamics – the entropy of a pure crystalline substance is zero as it is perfectly ordered. S (g) > S (l) > S (s)

44. Entropy Calculations (EX11)  S o rxn =  S o products -  S o reactants Calculate  S o rxn at 25C: 2NO 2 (g)  N 2 O 4 (g)

45. Spontaneity If  S is negative,  H is likely to be negative if the reaction is spontaneous.

46. Part V Free Energy Change (  G) and Spontaneity

47. Free Energy G relates heat change and entropy change and indicates the spontaneity of the process, not the speed. G (Gibbs Free Energy) – a measure of energy available to be used for work on the surroundings Affected by Temp, Pressure, and concentration of reactants (specific conditions)

48. Free Energy If  G < 0, the reaction is spontaneous, the system would be at a lower energy level after the reaction If  G = 0, the reaction is at equilibrium If  G > 0, the reaction is not spontaneous, the system will be at a higher energy level after the reaction.

49. Free Energy  G o = Σn  G o product - Σn  G o reactant  G =  H - T  S  G o =  H o - T  S o

50. Free Energy (EX12) Is the oxidation of Magnesium spontaneous at 25 o C?

51. Free Energy (EX13) Use thermodynamic data to estimate the normal b.p. of water. H 2 O (l)  H 2 O (g)

52. Summary HH SS GG Forward reaction spontaneous at all temperature -+- Forward reaction non spontaneous at all temperature +-+ Forward reaction spontaneous at low temperature --? Forward reaction spontaneous at high temperature ++?