1. Heat and Temperature 4. Heat is the TOTAL kinetic energy. 5. Temperature is the AVERAGE kinetic energy, measured by a thermometer. http://phet.colorado.edu/en/simulation/states -of-matter http://phet.colorado.edu/en/simulation/states -of-matter Heat is always transferred from objects at a HIGHER temperature to those at a LOWER temperature, until temperatures become equal. 1

2. Heat Units The heat required to raise the temperature of 1.00 g of water 1 o C is known as a calorie The SI unit for heat is the joule. It is based on the mechanical energy requirements. 1.00 calorie = 4.184 Joules 2

3. Calorimetry Calorimetry involves the measurement of heat changes that occur in chemical processes or reactions. The heat change that occurs when a substance absorbs or releases energy is really a function of three quantities: –The mass –The temperature change –The heat capacity of the material 3

4. Heat Capacity and Specific Heat The ability of a substance to absorb or retain heat varies widely. The heat capacity depends on the nature of the material. 6. The specific heat of a material is the amount of heat required to raise the temperature of 1 gram of a substance 1 degree C. 4

5. Substance C J g -1 K -1 C J mol -1 K -1 Water (liquid) 4.18475.327 Water (steam) 2.08037.47 Water (ice) 2.05038.09 Copper0.38524.47 Aluminum0.89724.2 Ethanol2.44112 Lead0.12726.4 Specific Heat values for Some Common Substances 5 The lower the specific heat the less energy required to raise the temperature...heat up faster. High values heat up slower.

6. Heat Changes The heat equation may be stated as Q = (m) (C) (  T) where: Q = Heat (J or cal) m = mass (grams) C = specific heat (J/g o C)  T = Temperature change ( o F or o C)  “Delta” or change 6

7. Heat Transfer Problem 1 Calculate the heat that would be required to heat an aluminum cooking pan whose mass is 400 grams, from 20 o C to 200 o C. The specific heat of aluminum is 0.902 J g -1 o C -1. Calculate the heat that would be required to heat an aluminum cooking pan whose mass is 400 grams, from 20 o C to 200 o C. The specific heat of aluminum is 0.902 J g -1 o C -1. Solution  Q = mC  T 0.902 J g -1 o C -1 )(200 o C – 20 o C ) = (400 g) (0.902 J g -1 o C -1 )(200 o C – 20 o C ) = 64,944 J = 64,944 J 7

8. Heat Exchange When two systems are put in contact with each other, there will be a net exchange of energy between them unless they are at thermal equilibrium, i.e. at the same temperature. 8 Heat will flow from the substance at the higher temperature to that at a lower temperature.

9. Heat Transfer Problem 2 What is the final temperature when 50 grams of water at 20 o C is added to 80 grams water at 60 o C? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 o C -1 What is the final temperature when 50 grams of water at 20 o C is added to 80 grams water at 60 o C? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 o C -1 Solution: Solution:  Q (Cold) =  Q ( hot ) mC  T= mC  T Let T = final temperature Let T = final temperature (50 g) (4.184 J g -1 o C -1 )(T- 20 o C) (50 g) (4.184 J g -1 o C -1 )(T- 20 o C) = (80 g) (4.184 J g -1 o C -1 )(60 o C- T) = (80 g) (4.184 J g -1 o C -1 )(60 o C- T) (50 g)(T- 20 o C) = (80 g)(60 o C- T) (50 g)(T- 20 o C) = (80 g)(60 o C- T) 50T -1000 = 4800 – 80T 50T -1000 = 4800 – 80T 130T =5800 130T =5800 T = 44.6 o C T = 44.6 o C 9

10. Phase Changes & Heat - Energy is required to change the phase of a substance. Separate the intermolecular forces. - Energy is required to change the phase of a substance. Separate the intermolecular forces. -The amount of heat necessary to melt a substance is called the Heat of fusion (  H fus ). The heat of fusion is expressed in terms of 1 mole or 1 gram It takes 6.00 kJ of energy to melt 1 mole (18 grams) of ice into liquid water. This is much greater than the energy needed to raise the temperature of water. 10

11. The amount of heat necessary to boil a substance is called the Heat of vaporization (  H vap ) It may be expressed in terms of 1 mole or 1 gram It takes 40.6 kJ of energy to turn one mole of water into a gas. This is how a swamp cooler works.

12. How much energy must be lost for 50.0 g of liquid wax at 85.0˚C to cool to room temperature at 25.0˚C? (C solid wax = 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, C liquid wax =2.31 J/g˚C; MM = 352.7 g/mol,  H fusion =70,500 J/mol) Heat Transfer Problem 3  Q total =  Q liquid wax +  Q solidification +  Q solid wax  Q total= = mC liquid wax  T +n(  Q fusion ) + mC solid wax  T 12

13.  Q tota = (50g)(2.31J g -1 ˚C -1 )(62˚C-85˚C) + (50g/352.7gmol -1 )(-70,500J mol -1 )+ (50g)(2.18J g -1 ˚C -1 )(25˚C-62˚C)  Q total = (-2656.5 J) + (-9994.3 J)+ (-4033 J)  Q total = -16,683.8 J

14. Chemical Reactions In a chemical reaction Chemical bonds are broken Atoms are rearranged New chemical bonds are formed These processes always involve energy changes 14

15. Energy Changes Breaking chemical bonds requires energy Breaking chemical bonds requires energy Forming new chemical bonds releases energy Forming new chemical bonds releases energy 15

16. Exothermic and Endothermic Processes Exothermic processes release energy C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4H 2 O (g) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4H 2 O (g) 2043 kJ Endothermic processes absorb energy C(s) + H 2 O (g) +113 kJ  CO(g) + H 2 (g) C(s) + H 2 O (g) +113 kJ  CO(g) + H 2 (g) 16

17. Energy Changes in endothermic and exothermic processes In an endothermic reaction there is more energy required to break bonds than is released when bonds are formed. ∆H increases Or is + 17

18. In an exothermic reaction there is less energy required to break bonds than is released when released when bonds are formed. ∆H decreases Or is -

19. Enthalpy and Hess’ Law

20. Enthalpy Enthalpy is the heat absorbed or released during a chemical reaction where the only work done is the expansion of a gas at constant pressure 20

21. Enthalpy Not all energy changes that occur as a result of chemical reactions are expressed as heat Energy = Heat + Work Work is a force applied over a distance. Most energy changes resulting from chemical reactions are expressed in a special term known as enthalpy 21

22. Enthalpy It is nearly impossible to set up a chemical reaction where there is no work performed. The conditions for a chemical reaction are often set up so that work is minimized. Enthalpy and heat are nearly equal under these conditions. 22

23. Enthalpy Changes The change in enthalpy is designated by the symbol  H. – If  H < 0 or (-) the process is exothermic. – If  H > 0 or (+) the process is endothermic. Sometimes the symbol for enthalpy (  H) is used for heat (  Q) Sometimes the symbol for enthalpy (  H) is used for heat (  Q) One must always remember that while they are closely related, heat and enthalpy are not the same thing 23

24. Energy and Enthalpy Changes It is impractical to measure absolute amounts of energy or enthalpy. Enthalpy is always measured relative to previous conditions. Hence we measure changes in enthalpy rather than total enthalpy Enthalpy is measured relative to the system. 24

25. Measuring Enthalpy The amount of heat absorbed or released during a chemical reaction depends on the conditions under which the reaction is carried out including: – the temperature – the pressure – the physical state of the reactants and products 25

26. Standard Conditions ∆H Ɵ For most thermodynamic measurements standard conditions are established as – 25 o C or 298 K – 1.0 atmosphere of pressure or 101.3 kPa Symbolized ∆H Ɵ 26

27. Standard State The pure form of a substance at standard conditions is said to be in the standard state. The most stable form of an element at standard conditions represents the standard state for that element. The most stable form of an element at standard conditions represents the standard state for that element. 27

28. Hess’ Law If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy change for the individual steps Hess’ Law provides a way to calculate enthalpy changes even when the reaction cannot be performed directly. 28

29. Hess’ Law: Example 1 N 2 (g) + O 2 (g)  2 NO (g)  H 1 = +181 kJ N 2 (g) + O 2 (g)  2 NO (g)  H 1 = +181 kJ 2 NO (g) + O 2 (g)  2 NO 2 (g)  H 2 = -113 kJ 2 NO (g) + O 2 (g)  2 NO 2 (g)  H 2 = -113 kJ Find the enthalpy change for Find the enthalpy change for N 2 (g) + 2 O 2 (g)  2 NO 2 (g) N 2 (g) + 2 O 2 (g)  2 NO 2 (g) 29

30. Hess’ Law: Example 1 Solution: N 2 (g) + O 2 (g)  2 NO (g)  H 1 = +181 kJ 2 NO (g) + O 2 (g)  2 NO 2 (g)  H 2 = -113 kJ ------------------------------------------------------------- ------------------------------------------------------------- N 2 (g) +2O 2 (g)+ 2 NO (g)  2 NO (g) + 2 NO 2 (g) N 2 (g) +2O 2 (g)  + 2 NO 2 (g)  H =  H 1 +  H 2 = +181 kJ +(-113) = + 68 kJ 30

31. Hess Law: Example 2 From the following reactions and enthalpy changes: From the following reactions and enthalpy changes: 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)  H = -196 kJ 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)  H = -196 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ Find the enthalpy change for the following reaction: Find the enthalpy change for the following reaction: S (s) + O 2 (g)  SO 2 (g) S (s) + O 2 (g)  SO 2 (g) Solution: 2 SO 3 (g)  2 SO 2 (g) + O 2 (g)  H = +196 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ-------------------------------------------------------------------------------------------------------------- Reversing the order of the first equation reverses the sign of  H Reversing the order of the first equation reverses the sign of  H 31

32. Hess Law Example 2 From the following reactions and enthalpy changes: From the following reactions and enthalpy changes: 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)  H = -196 kJ 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)  H = -196 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ Find the enthalpy change for the following reaction: Find the enthalpy change for the following reaction: S (s) + O 2 (g)  SO 2 (g) S (s) + O 2 (g)  SO 2 (g) 2 SO 3 (g)  2 SO 2 (g) + O 2 (g)  H = +196 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ-------------------------------------------------------------------------------------------------------------- 2 SO 3 (g) +2 S(s) + 2 3 O 2 (g)  2 SO 3 (g)+2 SO 2 (g) + O 2 (g)  H = -594 kJ 2 SO 3 (g) +2 S(s) + 2 3 O 2 (g)  2 SO 3 (g)+2 SO 2 (g) + O 2 (g)  H = -594 kJ 2 S(s) + 2 O 2 (g)  2 SO 2 (g)  H = -594 kJ 2 S(s) + 2 O 2 (g)  2 SO 2 (g)  H = -594 kJ 32

33. Hess Law: Example 2 From the following reactions and enthalpy changes: From the following reactions and enthalpy changes: 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)  H = -196 kJ 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)  H = -196 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ Find the enthalpy change for the following reaction: Find the enthalpy change for the following reaction: S (s) + O 2 (g)  SO 2 (g) S (s) + O 2 (g)  SO 2 (g) 2 SO 3 (g)  2 SO 2 (g) + O 2 (g)  H = +196 kJ 2 SO 3 (g)  2 SO 2 (g) + O 2 (g)  H = +196 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g)  H = -790 kJ-------------------------------------------------------------------------------------------------------------- 2 SO 3 (g) +2 S(s) + 2 3 O 2 (g)  2 SO 3 (g)+2 SO 2 (g) + O 2 (g)  H = -594 kJ 2 SO 3 (g) +2 S(s) + 2 3 O 2 (g)  2 SO 3 (g)+2 SO 2 (g) + O 2 (g)  H = -594 kJ 2 S(s) + 2 O 2 (g)  2 SO 2 (g)  H = -594 kJ 2 S(s) + 2 O 2 (g)  2 SO 2 (g)  H = -594 kJ S(s) + O 2 (g)  SO 2 (g)  H = -297 kJ S(s) + O 2 (g)  SO 2 (g)  H = -297 kJ 33

34. Standard Enthalpy Changes The enthalpy change that occurs when the reactants are converted to products, both being in their standard states is known as the standard enthalpy change. It is designated as  H o.  H o reaction =   H o products +   H o reactants 34

35. Calculating Enthalpy from tables The symbol for the bond enthalpy of formation is  H f Enthalpies of formation have been measured and tabulated for a large number of compounds. 35

36. Enthalpies of Formation Some enthalpies of formation for common compounds BaCO 3 -1219 H 2 O (g) -242 HCl (g) -93 Ba(OH) 2 -998 H 2 O (l) -286 HCl (aq) -167 BaO -554 H2O2H2O2H2O2H2O2-188 NH 3 (g) -46 CaCO 3 -1207 C3H8C3H8C3H8C3H8-104NO+90 CaO-636 C 4 H 10 -126 NO 2 +33.8 Ca(OH) 2 -987CO-110 SO 2 -297 CaCl 2 -796 CO 2 -394 Al 2 O 3 (s) -1670 36 Notice all values are negative since these molecules are being formed

37. Calculating Enthalpy from tables Enthalpies of formation represent the enthalpy changes when a compound forms from its elements. The enthalpy of formation for an uncombined element is therefore = 0 The enthalpy of formation for a chemical reaction can be expressed as the difference between the enthalpy state of the products and that of the reactants  Hreaction =   H o products +   H o reactants 37

38. Sample Problem 1 Calcium carbonate reacts with hydrochloric acid according to the following equation: CaCO 3 (s) + 2HCl (aq)  CaCl 2 (aq) + H 2 O (l) + CO 2 (g) Calculate the enthalpy change for this reaction  H o reaction =   H o products +  H o reactants  H o CaCO 3 -1207  H o HCl (aq) -167  H o CaCl 2 -796  H o H 2 O (l) -286  H o CO 2 (g) -394 Solution  H o products =(-796)+(-286)+(-394) = -1476 kJ = -1476 kJ  H o reactants =( +1207 )+(2)(+167) = +1541 kJ = +1541 kJ  H o reaction = -1476+1541 = +75 kJ 38

39. Sample Problem 2 Calculate the enthalpy change for the burning of 11 grams of propane C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g)  H o reaction =   H o products –  H o reactants  H o C 3 H 8 -104  H o O 2 (g) 0  H o H 2 O (g) -242  H o CO 2 (g) -394Solution  H o products =(3)(-394)+(4)(-242) = -2150 kJ = -2150 kJ  H o reactants =( -104 )+(5)(0) = -104 kJ = -104 kJ  H o reaction = -2150-(-104) = -2046 kJmol -1 39 Now 11 grams = 0.25 mole of propane (11 g/44 g mol -1 ) (0.25 mol )(-2046 kJ mol -1 ) = 511.5 kJ

40. Bond Enthalpies 40

41. Bond Enthalpies Another approach to determining an enthalpy change for a chemical reaction The energy required to break a covalent bond in the gaseous phase is called a bond enthalpy. Bond enthalpy tables give the average energy to break a chemical bond. Actually there are slight variations depending on the environment in which the chemical bond is located 41

42. Bond Enthalpy Table The average bond enthalpies for several types of chemical bonds are shown in the table below: 42

43. Bond Enthalpies Energy is required to break chemical bonds. Therefore when a chemical bond is broken its enthalpy change carries a positive sign. Energy is released when chemical bonds form. When a chemical bond is formed its enthalpy change is expressed as a negative value By combining the enthalpy required and the enthalpy released for the breaking and forming chemical bonds, one can calculate the enthalpy change for a chemical reaction 43

44. Bond Enthalpy Calculations Example 1: Calculate the enthalpy change for the reaction N 2 + 3 H 2  2 NH 3 44 Bonds broken 1N = N: = 945 3H-H: 3(435) = 1305 Total = 2250 kJ Bonds formed 2x3 = 6 N-H: 6 (390) = - 2340 kJ Net enthalpy change = + 2250 + -2340 = - 90 kJ

45. Entropy

46. Entropy Entropy is defined as a state of disorder or randomness. Entropy is defined as a state of disorder or randomness. In general the universe tends to move toward release of energy and greater entropy. In general the universe tends to move toward release of energy and greater entropy. 46

47. Entropy Spontaneous chemical processes often result in a final state is more Disordered or Random than the original. The Spontaneity of a chemical process is related to a change in randomness. Entropy is a thermodynamic property related to the degree of randomness or disorder in a system. Reaction of potassium metal with water. The products are more randomly distributed than the reactants 47

48. The entropy of a substance depends on its state: S (gases) > S (liquids) > S (solids) Entropy, S S o (J/K -1 mol - 1 ) S o (J/K -1 mol - 1 ) H 2 O (liquid)69.95 H 2 O (gas)188.8 S o (J/K -1 mol - 1 ) S o (J/K -1 mol - 1 ) H 2 O (liquid)69.95 H 2 O (gas)188.8 48

49. Entropy, Phase & Temperature S increases slightly with T S increases a large amount with phase changes 49

50. Factors That Determine Entropy States The greater the disorder or randomness in a system the larger the entropy. Some generalizations 1. The entropy of a substance always increases as it changes from solid to liquid to gas and vice versa. 2. When a pure solid or liquid dissolves in a solvent, the entropy of the substance increases. 3. When gas molecules escape from a solvent, the entropy increases. 4. Entropy generally decreases with increasing molecular complexity