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Energy and Chemical Reactions L. Scheffler 1. Heat and Temperature Heat is energy that is transferred from one object to another due to a difference in.

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Presentation on theme: "Energy and Chemical Reactions L. Scheffler 1. Heat and Temperature Heat is energy that is transferred from one object to another due to a difference in."— Presentation transcript:

1 Energy and Chemical Reactions L. Scheffler 1

2 Heat and Temperature Heat is energy that is transferred from one object to another due to a difference in temperature Temperature is a measure of the average kinetic energy of a body Heat is always transferred from objects at a higher temperature to those at a lower temperature 2

3 Factors Affecting Heat Quantities The amount of heat contained by an object depends primarily on three factors: –The mass of material –The temperature –The kind of material and its ability to absorb or retain heat. 3

4 Heat Quantities 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 calorie = Joules The energy required to raise 1 pound of water of 1 o F is called a British Thermal Unit or BTU The BTU is widely used in the USA to compute energy capacities of heating and air conditioning equipment 4

5 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 5

6 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. The specific heat of a material is the amount of heat required to raise the temperature of 1 gram of a substance 1 o C (or Kelvin) 6

7 Substance C J g -1 K -1 C J mol -1 K -1 Water (liquid) Water (steam) Water (ice) Copper Aluminum Ethanol Lead Specific Heat values for Some Common Substances 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 Changes The heat equation may be stated as  Q = m C  T where:  Q = Change in heat m = mass in grams C = specific heat in J g -1 o C -1  T = Temperature change 9

10 Temperature Changes 10 Measuring the temperature change in a calorimetry experiment can be difficult since the system is losing heat to the surroundings even as it is generating heat. By plotting a graph of time v temperature it is possible to extrapolate back to what the maximum temperature would have been had the system not been losing heat to the surroundings. A time v temperature graph

11 Heat Transfer Problem 1 Calculate the heat that would be required an aluminum cooking pan whose mass is 400 grams, from 20 o C to 200 o C. The specific heat of aluminum is J g -1 o C -1. Calculate the heat that would be required an aluminum cooking pan whose mass is 400 grams, from 20 o C to 200 o C. The specific heat of aluminum is J g -1 o C -1. Solution  Q = mC  T 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 11

12 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 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 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 = 4800 – 80T 50T = 4800 – 80T 130T = T =5800 T = 44.6 o C T = 44.6 o C 12

13 Phase Changes & Heat Energy is required to change the phase of a substance 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 equivalent to about 335 J per gram 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 boil away 1 mole (18 grams)of water. This is equivalent to about 2240 J per gram. It takes 40.6 kJ of energy to boil away 1 mole (18 grams) of water. This is equivalent to about 2240 J per gram. 13

14 Substance  Q fus  Q vap Mercury, Hg 2.29kJ/mol59.1kJ/mol Ethanol, C 2 H 5 OH 5.02kJ/mol38.6kJ/mol Water, H 2 O 6.00kJ/mol40.6kJ/mol Ammonia, NH kJ/mol23.4kJ/mol Helium, He 0.02kJ/mol0.08kJ/mol Acetone5.72kJ/mol29.1kJ/mol Methanol, CH 3 OH 3.16kJ/mol35.3kJ/mol Molar Heat Data for Some Common Substances 14

15 mC liquid wax  T  Q tota = (50g)(2.31J g -1 ˚C -1 )(62˚C-85˚C) (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) 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 = g/mol,  H fusion =70,500 J/mol) Heat Transfer Problem 3  Q =  Q total =  Q liquid wax +  Q solidification +  Q solid wax + n(  Q fusion ) + mC solid wax  T  Q total= = mC liquid wax  T +n(  Q fusion ) + mC solid wax  T  Q total = ( J) + ( J)+ (-4033 J)  Q total = -16,683.8 J 15

16 Steam at 175°C that occupies a volume of dm 3 and a pressure of 2.60 atm. How much energy would it need to lose to end as liquid water at 20 o C? Heat Transfer Problem 4 Solution: n = PV/RT = (2.60 atm)(32.75 dm 3 ) ( dm 3 atm mol -1 K -1 )(448 K -1 ) = mol  Q = (2.315 mol) (37.47 J mol -1 K -1 )(175 o C-100 o C) +(2.315 mol)(40600 J mol -1 ) +(2.315 mol)( J mol -1 K -1 )(100 o C-20 o C)  Q = J J J = J = kJ

17 Chemical Reactions In a chemical reaction Chemical bonds are broken Atoms are rearranged New chemical bonds are formed These processes always involve energy changes 17

18 Energy Changes Breaking chemical bonds requires energy Breaking chemical bonds requires energy Forming new chemical bonds releases energy Forming new chemical bonds releases energy 18

19 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) 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) 19

20 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. The opposite is true in an exothermic reaction. 20

21 Enthalpy and Hess’ Law

22 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 22

23 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 23

24 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 in minimized. Enthalpy and heat are nearly equal under these conditions. 24

25 Enthalpy Changes The change in enthalpy is designated by the symbol  H. – If  H < 0 the process is exothermic. – If  H > 0 the process is endothermic. Sometimes the symbol for enthalpy (  H) is used for heat (  Q) In many cases where work is minimal heat is a close approximation for enthalpy. One must always remember that while they are closely related, heat and enthalpy are not the same thing Sometimes the symbol for enthalpy (  H) is used for heat (  Q) In many cases where work is minimal heat is a close approximation for enthalpy. One must always remember that while they are closely related, heat and enthalpy are not the same thing 25

26 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. 26

27 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 27

28 Standard Conditions For most thermodynamic measurements standard conditions are established as – 25 o C or 298 K – 1.0 atmosphere of pressure 28

29 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. 29

30 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. 30

31 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) 31

32 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 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) 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 33

34 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 SO 3 (g) +2 S(s) O 2 (g)  2 SO 3 (g)+2 SO 2 (g) + O 2 (g)  H = -594 kJ 2 SO 3 (g) +2 S(s) 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 34

35 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 SO 3 (g) +2 S(s) O 2 (g)  2 SO 3 (g)+2 SO 2 (g) + O 2 (g)  H = -594 kJ 2 SO 3 (g) +2 S(s) 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 35

36 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 36

37 Calculating Enthalpy from tables The enthalpy of formation for compound is equal to the enthalpy change that occurs when a compound is formed from its elements 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 37

38 Enthalpies of Formation Some enthalpies of formation for common compounds BaCO H 2 O (g) -242 HCl (g) -93 Ba(OH) H 2 O (l) -286 HCl (aq) -167 BaO -554 H2O2H2O2H2O2H2O2-188 NH 3 (g) -46 CaCO C3H8C3H8C3H8C3H8-104NO+90 CaO-636 C 4 H NO Ca(OH) CO-110 SO CaCl CO Al 2 O 3 (s) See your text: Brown, LeMay and Bursten, Chemistry the Central Science, 7 th edition pages for addition values

39 Calculating Enthalpy from tables Enthalpies of formation represent the enthalpy changes when 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 39

40 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  H o HCl (aq) -167  H o CaCl  H o H 2 O (l) -286  H o CO 2 (g) -394 Solution  H o products =(-796)+(-286)+(-394) = kJ = kJ  H o reactants =( )+(2)(-167) = kJ = kJ  H o reaction = (-1541) = +75 kJ 40

41 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  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) = kJ = kJ  H o reactants =( -104 )+(5)(0) = -104 kJ = -104 kJ  H o reaction = (-104) = kJmol Now 11 grams = 0.25 mole of propane (11 g/44 g mol -1 ) (0.25 mol )(-2046 kJ mol -1 ) = kJ

42 Bond Enthalpies 42

43 Bond Enthalpies Another approach to determining an enthalpy change for a chemical reaction The energy to 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 43

44 Bond Enthalpy Table The average bond enthalpies for several types of chemical bonds are shown in the table below: 44

45 Bond Enthalpies Bond enthalpies can be used to calculate the enthalpy change for a chemical reaction. 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 45

46 Bond Enthalpy Calculations Example 1: Calculate the enthalpy change for the reaction N H 2  2 NH 3 46 Bonds broken 1N = N: = 945 3H-H: 3(435) = 1305 Total = 2250 kJ Bonds formed 2x3 = 6 N-H: 6 (390) = kJ Net enthalpy change = = - 90 kJ

47 Born Haber Cycle 47

48 Born-Haber Cycle Born-Haber Cycles are energy cycles for the formation of certain ionic compounds Application of Hess’ Law A Born-Haber cycle can be used to calculate quantities that are difficult to measure directly such as lattice energies The formation of an ionic compound as a sequence of steps whose energies can be determined. 48

49 Some Definitions The enthalpy of atomization is the enthalpy change that occurs when one mole of gaseous atoms is formed from the element in the standard state under standard conditions Example: ½ Cl 2 (g)  Cl (g)  H o at = 121 kJ mol -1 Example: ½ Cl 2 (g)  Cl (g)  H o at = 121 kJ mol -1 The electron affinity is the enthalpy change that occurs when an electron is added to an isolated atom in the gaseous state: O (g) + e-  O - (g)  H o = -142 kJ mol -1 O - (g) + e-  O 2- (g)  H o = +844 kJ mol -1 The lattice enthalpy is the enthalpy change that occurs from the conversion of an ionic compound in the gaseous state into its gaseous ions LiCl (g)  Li + (g) + Cl - (g)  H o = +846 kJ mol -1 LiCl (g)  Li + (g) + Cl - (g)  H o = +846 kJ mol -1 49

50 Born Haber Cycle Diagram The stepwise energy changes for the formation of NaCl 50

51 Born Haber Cycle for NaCl The formation of NaCl can be considered as a five step process Na (s) + 1/2 Cl 2 (g)  NaCl (s) 1. The vaporization of sodium metal to form the gaseous element. 2. The dissociation of chlorine gas to gaseous chlorine atoms is equal to one half of the bond energy for a Cl-Cl covalent bond 3. The ionization of gaseous sodium atoms to Na(g)  Na + 4. The ionization of chlorine atoms. (This quantity is the negative electron affinity for the element chlorine.) 5. The lattice energy on the formation of sodium chloride from the gaseous ions 51

52 Born-Haber Cycle for NaCl The stepwise energy changes for the formation of NaCl: The vaporization of sodium metal to form the gaseous element. Na (s)  Na (g) ∆H° sublimation = kJ mol -1 Na (s)  Na (g) ∆H° sublimation = kJ mol -1 The dissociation of chlorine gas to gaseous chlorine atoms is equal to one half of the bond energy for a Cl- Cl covalent bond 1/2 Cl 2 (g)  Cl (g) ∆H° diss = kJ mol -1 1/2 Cl 2 (g)  Cl (g) ∆H° diss = kJ mol -1 The ionization of gaseous sodium atoms to: Na (g)  Na + (g) + e - ∆H° ionization = kJ mol -1 Na (g)  Na + (g) + e - ∆H° ionization = kJ mol -1 The ionization of chlorine atoms. (This quantity is the negative electron affinity for the element chlorine.) Cl (g)+ e -  Cl - (g) ∆H° elect.affinity = kJ mol -1 Cl (g)+ e -  Cl - (g) ∆H° elect.affinity = kJ mol -1 The lattice energy on the formation of sodium chloride from the gaseous ions Na + (g) + Cl - (g)  NaCl (s) ∆H° lattice = kJ mol -1 Na + (g) + Cl - (g)  NaCl (s) ∆H° lattice = kJ mol -1 52

53 Entropy

54 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. 54

55 Entropy The statistical interpretation of thermodynamics was pioneered by James Clerk Maxwell (1831– 1879) and brought to fruition by the Austrian physicist Ludwig Boltzmann (1844–1906). The statistical interpretation of thermodynamics was pioneered by James Clerk Maxwell (1831– 1879) and brought to fruition by the Austrian physicist Ludwig Boltzmann (1844–1906). 55

56 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 56

57 Entropy is Disorder Disorder in a system can take many forms. Each of the following represent an increase in disorder and therefore in entropy: Disorder in a system can take many forms. Each of the following represent an increase in disorder and therefore in entropy: 1.Mixing different types of particles. i.e. dissolving salt in water. 2.A change is state where the distance between particles increases. Evaporation of water. 3.Increased movement of particles. Increase in temperature. 4.Increasing numbers of particles. Ex. 2 KClO 3  2 KCl + 3O 2 2 KClO 3  2 KCl + 3O 2 57

58 Entropy States The greatest increase in entropy is usually found when there is an increase of particles in the gaseous state. The symbol for the change in disorder or entropy is given by the symbol,  S. The more disordered a system becomes the more positive the value for  S will be. Systems that become more ordered have negative  S values. 58

59 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)

60 Entropy and States of Matter S˚(Br 2 liquid) < S˚(Br 2 gas) S˚(H 2 O solid) < S˚(H 2 O liquid) 60

61 Entropy, Phase & Temperature S increases slightly with T S increases a large amount with phase changes 61

62 The Entropy of a substance increases with temperature. The Entropy of a substance increases with temperature. Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps. Entropy and Temperature 62

63 Entropy - Boltzman Ludwig Boltzman S = k Ln W S = k Ln W Entropy is proportional to the number of degrees of freedom or possible configurations in a system. 63

64 Standard Entropy Values The standard entropy,  S o, of a substance is the entropy change per mole that occurs when heating a substance from 0 K to the standard temperature of 298 K. Unlike enthalpy, absolute entropy changes can be measured. Like enthalpy, entropy is a state function. The change in entropy is the difference between the products and the reactants  S o =  S o (products) -  S o (reactants) 64

65 Standard Entropy Values 65 The amount of entropy in a pure substance depends on the temperature, pressure, and the number of molecules in the substance. Values for the entropy of many substances at have been measured and tabulated. Values for the entropy of many substances at have been measured and tabulated. The standard entropy is also measured at 298 K. The standard entropy is also measured at 298 K. Some standard enthalpy values

66 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

67 Gibbs Free Energy 67

68 Spontaneity A chemical reaction is spontaneous if it results in the system moving form a less stable to a more stable state. Decreases in enthalpy and increases in entropy move a system to greater stability The combination of the enthalpy factor and the entropy factor can be expressed as the Gibbs Free Energy 68

69 Gibbs Free Energy The standard free energy change is defined by this equation  G o =  H o – T  S o Where  H o = the enthalpy change  S o = the entropy change T = Kelvin temperature T = Kelvin temperature A chemical reaction is spontaneous if it results in a negative free energy change. 69

70 Gibbs Free Energy Possible Combinations for free energy change:  G o =  H o – T  S o GGGG HHHH SSSS  H-T  S Always Spontaneous < 0 (-) > 0 (+) Always (-) Never Spontaneous > 0 (+) < 0 (-) Always (+) Spontaneous at High Temperature 0 (+) 0 (+) > 0 (+) (-) if T large (+) if T small Spontaneous at Low Temperature > 0 (+) < 0 (-) (+) if T large (-) if T small 70

71 Free Energy Problem 1 A certain chemical reaction is exothermic with a standard enthalpy of kJ mol -1. The entropy change for this reaction is +44 J mol -1 K -1. Calculate the free energy change for this reaction at 25 o C. Is the reaction spontaneous? Solution Convert the entropy value to kJ. 44 J mol -1 K -1 = kJ mol -1 K -1  G = kJ mol -1 – (298 K)(0.044 kJ mol -1 K -1 )  G = kJ mol -1 – (298 K)(0.044 kJ mol -1 K -1 )  G = kJ mol -1 – 13.1 kJ mol -1  G = kJ mol -1 – 13.1 kJ mol -1  G = kJ mol -1. Since  G is negative the reaction is spontaneous. reaction is spontaneous. Note. Because  H 0, this reaction is spontaneous at all temperatures. Note. Because  H 0, this reaction is spontaneous at all temperatures. 71

72 Free Energy Problem 2 A certain chemical reaction is endothermic with a standard enthalpy of +300 kJ mol -1. The entropy change for this reaction is +25 J mol -1 K -1. Calculate the free energy change for this reaction at 25 o C. Is the reaction spontaneous? Solution Convert the entropy value to kJ. 25 J mol -1 K -1 = kJ mol -1 K -1  G = kJ mol -1 – (298 K)(0.025 kJ mol -1 K -1 )  G = kJ mol -1 – (298 K)(0.025 kJ mol -1 K -1 )  G = kJ mol -1 – 7.45 kJ mol -1  G = kJ mol -1 – 7.45 kJ mol -1  G = kJ mol -1. Since  G is positive the reaction is non-spontaneous. reaction is non-spontaneous. Note. Because  H >0 and  S >0, this reaction is non- spontaneous at low temperatures. It the temperature were substantially increased it would become spontaneous. Note. Because  H >0 and  S >0, this reaction is non- spontaneous at low temperatures. It the temperature were substantially increased it would become spontaneous. 72


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