1. Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. The following items will be discussed : 6-1- The nature of energy and types of energy 6-2- Energy changes in chemical reactions 6-3- Introduction to thermodynamics 6-4- Enthalpy of chemical reactions 6-5-Calorimetry; the measurement of heat change. 6-6- Standard enthalpy of formation and reaction 6-7- Heat of solution and dilution

3. A LOOK AHEAD 1-The study of the nature and the different types of energy. 2- The study of heat change in chemical reactions, which is thermo chemistry. The vast majority of reactions are either endothermic (absorbing heat) or exothermic (releasing heat). 3-The thermo chemistry is a part of a broad subject called the first law of thermodynamics, which is based on the law of conservation of energy. The change in internal energy can be expressed in terms of the changes in heat and work done of a system. 4- The measure of the heat of reaction or calorimetry under constant-volume and constant-pressure, and the meaning of specific heat and heat capacity, quantities used in experimental work. Tro, Chemistry: A Molecular Approach3

4. A LOOK AHEAD 5- Knowing the standard enthalpies of formation of reactants and products enables us to calculate the enthalpy of a reaction. Discuss ways to determine these quantities either by the direct method from the elements or by the indirect method, which is based on Hess’s law of heat summation. 6- Finally, the study the heat changes when a solute dissolves in a solvent (heat of solution) and when a solution is diluted (heat of dilution). Every chemical reaction obeys two fundamental laws; the law of conservation of mass and the law of conservation of energy. The mass relationships between reactants and products are discussed. Tro, Chemistry: A Molecular Approach4

5. 6-1- The nature of energy and types of energy Energy is : the capacity to do work Work “ directed energy change resulting from a process “ work = force x distance Types of energy :. Radiant energy Thermal energy Chemical energy Nuclear energy Potential energy Kinetic energy

6. 6-1- The nature of energy and types of energy Energy is defined as the capacity to do work. Work = force x distance Chemists define work as directed energy change resulting from a process. Kinetic energy: is the energy produce by a moving object : is one form of energy that is of particular interest to chemists. Other include radiant energy, thermal energy, chemical energy and potential energy. Tro, Chemistry: A Molecular Approach6

7. 6-1- Types of energy Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position Kinetic energy consists of various types of molecular motion and the movement of electrons within molecules. 6.1

8. Heat is the transfer of thermal energy between two bodies that are at different temperatures. 6-2- Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy. 90 0 C 40 0 C greater thermal energy 6.2 Temperature = Thermal Energy All Chemical reactions absorb or produce (release) energy in the form of heat.

9. Some examples of energy conversion The workForm of the energyThe converted energy form Stand in sunlightRadiant energyThermal energy Do exercisesChemical energy stored in the bodies Kinetic energy A ball starts to roll downhill Potential energyKinetic energy It is impossible to measure the total energy of a system certainly but the change in energy can be determined experimentally.

10. 6-3- Introduction to thermodynamics Thermochemistry is part of a broader subject called thermodynamics The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing 6.2

11. Thermodynamics is the scientific study of the inter-conversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, pressure, volume, temperature 6.3  E = E final - E initial  P = P final - P initial  V = V final - V initial  T = T final - T initial

12. Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) 6.2 energy + H 2 O (s) H 2 O (l)

13. ExothermicEndothermic 6.2

14. First law of thermodynamics : energy can be converted from one form to another, but cannot be created or destroyed.  E system +  E surroundings = 0 or  E system = -  E surroundings C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Exothermic chemical reaction! 6.3 Chemical energy lost by combustion = Energy gained by the surroundings system surroundings The internal energy of a system has two components; kinetic energy and potential energy. The kinetic energy component consists of various types of molecular motion and the movement of electrons within molecules. Potential energy is determined by the attractive interactions between electrons and nuclei and by repulsive interactions between electrons and between nuclei in individual molecules, as well as by interaction between molecules.

15. Example Consider the reaction between 1 mole of sulfur and 1 mole of oxygen gas to produce 1 mole of sulfur dioxide: S(s) + O 2 (g) SO 2 (g) This system composed of the reactant molecules S and O 2 and the product molecules SO 2. We don’t know the internal energy of the reactants and the products, but we can measure the change in energy content, ΔΕ, ΔΕ = E (product) – E (reactant) = energy content of 1 mol SO 2 (g) – energy content of [1 mol S(s) + 1 mol O 2 (g)] Tro, Chemistry: A Molecular Approach15

16. *Conservation of Energy *For an exothermic reaction, “lost” heat from the system goes into the surroundings *two ways energy “lost” from a system, 1 - converted to heat, q 2 - used to do work, w *Energy conservation requires that the energy change in the system equal the heat released + work done  E = q + w  E =  H + P  V  E is a state function * internal energy change independent of how done Tro, Chemistry: A Molecular Approach16 First Law of Thermodynamics

17. 6.3  E = q + w  E is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -P  V when a gas expands against a constant external pressure Another form of the first law for  E system Although neither heat q nor work w is a state function, their sum (q + w) is equal to  E where E is a state function.

18. Work and Heat work can be defined as force F multiplied by distance d: w = Fd One way to illustrate mechanical work is to study the expansion or compression of a gas (see Fig.(6.5)). The work done by the surroundings is: w = - PΔV where ΔV, the change in volume, is given by V f - V i Tro, Chemistry: A Molecular Approach18

19. Work Done On the System 6.3 w = F x d W = -P A d = - P  V P x V = x d 3 = F x d = w F d2d2  V > 0 -P  V < 0 w sys < 0 Work and heat are not a state functions!  w = w final - w initial initial final  q = q final - q initial

20. A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = -P  V (a)  V = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 Latm = 0 joules (b)  V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = -14.1 Latm w = -14.1 Latm x 101.3 J 1Latm = -1430 J 6.3

21. Worked Example 6.1

23. 6-4- Enthalpy of chemical reactions Enthalpy and the First Law of Thermodynamics 6.4  E = q + w  E =  H - P  V  H =  E + P  V q p =  H and w = -P  V At constant pressure: At constant volume  V = 0 W = - P  V W = P X0 = 0  E = q v – 0 = q v q v =   E = 

24. * The two factors that determine the thermodynamic favorability are the enthalpy and the entropy. * The enthalpy is a comparison of the bond energy of the reactants to the products. * bond energy = amount needed to break a bond.  H * The entropy factors relates to the randomness/orderliness of a system  S * The enthalpy factor is generally more important than the entropy factor Tro, Chemistry: A Molecular Approach24

25. Enthalpy of Reactions Because most of reactions are constant pressure processes, we can equate the heat change to the change in enthalpy. For any reaction such as: A (reactant) B (product) the change in enthalpy, called the enthalpy of reaction, ΔH = H (products) – H (reactants) Tro, Chemistry: A Molecular Approach25

26. Entropy Tro, Chemistry: A Molecular Approach26 entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S  S generally J/mol S = k ln W k = Boltzmann Constant = 1.38 x 10 -23 J/K W is the number of energetically equivalent ways, unitless

27. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H < 0 H products > H reactants  H > 0 6.4

28. Thermochemical Equations H 2 O (s) H 2 O (l)  H = 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. 6.4

29. Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ Is  H negative or positive? System gives off heat Exothermic  H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. 6.4

30. H 2 O (s) H 2 O (l)  H = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = - 6.01 kJ If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ 6.4

31. H 2 O (s) H 2 O (l)  H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations 6.4 H 2 O (l) H 2 O (g)  H = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = -3013 kJ 266 g P 4 1 mol P 4 123.9 g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ

32. Worked Example 6.3

33. A Comparison of  H and  E 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2 (g)  H = -367.5 kJ/mol  E =  H - P  V At 25 0 C, 1 mole H 2 = 24.5 L at 1 atm P  V = 1 atm x 24.5 L = 24.5 L.atm = 2.5 kJ  E = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol In case of ideal gases and at constant temperature  E =  H – RT  n 6.4

34. Worked Example 6.4

35. The specific heat (s) Heat capacity (C) Heat (q) absorbed or released q = m x s x  t q = C x  t 6-5- Calorimetry

36. 6-5- Calorimetry : the measurement of heat change. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released: q = m x s x  t q = C x  t  t = t final - t initial 6.5

37. How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = 0.444 J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = ms  t = 869 g x 0.444 J/g 0 C x –89 0 C= -34,000 J 6.5

38. Worked Example 6.5

39. Worked Example 6.6

40. 6.5

41. Chemistry in Action: Fuel Values of Foods and Other Substances C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l)  H = -2801 kJ/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J Substance  H combustion (kJ/g) Apple-2 Beef-8 Beer-1.5 Gasoline-34

42. 6-6- Standard enthalpy of formation and reaction Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (  H 0 ) as a reference point for all enthalpy expressions. Standard enthalpy of formation (  H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero.  H 0 (O 2 ) = 0 f  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f 6.6

44. The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm which can be determined by: The direct method rxn aA + bB cC + dD H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - 6.6 Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) The indirect method

45. C (graphite) + 1/2O 2 (g) CO (g) CO (g) + 1/2O 2 (g) CO 2 (g) C (graphite) + O 2 (g) CO 2 (g) 6.6

46. Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ rxn S (rhombic) + O 2 (g) SO 2 (g)  H 0 = -296.1 kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g)  H 0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g)  H 0 = -296.1x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g)  H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l)  H 0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 6.6

47. Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C 6 H 6 6.6

48. Worked Example 6.9

49. Worked Example 6.10

50. 6-7- Heat of solution and dilution The enthalpy of solution (  H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.  H soln = H soln – H components 6.7 Which substance (s) could be used for melting ice? Which substance (s) could be used for a cold pack?

51. The Solution Process for NaCl  H soln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.7

52.  H soln = U – H hydration NaCl (s) Na+ (g) + Cl- (g) U = 788 kJ / mol Na+ (g) + Cl- (g) Na+ (aq) + Cl- (aq)  H = - 784 kJ/mol ______________________________________________________ NaCl (s) Na+ (aq) + Cl- (aq)  H = 4 kJ/mol

53. The heat of dilution is the heat change associated with the dilution process, if a certain solution process is endothermic and the solution is diluted, more heat will be absorbed by the same solution from the surrounding. In case of dilution H 2 SO 4 this process is so exothermic that you must never attempt to dilute the concentrated acid by adding water to it. The recommended procedure is to add the concentrated acid slowly to the water while constantly stirring.

54. Chemistry in Action: Bombardier Beetle Defense C 6 H 4 (OH) 2 (aq) + H 2 O 2 (aq) C 6 H 4 O 2 (aq) + 2H 2 O (l)  H 0 = ? C 6 H 4 (OH) 2 (aq) C 6 H 4 O 2 (aq) + H 2 (g)  H 0 = 177 kJ/mol H 2 O 2 (aq) H 2 O (l) + ½O 2 (g)  H 0 = -94.6 kJ/mol H 2 (g) + ½ O 2 (g) H 2 O (l)  H 0 = -286 kJ/mol  H 0 = 177 - 94.6 – 286 = -204 kJ/mol Exothermic!

55. Problems 1- 2 – 12 – 14 – 16 – 17 – 18 – 21 – 22 – 23 – 24 – 26 – 28 – 29 – 33 – 34 – 36 – 39 – 41 – 42 – 43 – 45 – 46 – 47 – 48 – 49 – 50 – 51 – 53 – 54 – 57 – 59 – 61 – 62 – 63 – 64 – 65 – 67 – 68 – 70.