1. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Thermochemistry Chapter 6

2. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Overview Introduce the nature of energy and the general topics related to energy problems. Familiarize with the experimental procedures for measuring heats of reactions. Hess’s law and its applications based on enthalpies. Read the present and new sources of energy.

3. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 3 Law of Conservation of Energy Energy can be converted from one form to another but can neither be created nor destroyed. (E universe is constant)

4. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4 The Two Types of Energy Potential: due to position or composition - can be converted to work (water before falling from dam, gasoline, etc.) Kinetic: due to motion of the object (water falling and doing work, gasoline burning and driving engine, etc.) KE = 1 / 2 mv 2 (m = mass, v = velocity)

5. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 Figure 6.1: (a) In the initial positions, ball A has a higher potential energy than ball B. (b) After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B.

6. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 Total Energy = (PE) A + (KE) B + frictional heat

7. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Energy is a State Function Depends only on the present state of the system - not how it arrived there. It is independent of pathway. Internal Energy or Total Energy  E, Enthalpy  H, V, P, T are State Functions Heat and Work are not state functions

8. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 Thermodynamics 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

9. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Temperature v. Heat Temperature reflects random motions of particles, therefore related to kinetic energy of the system. Heat involves a transfer of energy between 2 objects due to a temperature difference

10. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 Heat is the transfer of thermal energy between two bodies that are at different temperatures. Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy. Temperature = Thermal Energy 90 0 C 40 0 C greater thermal energy

11. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing SYSTEM SURROUNDINGS

12. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 System and Surroundings System: That on which we focus attention Surroundings: Everything else in the universe Universe = System + Surroundings

13. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 Exo and Endothermic Heat exchange accompanies chemical reactions. Exothermic: Heat flows out of the system (to the surroundings). Endothermic: Heat flows into the system (from the surroundings).

14. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. PE stored in chemical bonds are the source of chemical energy. 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) energy + H 2 O (s) H 2 O (l) Energy lost by system = Energy gained by surrounding

15. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15 Exothermic versus endothermic.

16. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 The study of energy and its conversion is called thermodynamics. The law of conservation of energy is called First Law of Thermodynamics.

17. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 17 First Law First Law of Thermodynamics: The energy of the universe is constant.

18. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 Mathematical Definition of the First Law  E = q + w  E = change in system’s internal energy q = heat w = work

19. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 19 S.Ex. 6.1 Calc. ΔE, for an endothermic process of 15.6 kJ and work done on the system is 1.4 kJ. q = +15.6 kJw = +1.4 kJ ΔE = q + w = +15.6 kJ + (+1.4 kJ) = 17.0 kJ

20. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 Figure 6.4: (a) The piston, moving a distance ∆h against a pressure P, does work on the surroundings. (b) Since the volume of a cylinder is the area of the base times its height, the change in volume of the gas is given by ∆h x A = ΔV. Expansion

21. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21 Work work = force  distance = F x Δh = PxA x Δh = PΔV ΔV = V f - V i since pressure = force / area, work = pressure  volume w system =  P  V

22. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 22 ‘w’ is work done ON the system w = -pΔV p = external pressure ΔV = V f – V i = + for expansion → w negative = - for compression → w positive

23. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 23 S. Ex. 6.2 Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. w = -pΔV = - 15 atm x (64 L – 46 L) = - 270 L.atm -ve, system doing work!!

24. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 24 S. Ex. 6.3 Hot air balloon- volume change 4.00x10 6 L to 4.50x10 6 L, by addition of 1.3x10 8 J of heat. q = 1.3x10 8 J ΔV = V f – V i = (4.50 – 4.00)x10 6 = 0.50x10 6 L w = -p ΔV = - 1.0 atm x 0.50x10 6 L = - 0.5x10 6 L.atm = -5.0x10 5 L.atm x (101.3 J/1 L.atm) = -5.1x10 7 J ΔE = q + w = 1.3x10 8 J + (- 5.1x10 7 J) = 8x10 7 J

25. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 25 Enthalpy Enthalpy = H = E + PV  E =  H  P  V = q P + w  H =  E + P  V At constant pressure, q P =  E + P  V, where q P =  H at constant pressure  H = energy flow as heat (at constant pressure) E, H, P, V are state functions

26. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 At constant pressure where PV work is allowed the change in enthalpy  H is equal to the energy flow as heat (q or Q).

27. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27 Signs Convention H eat Flowing to the system from surrounding (heating) :  E < 0, q or  H <0 H eat Flowing from the system to the surrounding (cooling) :  E > 0, q or  H >0 W ork done by the system on the surrounding : W > 0 (-) W ork done on the system by the surrounding : W < 0 (+)

28. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 28 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.

29. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 29 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

30. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 30 S.Ex. 6.4 When one mole of CH 4 is burnt at const.P, 890 kJ heat was released. Calc. ΔH when 5.8 g CH 4 is burnt at const P q p = ΔH = -890 kJ/molCH 4

31. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 31 Heat Capacity

32. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 32 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 = ms Heat (q) absorbed or released: q = ms  t q = C  t  t = t final - t initial

33. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 33

34. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 34 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

35. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 35 Some Heat Exchange Terms specific heat capacity heat capacity per gram = J/°C g or J/K g molar heat capacity heat capacity per mole = J/°C mol or J/K mol

36. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 36 Calorimetry The device used experimentally to determine the heat from chemical reaction is called a calorimeter. Calorimetry is the science to measure heat changes and it is based on the temperature change. Heat capacity is a measure of the body or system ability to absorb heat.

37. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 37 Two Types of Calorimeters 1.Constant Volume Calorimeter 2.Constant Pressure Calorimeter

38. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 38 Constant-Pressure or Coffee Cup Calorimetry q lost = q gained q rxn = - (q water + q calorimeter ) q water = ms  t q cal = C cal  t Reaction at Constant P  H = q rxn May be used to determine many Heats such as heat of solution, neutralization (acid-base), etc.

39. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 39 50.0 mL of 1.0M NaOH at 25 o C and 50.0 mL of 1.0M HCl at 25 o C are mixed Final temp = 31.9 o C; ∆H = ? q = m.s.∆T = 100 g x 4.18 J/g. o C x6.9 o C = 2.9x10 3 J H+ + OH- → H 2 O 0.050 mol 0.050 mol 0.050 mol ∆H = - 2.9x10 3 J/0.050 mol = -5.8x10 4 kJ/mol

40. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 40 S.Ex 6.5 1.00 L of 1.00 M Ba(NO 3 ) 2 solution at 25 o C is mixed with 1.00 L of 1.00M Na 2 SO 4 solution at 25 o C the temperature of the mixture increased to 28.1 o C. Calculate the enthalpy change per mole of BaSO4 formed.(s = 4.18 J/ o C.g) Ba 2+ (aq) + SO 4 2- (aq) → BaSO 4 (s) 1 mol1 mol1 mol q = msΔT = 2000 g x 4.18 J/ o C.g x 3.1 o C =2.6x10 4 J

41. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 41 Constant-Volume Calorimetry No heat enters or leaves! q sys = q water + q bomb + q rxn q sys = 0 q rxn = - (q water + q bomb ) q water = ms  t q bomb = C bomb  t Reaction at Constant V  H ~ q rxn  H = q rxn W = -p  V = 0  E = q + W  E = q V

42. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 42 Sample Exercise 6.6 Calculate the energy per gram for burning 1.50g of CH 4 in a constant-volume calorimeter (C=11.3 kJ/ 0 C) where  t = 7.3 0 C.. Energy released = C x  t = 11.3x10 3 x 7.3 = 83 kJ per 1.50g of CH 4 83 Energy released per g = = 1.50 55 kJ/g

43. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 43 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 Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack?

44. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 44 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

45. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 45 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 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 Q Q = -6470 kJ 1 mol P4 123.9g -3013 kJ 266gQ

46. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 46 Hess’s Law Reactants  Products The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

47. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 47 Calculations via Hess’s Law 1. If a reaction is reversed,  H is also reversed. N 2 (g) + O 2 (g)  2NO(g)  H = 180 kJ 2NO(g)  N 2 (g) + O 2 (g)  H =  180 kJ 2. If the coefficients of a reaction are multiplied by an integer,  H is multiplied by that same integer. 6NO(g)  3N 2 (g) + 3O 2 (g)  H =  540 kJ

48. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 48 S. Ex. 6.7 C(graphite-s)→ C(diamond-s) ΔH = ? Given, C g (s) + O 2 (g) → CO 2 (g) ΔH = -394 kJ C d (s) + O 2 (g) → CO 2 (g) ΔH = -396 kJ C g (s) + O 2 (g) → CO 2 (g) ΔH = -394 kJ CO 2 (g) → C d (s) + O 2 (g) ΔH = +396 kJ _________________________________ Cg(s) → Cd(s) ΔH = 2 kJ

49. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 49

50. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 50 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. f

51. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 51 Standard States Compound 4 For a gas, pressure is exactly 1 atmosphere. 4 For a solution, concentration is exactly 1 molar. 4 Pure substance (liquid or solid), it is the pure liquid or solid. Element 4 The form [N 2 (g), K(s)] in which it exists at 1 atm and 25°C.

52. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 52 Standard Enthalpies of Formation defined as Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. C(s)+ 2H 2 (g)+ ½ O 2 (g) → CH 3 OH (l) ΔH f o = -239 kJ/mol

53. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 53 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

54. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 54 Standard enthalpy of formation (  H 0 f ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. Example CO 2(g) is formed from C (s) and O 2(g) C (s) + O 2(g) CO 2(g) H0fH0f

55. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 55 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

56. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 56 The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. 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  - 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.)

57. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 57 CH 4 + 2O 2 → CO 2 + 2H 2 O

58. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 58 Change in Enthalpy Can be calculated from enthalpies of formation of reactants and products.  H rxn ° =  n p  H f  (products)   n r  H f  (reactants)

59. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 59

60. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 60 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

61. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 61 Keep in mind the following concepts.

62. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 62 S.Ex 6.9 4NH 3 (g) +7O 2 (g) → 4NO 2 (g) +6H 2 O(l) ΔH=? ΔH f o (NH 3 ) = -46 kJ/mol ΔH f o (NO 2 ) = 34 kJ/mol ΔH f o (H 2 O) = -286 kJ/mol ΔH = [6 ΔH f o (H 2 O) + 4 ΔH f o (NO 2 )] – [4 ΔH f o (NH 3 )+ 7ΔH f o (O 2 )] ΔH = [6x-286 + 4x34] – [4x-46 +7x0] =1396 kJ

63. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 63

64. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 64 S. Ex 6.11 2CH 3 OH + 3O 2 (g) → 2CO 2 + 4 H 2 O Calculate heat of combustion per gram of Methanol. ΔH= 2xΔH f (CO 2 ) + 4x ΔH f (H 2 O) – 2x ΔH f (CH 3 OH) = 2x(-394 kJ) + 4x(-286 kJ) -2x(-239kJ) = -1454 kJ for 2 mole methanol -1454 kJ/64.0 g = -22.7 kJ/g

65. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 65 Sample Exercise Calculate the work done when 50.0 g of magnesium (Mg) dissolve in excess acid at 1.00 atm and 25.0 0 C.

66. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 66 Sample Exercise Determine the heat of reaction at 298 K for the reaction which occurs in a welder's acetylene torch: H 2 (g) + 1/2 O 2 (g) H 2 O(l)H o /kJ = -285.8 2 C(s) + H 2 (g) C 2 H 2 (g)H o /kJ = +226.7 C(s) + O 2 (g) CO 2 (g)H o /kJ = -393.5 2 C 2 H 2 (g) + 5 O 2 (g) 4 CO 2 (g) + 2 H 2 O(l) Given the following equations and H o values

67. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 67 Sample Exercise A 15.0g nickel is heated to 100.0 0 C and dropped into 55.0g of water initially at 23 0 C. Assuming an isolated system, calculate the final temperature of the nickel and water. S nickel = 0.444 J/ 0 Cg. Q lost by nickel = - Q gained by water (mst) nickel = - (ms  t) water t = tf - ti t f is the same. t i is different.