Unit 3: Thermochemistry

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1 Unit 3: Thermochemistry
Print pp. 1-12, 15 – 20, 21 – 26,

2 Unit Outline Temperature and Kinetic Energy Heat/Enthalpy Calculation
Temperature changes (q = mc∆T) Phase changes (q = n∆H) Heating and Cooling Curves Calorimetry (q = C∆T & above formulas)

3 Unit Outline Chemical Reactions STSE: What Fuels You? PE Diagrams
Thermochemical Equations Hess’s Law Bond Energy STSE: What Fuels You?

4 Temperature and Kinetic Energy
Thermochemistry is the study of energy changes in chemical and physical changes eg. dissolving burning phase changes

5 Temperature - a measure of the average kinetic energy of particles in a substance
- a change in temperature means particles are moving at different speeds - measured in either Celsius degrees or degrees Kelvin Kelvin = Celsius

6 The Celsius scale is based on the freezing and boiling point of water
The Kelvin scale is based on absolute zero - the temperature at which particles in a substance have zero kinetic energy.

7 p. 628

8 K 50.15 450.15 °C 48 -200

9 300 K # of particles 500 K Kinetic Energy

10 Heat/Enthalpy Calculations
system - the part of the universe being studied and observed surroundings - everything else in the universe open system - a system that can exchange matter and energy with the surroundings eg. an open beaker of water a candle burning closed system - allows energy transfer but is closed to the flow of matter.

11 WorkSheet: Thermochemistry #1
isolated system – a system completely closed to the flow of matter and energy heat - refers to the transfer of kinetic energy from a system of higher temperature to a system of lower temperature. - the symbol for heat is q WorkSheet: Thermochemistry #1

12 Part A: Thought Lab (p. 631)

13 Part B: Thought Lab (p. 631)

14 Heat/Enthalpy Calculations
specific heat capacity – the energy , in Joules (J), needed to change the temperature of one gram (g) of a substance by one degree Celsius (°C). The symbol for specific heat capacity is a lowercase c

15 A substance with a large value of c can absorb or release more energy than a substance with a small value of c. ie. For two substances, the substance with the larger c will undergo a smaller temperature change with the same loss or gain of heat.

16 FORMULA q = mc∆T q = heat (J) m = mass (g) c = specific heat capacity
∆T = temperature change = T2 – T1 = Tf – Ti q = mc∆T

17 WorkSheet: Thermochemistry #2
eg. How much heat is needed to raise the temperature of g of water from 20.0 °C to 45.0 °C? Solve q = m c ∆T for c, m, ∆T, T2 & T1 p #’s 1 – 4 p #’s 5 – 8 WorkSheet: Thermochemistry #2

18 heat capacity - the quantity of energy , in Joules (J), needed to change the temperature of a substance by one degree Celsius (°C) The symbol for heat capacity is uppercase C The unit is J/ °C or kJ/ °C

19 WorkSheet: Thermochemistry #3
FORMULA C = heat capacity c = specific heat capacity m = mass ∆T = T2 – T1 C = mc q = C ∆T Your Turn p.637 #’s 11-14 WorkSheet: Thermochemistry #3

20 Enthalpy Changes enthalpy change - the difference between the potential energy of the reactants and the products during a physical or chemical change AKA: Heat of Reaction or ∆H

21 Endothermic Reaction Products PE ∆H Reactants Reaction Progress

22 Endothermic Reaction PE ∆H Enthalpy Reaction Progress Products ∆H
Reactants Reaction Progress

23 Products Enthalpy ∆H is + Reactants Endothermic

24 reactants ∆H is - Enthalpy products Exothermic

25 Enthalpy Changes in Reactions
All chemical reactions require bond breaking in reactants followed by bond making to form products Bond breaking requires energy (endothermic) while bond formation releases energy (exothermic) see p. 639

26

27 Enthalpy Changes in Reactions
endothermic reaction - the energy required to break bonds is greater than the energy released when bonds form. ie. energy is absorbed exothermic reaction - the energy required to break bonds is less than the energy released when bonds form. ie. energy is produced

28 Enthalpy Changes in Reactions
∆H can represent the enthalpy change for a number of processes Chemical reactions ∆Hrxn – enthalpy of reaction ∆Hcomb – enthalpy of combustion (see p. 643)

29 Formation of compounds from elements
∆Hof – standard enthalpy of formation The standard molar enthalpy of formation is the energy released or absorbed when one mole of a compound is formed directly from the elements in their standard states. (see p. 642) eg. C(s) + ½ O2(g) → CO(g) ΔHfo = kJ/mol

30 Use the equations below to determine the ΔHfo for CH3OH(l) and CaCO3(s)
2 C(s) + 4 H2(g) + O2(g) → 2 CH3OH(l) kJ 2 CaCO3(s) kJ → 2 Ca(s) + 2 C(s) + 3 O2(g)

31 Phase Changes (p.647) ∆Hvap – enthalpy of vaporization ∆Hfus – enthalpy of melting ∆Hcond – enthalpy of condensation ∆Hfre – enthalpy of freezing eg. H2O(l)  H2O(g) ΔHvap = Hg(l)  Hg(s) ΔHfre = +40.7 kJ/mol -23.4 kJ/mol

32 Solution Formation (p.647, 648) ∆Hsoln – enthalpy of solution
eg. ΔHsoln, of ammonium nitrate is kJ/mol. NH4NO3(s) kJ → NH4NO3(aq) ΔHsoln, of calcium chloride is −82.8 kJ/mol. CaCl2(s) → CaCl2(aq) kJ

33 Three ways to represent an enthalpy change:
1. thermochemical equation - the energy term written into the equation. 2. enthalpy term is written as a separate expression beside the equation. 3. enthalpy diagram.

34 thermochemical equation
eg. the formation of water from the elements produces kJ of energy. 1. H2(g) + ½ O2(g) → H2O(l) kJ 2. H2(g) + ½ O2(g) → H2O(l) ∆Hf = kJ/mol thermochemical equation

35 WorkSheet: Thermochemistry #4
enthalpy diagram H2(g) + ½ O2(g) 3. ∆Hf = kJ/mol Enthalpy (H) H2O(l) examples: pp questions p #’s 15-18 WorkSheet: Thermochemistry #4

36 Calculating Enthalpy Changes
FORMULA: q = n∆H q = heat (kJ) n = # of moles ∆H = molar enthalpy (kJ/mol)

37 eg. How much heat is released when 50.0 g of CH4 forms from C and H ?
(p. 642) q = nΔH = (3.115 mol)(-74.6 kJ/mol) = -232 kJ

38 eg. How much heat is released when 50
eg. How much heat is released when g of CH4 undergoes complete combustion? (p. 643) q = nΔH = (3.115 mol)( kJ/mol) = kJ

39 q = nΔH = (1.110 mol)(+40.7 kJ/mol) = +45.2 kJ
eg. How much energy is needed to change g of H2O(l) at 100 °C to steam at 100 °C ? Mwater = g/mol ΔHvap = kJ/mol q = nΔH = (1.110 mol)(+40.7 kJ/mol) = kJ

40 ∆Hfre and ∆Hcond have the opposite sign of the above values.

41 q = nΔH = (0.4996 mol)(+25.7 kJ/mol) = +12.8 kJ
eg. The molar enthalpy of solution for ammonium nitrate is kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves? q = nΔH = ( mol)(+25.7 kJ/mol) = kJ

42 What mass of ethane, C2H6, must be burned to produce 405 kJ of heat?
ΔH = kJ q = kJ q = nΔH n = mol m = n x M = ( mol)(30.08 g/mol) = g

43 Complete: p. 643 #’s 15 - 18 p. 645; #’s 19 – 23
pp. 648 – 649; #’s 24 – 29 19. (a) kJ (b) kJ x103 kJ 21. (a) x 103 kJ 21. (b) x104 kJ 21. (c) x 106 kJ kJ x103 g

44 kJ 25.(a) kJ (b) kJ 26.(a) absorbed (b) kJ 27.(a) NaCl(s) kJ/mol → NaCl(aq) (b) kJ (c) cool; heat absorbed from water g x 104 kJ

45 WorkSheet: Thermochemistry #5
p #’ 4 – 8 pp. 649, #’s 3 – 8 p. 657, 658 #’s WorkSheet: Thermochemistry #5

46 Heating and Cooling Curves
Demo: Cooling of p-dichlorobenzene Time (s) Temperature (°C)

47 Cooling curve for p-dichlorobenzene
KE 80 Temp. (°C ) liquid PE 50 freezing KE solid 20 Time

48 Heating curve for p-dichlorobenzene
80 Temp. (°C ) KE 50 PE KE 20 Time

49 What did we learn from this demo??
During a phase change temperature remains constant and PE changes Changes in temperature during heating or cooling means the KE of particles is changing

50 p. 651

51 p. 652

52 p. 656

53 Heating Curve for H20(s) to H2O(g)
A 40.0 g sample of ice at -40 °C is heated until it changes to steam and is heated to 140 °C. Sketch the heating curve for this change. Calculate the total energy required for this transition.

54 Time 140 100 Temp. (°C ) -40 q = mc∆T q = n∆H q = mc∆T q = n∆H
q = mc∆T -40 Time

55 Data: cice = J/g.°C cwater = J/g.°C csteam = J/g.°C ΔHfus = kJ/mol ΔHvap = kJ/mol

56 warming ice: q = mc∆T = (40.0)(2.01)( ) = 3216 J warming water: = (40.0)(4.184)(100 – 0) = J warming steam: q = mc∆T = (40.0)(2.01)( ) = 3216 J

57 n = 40.0 g moles of water: 18.02 g/mol = 2.22 mol melting ice:
q = n∆H = (2.22 mol)(6.02 kJ/mol) = kJ boiling water: = (2.22 mol)(40.7 kJ/mol) = kJ

58 Total Energy kJ kJ 3216 J 16736 J 127 kJ

59 Practice p. 655: #’s 30 – 34 pp. 656: #’s 1 - 9 WorkSheet:
Thermochemistry #6 30.(b) 3.73 x103 kJ 31.(b) 279 kJ 32.(b) x10-3 kJ 33.(b) kJ kJ kJ

60 Law of Conservation of Energy (p. 627)
The total energy of the universe is constant ∆Euniverse = 0 Universe = system + surroundings ∆Euniverse = ∆Esystem + ∆Esurroundings ∆Euniverse = ∆Esystem + ∆Esurroundings = 0 OR ∆Esystem = -∆Esurroundings OR qsystem = -qsurroundings First Law of Thermodynamics

61 Calorimetry (p. 661) calorimetry - the measurement of heat changes during chemical or physical processes calorimeter - a device used to measure changes in energy 2 types of calorimeters 1. constant pressure or simple calorimeter (coffee-cup calorimeter) 2. constant volume or bomb calorimeter.

62 Simple Calorimeter p.661

63 a simple calorimeter consists of an insulated container, a thermometer, and a known amount of water
simple calorimeters are used to measure heat changes associated with heating, cooling, phase changes, solution formation, and chemical reactions that occur in aqueous solution

64 to calculate heat lost or gained by a chemical or physical change we apply the first law of thermodynamics: qsystem = -qcalorimeter Assumptions: the system is isolated c (specific heat capacity) for water is not affected by solutes heat exchange with calorimeter can be ignored

65 eg. A simple calorimeter contains g of water. A 5.20 g piece of aluminum alloy at 525 °C is dropped into the calorimeter causing the temperature of the calorimeter water to increase from 19.30°C to 22.68°C. Calculate the specific heat capacity of the alloy.

66 eg. The temperature in a simple calorimeter with a heat capacity of 1.05 kJ/°C changes from 25.0 °C to °C when a very cold 12.8 g piece of copper was added to it. Calculate the initial temperature of the copper. (c for Cu = J/g.°C)

67 Homework p. 664, #’s 1b), 2b), 3 & 4 p. 667, #’s 5 - 7

68 p # 4.b) (60.4)(0.444)(T2 – 98.0) = -(125.2)(4.184)(T2 – 22.3) 26.818(T2 – 98.0) = (T2 – 22.3) 26.818T = T 550.66T2 = T2 = 26.0 °C

69 qMg = -qcal nΔH = -mcΔT 6. System (Mg) m = 0.50 g = 0.02057 mol
Find ΔH Calorimeter v = 100 ml so m = 100 g c = 4.184 T2 = 40.7 T1 = 20.4 qMg = -qcal nΔH = -mcΔT 7. System ΔH = kJ/mol n = CV = (0.0550L)(1.30 mol/L) = mol Calorimeter v = 110 ml so m = 110 g c = 4.184 T1 = 21.4 Find T2

70 Bomb Calorimeter

71 Bomb Calorimeter used to accurately measure enthalpy changes in combustion reactions the inner metal chamber or bomb contains the sample and pure oxygen an electric coil ignites the sample temperature changes in the water surrounding the inner “bomb” are used to calculate ΔH

72 to accurately measure ΔH you need to know the heat capacity (kJ/°C) of the calorimeter.
must account for all parts of the calorimeter that absorb heat Ctotal = Cwater + Cthermom.+ Cstirrer + Ccontainer NOTE: C is provided for all bomb calorimetry calculations

73 eg. A technician burned 11. 0 g of octane in a steel bomb calorimeter
eg. A technician burned 11.0 g of octane in a steel bomb calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C. What is the enthalpy of combustion for octane?

74 eg. 1.26 g of benzoic acid, C6H5COOH(s), is burned in a bomb calorimeter. The temperature of the calorimeter and contents increases from °C to °C. Calculate the heat capacity of the calorimeter. (∆Hcomb = kJ/mol) Homework p #’s 8 – 10 WorkSheet: Thermochemistry #7

75 Hess’s Law of Heat Summation
the enthalpy change (∆H) of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products) ∆H is independent of the pathway and/or the number of steps in the process ∆H is the sum of the enthalpy changes of all the steps in the process

76 eg. production of carbon dioxide
Pathway #1: 2-step mechanism C(s) + ½ O2(g) → CO(g) ∆H = kJ CO(g) + ½ O2(g) → CO2(g) ∆H = kJ C(s) + O2(g) → CO2(g) ∆H = kJ

77 eg. production of carbon dioxide
Pathway #2: formation from the elements C(s) + O2(g) → CO2(g) ∆H = kJ

78 Using Hess’s Law We can manipulate equations with known ΔH to determine the enthalpy change for other reactions. NOTE: Reversing an equation changes the sign of ΔH. If we multiply the coefficients we must also multiply the ΔH value.

79 eg. Determine the ΔH value for: H2O(g) + C(s) → CO(g) + H2(g) using the equations below. C(s) + ½ O2(g) → CO(g) ΔH = kJ H2(g) + ½ O2(g) → H2O(g) ΔH = kJ

80 Switch Multiply by 5 Multiply by 4 eg. Determine the ΔH value for:
4 C(s) H2(g) → C4H10(g) using the equations below. ΔH (kJ) C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g) H2(g) + ½ O2(g) → H2O(g) C(s) + O2(g) → CO2(g) Switch Multiply by 5 Multiply by 4

81 Ans: -2672.5 kJ 5(H2(g) + ½ O2(g) → H2O(g) -241.8)
4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g) 5(H2(g) + ½ O2(g) → H2O(g) ) 4(C(s) + O2(g) → CO2(g) ) 4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g) 5 H2(g) + 2½ O2(g) → 5 H2O(g) 4C(s) O2(g) → 4 CO2(g) Ans: kJ

82 Practice pg #’s 11-14 WorkSheet: Thermochemistry #8

83 Review ∆Hof (p. 642, 684, & 848) The standard molar enthalpy of formation is the energy released or absorbed when one mole of a substance is formed directly from the elements in their standard states. ∆Hof = 0 kJ/mol for elements in the standard state The more negative the ∆Hof, the more stable the compound

84 Using Hess’s Law and ΔHf
Use the formation equations below to determine the ΔH value for: C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g) ΔHf (kJ/mol) 4 C(s) H2(g) → C4H10(g) H2(g) + ½ O2(g) → H2O(g) C(s) + O2(g) → CO2(g)

85 Using Hess’s Law and ΔHf
ΔHrxn = ∑ΔHf (products) - ∑ΔHf (reactants) eg. Use ΔHf , to calculate the enthalpy of reaction for the combustion of glucose. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)

86 Use the molar enthalpy’s of formation to calculate ΔH for the reaction below
Fe2O3(s) CO(g) → 3 CO2(g) Fe(s) p #’s 21 & 22

87 Eg. The combustion of phenol is represented by the equation below: C6H5OH(s) + 7 O2(g) → 6 CO2(g) H2O(g) If ΔHcomb = kJ/mol, calculate the heat of formation for phenol.

88 Bond Energy Calculations (p. 688)
The energy required to break a bond is known as the bond energy. Each type of bond has a specific bond energy (BE). (table p. 847) Bond Energies may be used to estimate the enthalpy of a reaction.

89 Bond Energy Calculations (p. 688)
ΔHrxn = ∑BE(reactants) - ∑BE (products) eg. Estimate the enthalpy of reaction for the combustion of ethane using BE. 2 C2H6(g) O2(g) → 4 CO2(g) H2O(g) Hint: Drawing the structural formulas for all reactants and products will be useful here.

90 → 4 O=C=O + 6 H-O-H + 7 O = O - [4(2)(745) + 6(2)(460)] = -3244 kJ
[2(347) + 2(6)(338) + 7(498)] - [4(2)(745) + 6(2)(460)] = kJ p #’s 23,24,& 26 p #’s 3, 4, 5, & 7

91 Energy Comparisons Phase changes involve the least amount of energy with vaporization usually requiring more energy than melting. Chemical changes involve more energy than phase changes but much less than nuclear changes. Nuclear reactions produce the largest ΔH eg. nuclear power, reactions in the sun

92 STSE What fuels you? (Handout)

93 BACK aluminum alloy water m = 5.20 g m = 150.0 g
T1 = 525 ºC T1 = ºC T2 = ºC T2 = ºC FIND c for Al c = J/g.ºC qsys = - qcal mcT = - mc T (5.20)(c)( ) = -(150.0)(4.184)(22.68 – 19.30) -2612 c = c = J/g.°C BACK

94 copper FIND T1 for Cu BACK calorimeter C = 1.05 kJ/°C m = 12.8 g
T2 = ºC c = J/g.°C FIND T1 for Cu qsys = - qcal mcT = - CT (12.8)(0.385)(23.94 – T1) = -(1050)(23.94 – 25.0) 4.928 (23.94 – T1) = 1113 23.94 – T1= 1113/4.928 23.94 – T1= 225.9 T1= ºC calorimeter C = 1.05 kJ/°C T1 = ºC T2 = ºC BACK

95 Specific heat capacity J/g.ºC C Heat capacity kJ/ ºC or J/ ºC ΔH
q heat J or kJ c Specific heat capacity J/g.ºC C Heat capacity kJ/ ºC or J/ ºC ΔH Molar heat or molar enthalpy kJ/mol