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Thermochemistry.

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Presentation on theme: "Thermochemistry."— Presentation transcript:

1 Thermochemistry

2 Objective Introduction to Thermochemistry Define Temperature
Potential Energy Enthalpy Endothermic and Exothermic Entropy?

3 Warm – up X2 What is potential Energy?
Energy can not be created or destroyed!! Where does the energy come from when you burn gasoline? How about an Explosion??

4 The Nobel Prize (Alfred Nobel)
How Noble is the Nobel Prize?? Found that when you mixed Nitroglycerin (shock sensitive) with diatomatious earth (inert) it was no longer nearly as shock sensitive. His fortune was made in Dynamite!

5 What happens after the material explodes?
Is an explosion exothermic or endothermic? Is that positive or negative? Does the material contain more or less potential energy after it explodes? Loosing potential energy in the bonds

6 What is an Thermochemistry (V)??
The study of the transfer of of energy, as heat that is accompanied by chemical reactions and physical changes.

7 Thermochemical Equation (v)
A Chemical equation that includes the enthalpy. 2H2(g) + O2(g)  2H2O(g) kJ The kJ is the amount of energy released (exothermic) 2H2O(g) kJ  2H2(g) + O2(g) The kJ is the amount of energy absorbed (endothermic)

8 Thermochemical Equation
The equations can also be written: 2H2(g) + O2(g)  2H2O(g); ΔH= kJ 483.6 kJ is negative (or exothermic) What side? 2H2O(g)  2H2(g) + O2(g); ΔH= kJ 483.6 kJ is positive (or endothermic) What side does it go on? Note the substances and moles formed

9 More Vocabulary for This Chapter
Calorimeter – a device that is intended to measure energy absorbed or released. Temperature – A measure of the average kinetic energy of the particles in matter. (always Kelvin). Joule – The SI unit for energy. (N-m)

10 More Vocabulary for This Chapter
Joule – The SI unit for energy. So if I said Joule per gram: What does that mean? If I said Joule per mole: What does that mean?

11 Enthalpy (ΔH) change Enthalpy has the symbol (H) and is the amount of energy contained within a system (bonds, state, temperature). Potential energy within the bonds lost or gained. Is it more or less stable after? If I set off some Trinitrotoluene (TNT) is it more or less stable after I set it off?

12 Vocabulary for This Chapter
Enthalpy change – Amount of energy, as heat absorbed or released by a system at constant pressure. ΔHrx = Hproducts – Hreactants ΔHmixing = Hsolution – (Hsolute + Hsolvent) Note Enthalpy is an measure of potential energy or energy contained within the bonds.

13 Enthalpy (ΔH) change Exothermic – release heat in a reaction!
Endothermic absorb heat in a reaction! Exothermic ΔH is negative. Endothermic ΔH is positive.

14 Vocabulary for This Chapter
Enthalpy change – Amount of energy, as heat absorbed or released by a system at constant pressure. Enthalpy of reaction – is the quantity of energy transferred as heat during a chemical reaction. Endothermic – When heat is absorbed by a system from the surroundings (+). Exothermic – When heat is released by the system to the surroundings (-)

15 The reason why exothermic is neg and endothermic is positive.
It is adding or subtracting potential energy from the energy stored in the bonds.

16 Compounds whose enthalpies of formation are highly negative
a. Do not exist. b. Are somewhat stable. c. Are very unstable. d. Are very stable. Why???

17 Entropy S is a measure of disorder
What is more random, has a higher entropy ΔS ???

18 Higher Entropy ΔS??? Type/more/less
                        

19 Entropy S What is more random, Solid, Liquid or Gas?
Measure of the degree of randomness in a system? (V) S What is more random, Solid, Liquid or Gas?

20 (ΔS positive or negative?)
Entropy S Entropy is a measure of the disorder in a system. If disorder increases the change in Entropy is positive. Change in Entropy is ΔS! If entropy increases disorder increases. CO2 (s)  CO2(g) (ΔS positive or negative?)

21 NaCl(s)  Na+(aq) + Cl-(aq) (ΔS positive or negative?)
Entropy S Entropy is a measure of the disorder in a system. If disorder increases the change in Entropy is positive. Change in Entropy is ΔS! If entropy increases disorder increases (In a Solution?) NaCl(s)  Na+(aq) + Cl-(aq) (ΔS positive or negative?)

22 Entropy S What is more random Reactant or Product?
2H2(g) + O2(g)  2H2O(l) S <0 Two molecules of gas converting into one liquid.

23 Two molecules of gas converting into one of gas.
Entropy S What is more random Reactant or Product? 3H2(g) + N2(g)  2NH3(g) S <0 Two molecules of gas converting into one of gas.

24 Entropy S What is more random Reactant or Product? S > 0
2C6H6(l) + 15O2(g) 6H2O(g) + 12CO2(g) S > 0 Two molecules of gas converting into one of gas. More or less favored? More!!

25 What is more random Reactant or Product? KNO3(s)  K+(aq) + NO3-(aq)
Entropy S What is more random Reactant or Product? KNO3(s)  K+(aq) + NO3-(aq) S > 0 Two molecules of solid converting into two moles of aqueous ions. More or less favored?

26 Entropy is also additive
ΔS = [Sum of Sproducts – Sum of Sreactants] Problem: Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g) ΔS = ? If entropy of Al2O3(s) =51.00kJ/mol·K; Al(s) = kJ/mol·K; H2O(g) is kJ/mol·K; H2(g) is kJ/mol·K .

27 (ΔS positive or negative?)
Entropy S Entropy is a measure of the disorder in a system. If disorder increases the change in Entropy is positive. Change in Entropy is ΔS! If entropy increases disorder increases. 2NO(g)  N2(g) + O2(g) (ΔS positive or negative?)

28 Day 2 (Warm-up) Does Entropy increase or decrease?
Does Enthalpy increase or decrease? Combustion of Liquid Octane (C8H18)? Evaporation of Water?

29 The Most Expensive Combustion Reaction!! (The Russians are coming!)
Cdiamond + O2  CO2 ΔH = kJ/mol

30 Enthalpy (ΔH) change is the
a. pressure change of a system at constant temperature. b. entropy change of a system at constant pressure. c. temperature change of a system at constant pressure. d. amount of energy absorbed or lost by a system as energy is the form of heat.

31 Objective Gibbs Free Energy (spontaneous reactions)
Introduction to Specific Heat calculations Energy Curve, Activation Energy Explain Hess’s Law

32 Spontaneous (v) reaction
A spontaneous process is the time-evolution of a system in which it releases free energy (most often as heat) and moves to a lower, more thermodynamically stable, energy state. Or the reaction happens…with no help Free Energy???? Thermodynamically stable??? Ahhhhhhhh……………………..

33 G = H - TS If Negative Spontaneous
At 300 K is this reaction spontaneous, not spontaneous, or can it not be determined? C2H4 (g) + H2(g)  C2H6(g) ΔH = kJ/mol; ΔS = kJ/mol·K G = H -TS G = kJ/mol – (300K)(-0.121kJ/mol·K) kJ/mol

34 Gibbs Free Energy (ΔG) ΔG = Gibbs free Energy
If negative the reaction is spontaneous If positive will not happen. ΔG = ΔH - TΔS

35 The Driving Force In the Nature of Reactions
Two factors effect whether a reaction will occur spontaneously. H – Enthalpy of formation if negative will favor Spontaneous Reaction -- WHY??

36 The Driving Force In the Nature of Reactions
Factor Number TWO effect whether a reaction will occur spontaneously. G = H - TS TS – Entropy Term (Entropy times temperature). If positive will favor Spontaneous Reaction -- WHY??

37 Spontaneous reactions are driven by
a. decreasing enthalpy and decreasing entropy. b. decreasing enthalpy and increasing entropy. c. increasing enthalpy and decreasing entropy. d. increasing enthalpy and increasing entropy.

38 Vocabulary for This Chapter
Enthalpy change – Amount of energy, as heat absorbed or released by a system at constant pressure. Enthalpy of reaction – is the quantity of energy transferred as heat during a chemical reaction. ΔH = Hproducts – Hreactants Note Enthalpy is an measure of potential energy or energy contained within the bonds.

39 Thermochemical Equation
A Chemical equation that includes the enthalpy. 2H2(g) + O2(g)  2H2O(g) kJ The kJ is the amount of energy released (exothermic) 2H2O(g) kJ  2H2(g) + O2(g) The kJ is the amount of energy absorbed (endothermic)

40 Thermochemical Equation
Write the Thermal Chemical Equation For: 2H2(g) + O2(g)  2H2O(g); ΔH= kJ kJ is negative (or __________) 2H2O(g)  2H2(g) + O2(g); ΔH= kJ 483.6 kJ is positive (or ____________) Note the substances and moles formed

41 Warm up Question Balance and write the thermochemical equation for the combustion of benzene (C6H6) ΔH = kJ/mol:

42 Objectives What is Activation Energy? Why does it exist?
Potential Energy Diagram? What is it and how to read it.

43 Activation Energy the minimum energy that must be put into a chemical system, containing potential for reactants to react! In order for the rock to get to the bottom of the hill it has to go over the hump

44 In Chemical reactions, What causes the hump?
Energy at which they collide (not all have the same energy). Why?? What surrounds an atom or molecule? Orientation (for most reactions to occur they have to have the correct orientation (meet the right way)

45 Collision Theory When particles of the reactant hit each other, only a certain percentage of the collisions cause a significant chemical change.

46 Potential Energy Diagram

47 Which is exothermic and which is endothermic
Which is exothermic and which is endothermic?. Which curve shows more storage at the end?

48 Catalyst (v) A substance that is not used up in a reaction that lowers the activation energy!!

49 Extra

50 Other Vocabulary for This Chapter
Specific Heat – amount of energy required to raise the temperature of 1 gram, K (1 °C ). Heat – Energy transferred between matter due to their temperature differences. Which way does heat flow.

51

52 How much energy is absorbed as heat by 20
How much energy is absorbed as heat by 20. g of gold when it is heated from 25°C to 55°C? The specific heat of gold is 0.13 J/g·°C.

53 Problem How much energy is needed to raise the temperature of a 55g piece of aluminum from 22.4°C to 94.6 °C. Given: Specific Heat = Cp = 0.897J/g-K

54 Problem Given: An amount of water weighting 500 grams is in an insulated container is being used to make tea. We want to get the water from a temperature of 25°C to 100°C. Assuming no boiling or vaporization and no loss of heat to surroundings. How long will it take to heat this in a 1000 Watt microwave oven. Specific heat of H2O = 4.18 J/g-K. 1000 watt microwave oven supplies = 3,600,000 J/hr. Calculate the time it takes to heat up the water.

55 How to manipulate thermochemical equations
2H2 + O2  2H2O ΔH = kJ H2 + ½O2  H2O ΔH = kJ 4H2 + 2O2  4H2O ΔH = kJ 2H2O  2H2 + O ΔH = kJ

56 How to manipulate thermochemical equations
Want: C(s) + 2H2(g)  CH4(g) ΔH = ? kJ Given: H2(g) + ½O2(g)  H2O(l) ΔH = kJ C(s) + O2(g)  CO2(g) ΔH = kJ CH4 + 2O2  2H2O + CO2 ΔH = kJ 1st Get reactants and products on the right sides of the equations: 2nd Balance them 3rd add them altogether

57 The enthalpy of formation of an element is?
Why??? Definition: The molar enthalpy of formation Hf is the enthalpy change that occurs when one mole of a compound is formed in their standard state (1 atm, 25°C) (V)

58 How to manipulate thermochemical equations
2H2 + O2  2H2O ΔH = kJ H2 + ½O2  H2O ΔH = kJ 4H2 + 2O2  4H2O ΔH = kJ 2H2O  2H2 + O ΔH = kJ

59 How to manipulate thermochemical equations
Want: C(s) + 2H2(g)  CH4(g) ΔH = ? kJ Given: H2(g) + ½O2(g)  H2O(l) ΔH = kJ C(s) + O2(g)  CO2(g) ΔH = kJ CH4 + O2  2H2O + CO2 ΔH = kJ 1st Get reactants and products on the right sides of the equations: 2nd Balance them 3rd add them altogether

60 Compounds whose enthalpies of formation are highly negative
a. do not exist. b. are somewhat stable. c. are very unstable. d. are very stable. Why???

61 Dr. Canale, Why are you doing this to me are you trying to drive me insane???
YES I AM!!!!!! But also…….. Hess’s Law – The overall enthalpy change in a reaction is the sum of enthalpy changes for the individual steps in the process.

62 Very Important Rules Enthalpy of Combustion Hc – The enthalpy change that occurs during the complete combustion of one mole of reactant. C6H6 + 7½O2  3H2O + 6CO2

63 Very Important Rules Enthalpy of Formation (heat of formation) – The enthalpy change that occurs during the formation of one mole of product from it’s elements. 6C + 2½H2  C6H5 6C + 3H2  C6H6

64 Problem The enthalpy of formation of ethane C2H6 is: Write the equation!

65 Problem The enthalpy of formation of Xylene C8H10 is: Write the equation!

66 Hess’s Law – The overall enthalpy change in a reaction is the sum of enthalpy changes for the individual steps in the process. or A bond is a bond and as long as it’s the same in the end no mater how it gets there.

67 How to manipulate thermochemical equations
2H2 + O2  2H2O ΔH = kJ H2 + ½O2  H2O ΔH = kJ 4H2 + 2O2  4H2O ΔH = kJ 2H2O  2H2 + O ΔH = kJ

68 How to manipulate thermochemical equations
Want: C(s) + 2H2(g)  CH4(g) ΔH = ? kJ Given: H2(g) + ½O2(g)  H2O(l) ΔH = kJ C(s) + O2(g)  CO2(g) ΔH = kJ CH4 + O2  2H2O + CO2 ΔH = kJ 1st Get reactants and products on the right sides of the equations: 2nd Balance them 3rd add them altogether

69 The enthalpy of formation of an element is?
zero Why???

70 How to manipulate thermochemical equations
Want: C(s) + 2H2(g)  CH4(g) ΔH = ? kJ Given: H2(g) + ½O2(g)  H2O(l) ΔH = kJ C(s) + O2(g)  CO2(g) ΔH = kJ CH4 + 2O2  2H2O + CO2 ΔH = kJ 1st Get reactants and products on the right sides of the equations: 2nd Balance them 3rd add them altogether

71 Very Important Rules Enthalpy of Combustion Hc – The enthalpy change that occurs during the complete combustion of one mole of reactant. C6H6 + 7½O2  3H2O + 6CO2

72 Very Important Rules 6C + 2½H2  C6H5 6C + 3H2  C6H6
Enthalpy of Formation – The enthalpy change that occurs during the Formation of one mole of product from it’s elements. 6C + 2½H2  C6H5 6C + 3H2  C6H6

73 Compounds whose enthalpies of formation are highly negative
a. Do not exist. b. Are somewhat stable. c. Are very unstable. d. Are very stable. Why???

74 Hess’s Law – The overall enthalpy change in a reaction is the sum of enthalpy changes for the individual steps in the process.

75 Graphite and diamond are different forms of carbon, one just a little more valuable than the other. Using these two equations, tell me how much energy does it take to convert graphite to diamond? Cgraphite  Cdiamond =? Cgraphite + O2  CO2 ΔH = kJ Cdiamond + O2  CO2 ΔH = kJ Yes they burned a diamond!!!!!!

76 Carbon and Carbon monoxide can be combusted in oxygen to form Carbon dioxide. Use Hess’s Law to compute the formation of carbon and oxygen to carbon monoxide? C + ½ O2  CO =? C + O2  CO2 ΔH = kJ CO + ½ O2  CO2 ΔH = kJ

77 Very Important Rules Enthalpy of Combustion Hc – The enthalpy change that occurs during the complete combustion of one mole of reactant. C6H6 + 7½O2  3H2O + 6CO2

78 Very Important Rules Enthalpy of Formation – The enthalpy change that occurs during the complete combustion of one mole of product. 6C + 2½H2  C6H5 6C + 3H2  C6H6

79 Compounds whose enthalpies of formation are highly negative
a. Do not exist. b. Are somewhat stable. c. Are very unstable. d. Are very stable. Why???

80 (Hf) for 2N2(g) + 5O2(g)  2N2O5(g)
H2(g) + ½O2(g)  H2O (l) H = kJ N2O5(g) + H2O(l) 2HNO3(l) H = -76.6kJ 1/2N2(g) +3/2O2(g) + 1/2H2(g)  HNO3(l) H = kJ

81 Do you Remember Free Energy?
You know Gibbs Free Energy?? ΔG = The reaction happens if…. ΔG = - (negative) ΔG = ΔH - TΔS

82 The Driving Force In the Nature of Reactions
Two factors effect whether a reaction will occur spontaneously. H – Enthalpy of formation if negative will favor Spontaneous Reaction -- WHY??

83 H – Enthalpy of formation if negative will favor Spontaneous Reaction -- WHY??
Because H is negative and when it is negative it has less potential energy and is more stable. What is more stable a rock at a steeper angle or a rock at a slight angle???

84 If the reaction is spontaneous the H of reaction will probably be negative
What is more stable, products or reactants? Two moles of H2 and one mole of O2 or two moles of H20 2H2(g) + O2(g)  2H20(g) H20 and the reaction actually has negative enthalpy or exothermic, and is spontaneous. (H = kJ)

85 If the reaction is spontaneous the H of reaction will probably be negative
What is more stable, products or reactants? Two moles of Iron and 3 moles of O2 or two moles of Fe203 (Red Iron Oxide = RUST) 4Fe(s) + 3O2(g)  2Fe203(s) Fe203 and the reaction actually has negative enthalpy or exothermic, and is spontaneous.

86 What is the H of reaction?
4Fe(s) + 3O2(g)  2Fe203(s) H = ? Formed from it’s elements?? Hf = − kJ Definition of Hf ?? H for reaction is kJ

87 If the reaction is spontaneous the H of reaction will probably be negative
What is more stable, products or reactants? Two moles of Butane and 13 moles of O2 or 10 moles of Water and 8 moles of CO2? 2C4H O2  10H20 + 8CO2 H20 and CO2 and the reaction actually has negative enthalpy or exothermic, and is spontaneous.

88 The Driving Force In the Nature of Reactions
Two factors effect whether a reaction will occur spontaneously. H – Enthalpy of formation if negative will favor Spontaneous Reaction -- WHY??

89 Enthalpy of Combustion Hc – The enthalpy change that occurs during the complete combustion of one mole of reactant. C6H6 + 7½O2  3H2O + 6CO2 Enthalpy of Formation – The enthalpy change that occurs during the complete combustion of one mole of product. 6C + 2½H2  C6H5

90 (Hf) for 2N2(g) + 5O2(g)  2N2O5(g)
H2(g) + ½O2(g)  H2O (l) H = kJ N2O5(g) + H2O(l) 2HNO3(l) H = -76.6kJ 1/2N2(g) +3/2O2(g) + 1/2H2(g)  HNO3(l) H = kJ

91 If the reaction is spontaneous the H of reaction will probably be negative
What is more stable, products or reactants? Two moles of H2 and one mole of O2 or two moles of H20 2H2(g) + O2(g)  2H20(g) H20 and the reaction actually has negative enthalpy or exothermic, and is spontaneous. (H = kJ)

92 If the reaction is spontaneous the H of reaction will probably be negative
What is more stable, products or reactants? Two moles of Iron and 3 moles of O2 or two moles of Fe203 (Red Iron Oxide = RUST) 4Fe(s) + 3O2(g)  2Fe203(s) Fe203 and the reaction actually has negative enthalpy or exothermic, and is spontaneous.

93 What is the H of reaction?
4Fe(s) + 3O2(g)  2Fe203(s) H = ? Formed from it’s elements?? Hf = − kJ Definition of Hf ?? H for reaction is kJ

94 H – Enthalpy of formation if negative will favor Spontaneous Reaction -- WHY??
Because H is negative and when it is negative it has less potential energy and is more stable. What is more stable a rock at a steeper angle or a rock at a slight angle???

95 If the reaction is spontaneous the H of reaction will probably be negative
What is more stable, products or reactants? Two moles of H2 and one mole of O2 or two moles of H20 2H2(g) + O2(g)  2H20(g) H20 and the reaction actually has negative enthalpy or exothermic, and is spontaneous. (H = kJ)

96 If the reaction is spontaneous the H of reaction will probably be negative
What is more stable, products or reactants? Two moles of Iron and 3 moles of O2 or two moles of Fe203 (Red Iron Oxide = RUST) 4Fe(s) + 3O2(g)  2Fe203(s) Fe203 and the reaction actually has negative enthalpy or exothermic, and is spontaneous.

97 Thermite I (red) 2Fe + 1½O2  Fe2O3; ΔH = -826 kJ
Fe2O3 + 2Al  Al2O3 + 2Fe; ΔH = ? 2Fe + 1½O2  Fe2O3; ΔH = -826 kJ 2Al + 1½O2Al2O3; ΔH = kJ Fe2O3 + 2Al  Al2O3 + 2Fe; ΔH= -850kJ ΔH= -425kJ/mol Fe; ΔH=-850kJ/mol Al2O3

98 3Fe3O4(s) + 8Al(s)4Al2O3(s)+9Fe(s); ΔH = ?
Thermite II (Black) 3Fe3O4(s) + 8Al(s)4Al2O3(s)+9Fe(s); ΔH = ? 3Fe + 2O2  Fe3O4; ΔH = kJ 2Al + 1½O2Al2O3; ΔH = kJ Fe3O4  3Fe + 2O2; ΔH = kJ

99 Thermite II (Black) 3Fe + 2O2  Fe3O4; ΔH = -1117 kJ
3Fe3O4(s) + 8Al(s)4Al2O3(s)+9Fe(s); ΔH = ? 3Fe + 2O2  Fe3O4; ΔH = kJ (4){2Al + 1½O2Al2O3; ΔH = kJ} (3){Fe3O4  3Fe + 2O2; ΔH = kJ}

100 3Fe3O4(s) + 8Al(s)4Al2O3(s)+9Fe(s); ΔH = ?
Thermite II (Black) 3Fe3O4(s) + 8Al(s)4Al2O3(s)+9Fe(s); ΔH = ? 8Al + 6O24Al2O3; ΔH = kJ} 3Fe3O4  9Fe + 6O2; ΔH = kJ}

101 Thermite II (Black) 3Fe3O4(s) + 8Al(s)4Al2O3(s)+9Fe(s); ΔH = ?
8Al(s) + 6O2(g)4Al2O3(s); ΔH = kJ} 3Fe3O4(s) 9Fe(s) + 6O2(g); ΔH = kJ} ____________________________________________________________________________________________ 3Fe3O4(s)+8Al(s)4Al2O3(s)+9Fe(s);ΔH kJ -373 kJ/mole Fe; kJ/mol Al2O3


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