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Thermodynamics. Energy is neither created or destroyed during chemical or physical changes, but it is transformed from one form to another.  E universe.

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Presentation on theme: "Thermodynamics. Energy is neither created or destroyed during chemical or physical changes, but it is transformed from one form to another.  E universe."— Presentation transcript:

1 Thermodynamics

2 Energy is neither created or destroyed during chemical or physical changes, but it is transformed from one form to another.  E universe = 0

3 TYPES of ENERGY KineticPotential MechanicalGravitational ThermalElectrostatic ElectricalChemical Radiant Energy Conversion Examples: 1. dropping a rock 2. using a flashlight 3. driving a car

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5 Endo: heat added to systemExo: heat released by system SYSTEMS and SURROUNDINGS System: The thing under study Surroundings: Everything else in the universe Energy transfer between system and surroundings:

6 HEAT: What happens to thermal (heat) energy? Three possibilities: Warms another object Causes a change of state Is used in an endothermic reaction

7 Example 1: 5 g wood at 0 o C + 5 g wood at 100 o C Example 2: 10 g wood at 0 o C + 5 g wood at 100 o C Example 3: 5 g copper at 0 o C + 5 g copper at 100 o C Example 4: 5 g wood at 0 o C + 5 g copper at 100 o C Choices: 1: 0 o C 2: 33 o C 3: 50 o C 4. 67 o C 5: 100 o C 6: other Temperature Changes from Heat Exchange

8 What happens to thermal (heat) energy? When objects of different temperature meet: Warmer object cools Cooler object warms Thermal energy is transferred q warmer = -q cooler

9 Quantitative: Calculating Heat Exchange: Specific Heat Capacity

10 Specific Heat Capacity The energy required to heat one gram of a substance by 1 o C. Usefulness: #J transferred = S.H. x #g x  T How much energy is used to heat 250 g water from 17 o C to 100 o C?

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12 What happens to thermal (heat) energy? When objects of different temperature meet: Warmer object cools Cooler object warms Thermal energy is transferred q warmer = -q cooler specific heat x mass x  T = specific heat x mass x  T warmer object cooler object

13 Heat transfer between substances: Specific heats: Cu = 0.385 J/g  o C Wood = 1.8 J/g  o C

14 Conceptually Easy Example with Annoying Algebra: If we mix 250 g H 2 O at 95 o C with 50 g H 2 O at 5 o C, what will the final temperature be?

15 Thermal Energy and Phase Changes First: What happens?

16 Thermal Energy and Phase Changes First: What happens?

17 Thermal Energy and Phase Changes First: What happens?

18 Warming: Molecules move more rapidly Kinetic Energy increases Temperature increases Melting/Boiling: Molecules do NOT move more rapidly Temperature remains constant Intermolecular bonds are broken Chemical potential energy (enthalpy) increases But what’s really happening?

19 Energy and Phase Changes: Quantitative Treatment Melting: Heat of Fusion (  H fus ) for Water: 333 J/g Boiling: Heat of Vaporization (  H vap ) for Water: 2256 J/g

20 Total Quantitative Analysis Convert 40.0 g of ice at –30 o C to steam at 125 o C Warm ice: (Specific heat = 2.06 J/g- o C) Melt ice: Warm water (s.h. = 4.18 J/g- o C)

21 Total Quantitative Analysis Convert 40.0 g of ice at –30 o C to steam at 125 o C Boil water: Warm steam (s.h. = 1.92 J/g- o C)

22 Lots of different types of energy. We use Enthalpy: Heat exchanged under constant pressure. Energy and Chemical Reactions

23 Energy/Enthalpy Diagrams

24 Some Examples of Enthalpy Change 2 C(s) + 2 H 2 (g)  C 2 H 4 (g)  H = +52 kJ

25 Enthalpy Change and Chemical Reactions  H is usually more complicated, due to solvent and solid interactions. So, we measure  H experimentally. Calorimetry Run reaction in a way that the heat exchanged can be measured. Use a “calorimeter.”

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27 Bomb Calorimetry Experiment N 2 H 4 + 3 O 2  2 NO 2 + 2 H 2 O Energy released = E absorbed by water + E absorbed by calorimeter E water = E calorimeter = Total E =  H = energy/moles = 0.500 g N2H4 600 g water 420 J/ o C

28 Enthalpy Change and Bond Energies  H = energy needed to break bonds – energy released forming bonds Example: formation of water:  H = [498 + (2 x 436)] – [4 x 436] kJ = -482 kJ

29 General Rule:

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31 SO 2 + ½ O 2  SO 3 dH = -98.9 kJ 2 SO 3  2 SO 2 + O 2 dH = ?

32 Hess’s Law Enthalpy is a State Function.

33 Thermochemistry Lab Calculations Goal: What is  H for the formation of MgO from Mg(s) and O 2 (g)? Mg(s) + ½ O 2 (g)  MgO(s)  H = ? kJ/mol Data: From lab measurements: Mg(s) + 2 H + (aq)  Mg 2+ (aq) + H 2 (g)  H 1 = ___________ kJ/mol MgO(s) + 2 H + (aq)  Mg 2+ (aq) + H 2 O(l)  H 2 = ___________ kJ/mol From a table: H 2 (g) + ½ O 2 (g)  H 2 O(l)  H 3 = -285.8 kJ/mol Task: Find a way to add these three reactions to get the desired reaction. Manipulate the  H values as needed, and add them.

34 Calculating Heat Production

35 Heat of Formation

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37 Heat of Formation: The general idea

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39 Find the enthalpy change for burning ethyl alcohol


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