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Chapter 6: 

Chapter 6 Thermochemistry

Terminology: 

Terminology Energy capacity to do work Kinetic Energy energy that something has because it is moving Potential Energy energy that something has because of its position

Internal Energy: 

Internal Energy The sum of the individual energies of all nanoscale particles (atoms, ions, or molecules) in that sample. E = 1/2mc2 The total internal energy of a sample of matter depends on temperature, the type of particles, and how many of them there are in the sample.

Energy Units: 

Energy Units calorie - energy required to heat 1-g of water 1oC Calorie - unit of food energy; 1 Cal = 1-kcal = 1000-cal Joule - 1-cal = 4.184 J = 1-kg*m2/sec2

Law of Conservation of Energy: 

Law of Conservation of Energy energy can neither be created nor destroyed the total amount of energy in the universe is a constant energy can be transformed from one form to another

First Law of Thermodynamics: 

First Law of Thermodynamics the amount of heat transferred into a system plus the amount of work done on the system must result in a corresponding increase of internal energy in the system

Thermochemistry Terminology: 

Thermochemistry Terminology system => that part of the universe under investigation surroundings => the rest of the universe universe = system + surroundings

Thermodynamic System: 

Thermodynamic System

Energy Transfer: 

Energy Transfer Energy is always transferred from the hotter to the cooler sample Heat – the energy that flows into or out of a system because of a difference in temperature between the thermodynamic system and its surroundings

Thermochemistry Terminology: 

Thermochemistry Terminology state properties => properties which depend only on the initial and final states => properties which are path independent non-state properties => properties which are path dependent state properties => E non-state properties => q & w

Thermochemistry Terminology: 

Thermochemistry Terminology exothermic - reaction that gives off energy endothermic - reaction that absorbs energy chemical energy - energy associated with a chemical reaction thermochemistry - the quantitative study of the heat changes accompanying chemical reactions thermodynamics - the study of energy and its transformations

Enthalpy: 

Enthalpy heat at constant pressure qp = DH = Hproducts - Hreactants Exothermic Reaction DH = (Hproducts - Hreactants) < 0 H2O(l) -----> H2O(s) DH < 0 Endothermic Reaction DH = (Hproducts - Hreactants) > 0 H2O(l) -----> H2O(g) DH > 0

Enthalpy: 

Enthalpy H = E + PV DH = DE + PDV DE = DH – PDV Where text uses U for internal energy

Pressure-Volume Work: 

Pressure-Volume Work

First Law of Thermodynamics: 

First Law of Thermodynamics heat => q internal energy => E internal energy change =>DE work => w DE = q - w (Engineering convention)

Specific Heat: 

Specific Heat the amount of heat necessary to raise the temperature of 1 gram of the substance 1oC independent of mass substance dependent s.h. Specific Heat of Water = 4.184 J/goC

Heat: 

Heat q = m * s.h. * Dt where q => heat, J m => mass, g s.h. => specific heat, J/g*oC Dt = change in temperature, oC

Molar Heat Capacity: 

Molar Heat Capacity the heat necessary to raise the temperature of one mole of substance by 1oC substance dependent C

Heat Capacity: 

Heat Capacity the heat necessary to raise the temperature 1oC mass dependent substance dependent C

Heat Capacity: 

Heat Capacity C = m X s.h. where C => heat capacity, J/oC m => mass, g s.h. => specific heat, J/goC

Slide21: 

Plotted are graphs of heat absorbed versus temperature for two systems. Which system has the larger heat capacity? A, B

Heat Transfer: 

Heat Transfer qlost = - qgained (m X s.h. X Dt)lost = - (m X s.h. X Dt)gained

EXAMPLE If 100. g of iron at 100.0oC is placed in 200. g of water at 20.0oC in an insulated container, what will the temperature, oC, of the iron and water when both are at the same temperature? The specific heat of iron is 0.106 cal/goC.: 

EXAMPLE If 100. g of iron at 100.0oC is placed in 200. g of water at 20.0oC in an insulated container, what will the temperature, oC, of the iron and water when both are at the same temperature? The specific heat of iron is 0.106 cal/goC. (100.g*0.106cal/goC*(Tf - 100.)oC) = qlost - qgained = (200.g*1.00cal/goC*(Tf - 20.0)oC) 10.6(Tf - 100.oC) = - 200.(Tf - 20.0oC) 10.6Tf - 1060oC = - 200.Tf + 4000oC (10.6 + 200.)Tf = (1060 + 4000)oC Tf = (5060/211.)oC = 24.0oC

EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?: 

EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC? q = DHice + DHfusion + DHwater + DHboil. + DHsteam q = DHice + DHfusion + DHwater + DHboil. + DHsteam

EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC? q = DHice + DHfusion + DHwater + DHboil. + DHsteam: 

EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC? q = DHice + DHfusion + DHwater + DHboil. + DHsteam q = (10.0g*2.09J/goC*15.0oC) + (10.0g*333J/g) + (10.0g*4.18J/goC*100.0oC) + (10.0g*2260J/g) + (10.0g*2.03J/goC*27.0oC) q = (314 + 3.33X103 + 4.18X103 + 2.26X104 + 548)J = 23.3 kJ

Bomb Calorimeter: 

Bomb Calorimeter Parr calorimeter

EXAMPLE: 

EXAMPLE A 1.000g sample of a particular compound produced 11.0 kJ of heat. The temperature of the calorimeter and 3000 g of water was raised 0.629oC. How much heat is gained by the calorimeter? heat gained = - heat lost heatcalorimeter + heatwater = heatreaction heatcalorimeter = heatreaction - heatwater

EXAMPLE: 

EXAMPLE A 1.000g sample of a particular compound produced 11.0 kJ of heat. The temperature of the calorimeter and 3000 g of water was raised 0.629oC. How much heat is gained by the calorimeter? heatcalorimeter = heatreaction - heatwater heat = 11.0 kJ - ((3.00kg)(0.629oC)(4.184kJ/kgoC)) = 3.1 kJ

Example: 

Example What is the mass of water equivalent of the heat absorbed by the calorimeter? #g = (3.1 kJ/0.629oC)(1.00kg*oC/4.184kJ) = 6.5 x 102 g

Example: 

Example A 1.000 g sample of ethanol was burned in the sealed bomb calorimeter described above. The temperature of the water rose from 24.284oC to 26.225oC. Determine the heat for the reaction. m = (3000 + "647")g H2O q = m X s.h. X Dt = (3647g)(4.184J/goC)(1.941oC) = 29.61 kJ

Slide31: 

When graphite is burned to yield CO2, 394 kJ of energy are released per mole of C atoms burned. When C60 is burned to yield CO2 approximately 435 kJ of energy is released per mole of carbon atoms burned. Would the buckyball-to-graphite conversion be exothermic or endothermic? exothermic, endothermic

Laws of Thermochemistry: 

Laws of Thermochemistry 1. The magnitude of DH is directly proportional to the amount of reactant or product. s --> l DH => heat of fusion l --> g DH => heat of vaporization

Laws of Thermochemistry: 

Laws of Thermochemistry 2. DH for a reaction is equal in magnitude but opposite in sign to DH for the reverse reaction. H2O(l) -----> H2O(s) DH < 0 H2O(s) -----> H2O(l) DH > 0

Laws of Thermochemistry: 

Laws of Thermochemistry 3. The value of H for the reaction is the same whether it occurs directly or in a series of steps. DHoverall = DH1 + DH2 + DH3 + · · ·

Hess' Law: 

Hess' Law a relation stating that the heat flow in a reaction which is the sum of a series of reactions is equal to the sum of the heat flows in those reactions

EXAMPLE CH4(g) + 2 O2(g) -----> CO2(g) + 2 H2O(l): 

EXAMPLE CH4(g) + 2 O2(g) -----> CO2(g) + 2 H2O(l) CH4(g) -----> C(s) + 2 H2(g) DH1 2 O2(g) -----> 2 O2(g) DH2 C(s) + O2(g) -----> CO2(g) DH3 2 H2(g) + O2(g) -----> 2 H2O(l) DH4 --------------------------------------------- CH4(g) + 2 O2(g) -----> CO2(g) + 2 H2O(l) DHoverall = DH1 + DH2 + DH3 + DH4

Standard Enthalpy of Formation: 

Standard Enthalpy of Formation the enthalpy associated with the formation of a substance from its constituent elements under standard state conditions

Calculation of DHo: 

Calculation of DHo DHo = Sc*DHfoproducts - Sc*DHforeactants

Example What is the value of DHrx for the reaction: 2 C6H6(l) + 15 O2(g) --> 12 CO2(g) + 6 H2O(g): 

Example What is the value of DHrx for the reaction: 2 C6H6(l) + 15 O2(g) --> 12 CO2(g) + 6 H2O(g) from Appendix J Text C6H6(l) DHfo = + 49.0 kJ/mol O2(g) DHfo = 0 CO2(g) DHfo = - 393.5 H2O(g) DHfo = - 241.8 D Hrx = [S c* D Hfo]product - [S c* D Hfo]reactants

Example What is the value of DHrx for the reaction: 2 C6H6(l) + 15 O2(g) --> 12 CO2(g) + 6 H2O(g) from Appendix J Text C6H6(l) DHfo = + 49.0 kJ/mol; O2(g) DHfo = 0 CO2(g) DHfo = - 393.5; H2O(g) DHfo = - 241.8 D Hrx = [S c* D Hfo]product - [S c* D Hfo]reactants: 

Example What is the value of DHrx for the reaction: 2 C6H6(l) + 15 O2(g) --> 12 CO2(g) + 6 H2O(g) from Appendix J Text C6H6(l) DHfo = + 49.0 kJ/mol; O2(g) DHfo = 0 CO2(g) DHfo = - 393.5; H2O(g) DHfo = - 241.8 D Hrx = [S c* D Hfo]product - [S c* D Hfo]reactants D Hrx = [12(- 393.5) + 6(- 241.8)]product - [2(+ 49.0 ) + 15(0)]reactants kJ/mol = - 6.271 x 103 kJ

Fossil Fuels: 

Fossil Fuels coal petroleum natural gas

Energy Sources: 

Energy Sources

Based on 1998 Data: 

Based on 1998 Data

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