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Chapter 6: Thermochemistry. AP Chemistry. Energy. work. The capacity to do WORK or produce HEAT. No work. Forms of Energy. Potential Energy. Stored energy due to position or composition. Gravitational Potential Energy ( E g ).
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Chapter 6: Thermochemistry AP Chemistry
Energy work The capacity to do WORK or produce HEAT No work
Forms of Energy Potential Energy Stored energy due to position or composition
Gravitational Potential Energy (Eg) Stored energy due to an object’s height above the surface of the earth
Chemical Potential Energy (Echem) Methane CH4 Landfill Gas Flare Energy Stored in the bonds of molecules
Forms of Energy Kinetic Energy Energy due to motion
Mechanical Kinetic Energy (Ek) Ek = ½ mv2 Energy due to an object’s motion
Thermal Kinetic Energy • Energy due to molecular motion
First Law of Thermodynamics AKA The Law of Conservation of Energy Energy cannot be created or destroyed; it can just change forms
James Prescott Joule Wealthy English Beer Baron (1818 - 1889)
Joule’s Chamonix Falls Experiment Potential Energy (mechanical) P.E. = m g h Kinetic Energy (mechanical and thermal) K.E. = ½ m v2 Q = m c DT Law of Conservation of Energy Energy cannot be created or destroyed in only changes forms
How we look at reactions • system: The Reactants and Products • surroundings: The room, container, etc. http://www.weldonryan.com/new_page_1.htm
DE = Efinal- Einitial DE E R P R P Internal Energy, E • The TOTAL energy of the system (potential + kinetic energy) DE E Energy GAINED by system Energy LOST by system
Temperature vs. Heat Temperature— Measure of the average kinetic energy of molecules Cl2, N2, He 100, 300, 700K
Heat (q) Transfer of thermal energy between the system and surroundings from high to low temps
Work (w) A force exerted over a distance w = F•Dd Examples: syringe, hatchback
Internal Energy surroundings • The energy of a system can be changed by a flow of heat (q), work (w) or both System DE = q + w Note: q and w must have a magnitude (#) and a sign (+ or -)
Energy transfer of heat only (no work) EXOTHERMIC ENDOTHERMIC heat flows INTO the system heat EXITS the system the temperature _________ the temperature ________ surroundings system system
ENDOTHERMIC +DE E R P R P Energy transfer of heat only EXOTHERMIC -DE E Energy GAINED by system (+q) Energy LOST by system (-q)
demos: Identify the following as endo- or exothermic. 1) 2 H2 (g) + O2 (g) 2) Ba(OH)2•8H2O + NH4Cl BaCl2 •2H2O + 2NH3 + 8H2O D 3) CuSO4 • 5 H2O (s) 4) CuSO4 (s) + H2O 5) NaC2H3O2 (aq) Exo H2O(g) Endo Endo Exo NaC2H3O2 (s) Exo
P =1 atm cnst Dd When work is done by a gas F P = w = FDd DV =ADd surroundings A F = PA System wout = PADd wout= PDV Sign of w? Why? -w= PDV w= -PDV
Sample Problem: • Calculate DE for an exothermic reaction in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. DE = q + w DE = -15.6 kJ + 1.4 kJ = -14.2 kJ
Sample Problem A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 X 106 L to 4.50 X 106 L by the addition of 1.3 X 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate DE for the process. DE = q + w w = -PDV q = +1.3 X 108 J w = - 1.0 atm (4.5 X 106 – 4.0 X 106 L) w = -5 X 107atm • L 5 X 107 atm • L = ? J DE = 1.3 X 108 J + -5.1 X 10-7 J = 8.0 X 107 J
Units of Energy • calorie • Joule • 1 cal = 4.18 J
State Function—UPPER CASE depends on the initial and final positions; not the path depends on the current state; not how it got there 6 pts either way Panthers Visitors Explain why Energy is a state function
start finish Path Function—lower case depends on the pathbecause they have to do with a transition. Explain why work is a path function Explain why heat is a path function
STATE path • Temperature • Pressure • Volume • Heat Capacity • Density • Energy • Displacement • Altitude • Heat • Work • Distance
H2O2 Ep H2O + O2 Demo: State or Path Function? control add KI add MnO4
Sign Conventions for q & w— Take the point of view of the system! • +q: • - q: • +w: • -w: heat flows into the system heat flows out of the system work is done ON the system work is done BY the system
Types of Reaction systems OPENCLOSEDINSULATED Enthalpy; Endo/Exo Problems Calorimetry HARD! Engineering/Thermo Classes
Pressure • volume = work Energy = q + w Enthalpy (H) Enthalpy a property defined as internal energy + product of pressure and volume H = E + PDV DH = DE + PDV
Benjamin Thomson AKA – Count Rumford American?/English Scientist (1753-1814) Heat is produced by friction that is generated by the process of doing work
Comparing DE and DH • Reactions that do not involve gases 2KOH (aq) + H2SO4(aq) K2SO4 (aq) + 2H2O(l) DH = DE + PDV What is DV? With no gases, DV = 0 sooooooo DH = DE
Comparing DE and DH • Reactions in which the moles of gas does NOT change N2 (g) + O2 (g) 2NO(g) DH = DE + PDV What is DV? same moles of gas, DV = 0 sooooooo DH = DE
Comparing DE and DH—redo this; add derivation • Reactions in which the moles of gas DOES change 2H2 (g) + O2(g) 2H2O (g) DH = DE + PDV What is DV? waaaayyyy smaller than DH so that it’s insignificant and DH ≈ DE
Heat absorbed q J C = = DT DT oC How can we calculate heat (q)? • Heat Capacity (C): quantity of heat needed to change it’s temperature by 1 oC an extensive property— depends on amount of the substance
Heat q J c = = m • DT g • oC m • DT How can we calculate heat (q)? 2) specific heat capacity (c) intensive property does not depend on amount can be used to identify a substance The amount of heat needed to raise the temperature of 1 g of substance 1oC cH2O(l) = 4.18 J/g •oC
Demos 1) Which metal will melt more ice? How do you know? cCu = 0.385 J/g oC cPb = 0.128 J/g oC cAl = 0.900 J/g oC cSn = 0.21 J/g oC All metals have the same mass (70 g) and Tinitial (100oC) 2) How much energy is in a match? Heat Lost = Heat Gained 3) Dollar Bill
Calorimetry: A calorimeter is a device used to experimentally determine the amount of heat associated with a chemical reaction (a device that measures calories)
Consider a system at a constant pressure (an insulated system)—check/redo this derivation Recall: E = q + w and E = H -PDV E = q - PDV q - PDV = H - PDV q - PDV = H - PDV H = q Conclusion: Enthalpy = heat AT A CONSTANT PRESSURE DH = qp
Bomb Calorimetry: used to determine heat content in food/fuels volume constant __________ E = q + w E = q - PDV E = qv w = 0 since constant volume
Hess’s Law: In going from a set of reactants to a set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps
Hess’s Law Problems Use the thermochemical equations to determine the enthalpy for the reaction: C2H4O(l) + 5/2O2(g) 2CO2(g) + 2H2O(g) 2C2H4O(l) + 2H2O(l)2C2H6O(l) + O2(g) DH=610.5KJ 2CO2(g) + 3H2O(l)C2H6O(l) + 3O2(g) DH= 2056.5KJ
NO2(g) + 2H2(g) 1/2N2(g) + 2H20(l) 2NH3(g)N2(g) + 3H2(g) DH= 23KJ 2H2O(l)+NH3(g)NO2(g) + 7/2H2(g) DH=54KJ
Things to Remember 1) If you reverse a reaction, you must change the sign of DH 2) If you multiply the coefficients by an integer, do the same to DH
Standard State • Symbol = o • gases are at 1 atm ex: NO(g) • solutions are at 1.0 M ex: NaCl(aq), Ba2+ • Pure liquids and solids are in their condensed states ex: H2O (l)