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CH 7_student_6_per_page_sp07 - Chemistry 140 Fall 2002...

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Unformatted text preview: Chemistry 140 Fall 2002 Contents 7-1 7-2 7-3 7-4 7-5 7-6 7-7 7-8 7-9 Getting Started: Some Terminology Heat Heats of Reaction and Calorimetry Work The First Law of Thermodynamics Heats of Reaction: ∆U and ∆H The Indirect Determination of ∆H: Hess’s Law Standard Enthalpies of Formation Fuels as Sources of Energy Chapter 7: Thermochemistry CHM 25 Spring 2007 Dr. Berk & Dr. Koel Slide 1 of 53 SP 07 LU Slide 2 of 53 SP 07 LU Some Terminology ♦SYSTEM • The object under study Energy & Chemistry ENERGY is the capacity to do work (w) or transfer heat (q). Other forms of energy — ♦SURROUNDINGS • Everything outside the system light electrical kinetic and potential SP 07 LU Slide 4 of 53 SP 07 LU Slide 3 of 53 Potential & Kinetic Energy Potential ♦Potential Energy — energy a motionless body has by virtue of its position. ♦Kinetic Energy — energy of motion Slide 5 of 53 SP 07 LU Potential Energy on the Atomic Scale ♦ Positive and negative particles (ions) attract one another. ♦ Two atoms can bond ♦ As the particles attract they have a lower potential energy Slide 6 of 53 NaCl — composed composed of Na+ and Cl- ions. of and Cl SP 07 LU 1 Chemistry 140 Fall 2002 Thermodynamics Thermodynamics ♦Thermodynamics is the science of heat (energy) transfer. Heat (q) Energy transferred between a system and its surroundings as a result of a temperature difference. ♦Heat flows from hotter to colder. • Temperature may change. • Phase may change (an isothermal process). Slide 7 of 53 Heat transfers until is established. T measures energy transferred. SP 07 LU Slide 8 of 53 SP 07 LU Conservation of Energy All of thermodynamics depends on the law of CONSERVATION OF ENERGY. ♦ In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed. qsystem + qsurroundings = 0 qsystem = -qsurroundings Slide 9 of 53 SP 07 LU Slide 10 of 53 ♦ to thermic: heat transfers from . • Change in T is negative for the system T(system) goes down T(surr) goes up SP 07 LU ♦ thermic: heat transfers from . • Change in T is positive for the system 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. temp. by 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) kcal But we use the unit called the But T(system) goes up T (surr) goes down 1 cal = 4.184 joules cal Slide 12 of 53 James Joule 1818-1889 Slide 11 of 53 SP 07 LU SP 07 LU 2 Chemistry 140 Fall 2002 Heat Capacity ♦ The quantity of heat required to change the temperature of a system by one degree. • Molar heat capacity. System is one mole of substance. Specific Heat Capacity Specific Heat Capacity How much energy is transferred due to T difference? The heat !" “lost” or “gained” is related to a) sample mass b) change in T and c) specific heat capacity (c) Specific heat capacity = heat lost or gained by substance (J) (mass, g)(T change, K) Slide 14 of 53 SP 07 LU • Specific heat capacity, (c). System is one gram of substance q = mc∆T q = C∆T SP 07 LU • Heat capacity (C) Mass * specific heat. Slide 13 of 53 # # $# $# # # $# $# Substance Spec. Heat (J/g•K) H2O 4.184 Ethylene glycol 2.39 Al 0.897 glass 0.84 Aluminum Slide 15 of 53 SP 07 LU If 25.0 g of Al cool from 310 oC to 37 oC, how to how many joules of heat energy are lost by the Al? Al? Specific heat capacity = heat lost or gained by substance (J) (mass, g)(T change, K) Slide 16 of 53 SP 07 LU % $ & & ! Slide 17 of 53 '( ") "* " SP 07 LU Slide 18 of 53 ♦ Use heat transfer as a way to find specific heat capacity, c ♦ 55.0 g Fe at 99.8˚C ♦ Drop into 225 g water at 21.0˚C ♦ Water and metal come to 23.1˚C ♦ What is the specific heat capacity of the metal? SP 07 LU 3 Chemistry 140 Fall 2002 Chemical Reactions Involve Energy Heats of Reaction and Calorimetry ♦ Chemical energy. • Contributes to the internal energy of a system. ♦ Heat of reaction, qrxn. • The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system, at constant temperature. Slide 19 of 53 SP 07 LU Slide 20 of 53 SP 07 LU How do you determine the Heat of a Reaction? ♦Calculations • These are only an estimate Bond Energies ♦ Think of a chemical reaction = the making and breaking of bonds ... it all involves ENERGY ♦ Therefore bond energies can be used to approximate energies of reaction Think of it like this: reactants =(break apart)=> atoms =(come together)=> products costs energy releases energy Slide 22 of 53 SP 07 LU ♦Experiments - Calorimetry • Studio #7: Fuels and Studio #8 Hess’s Law will allow you to compare calculated versus experimental values for some common chemical reactions. Slide 21 of 53 SP 07 LU Bond Energies Bond Energies ♦ An example: C3H8(g) + 5O2 (g) 3CO2 (g) + 4 H2O (g) ♦ Just tally up the energies in the bonds … Hrxn = BE (bonds broken) – BE (bonds formed) NOTE: additional value for bonds in CO2 in footnote (b) Slide 23 of 53 SP 07 LU Slide 24 of 53 SP 07 LU 4 Chemistry 140 Fall 2002 Fuels as Sources of Energy ♦ Fossil fuels. • Combustion is exothermic. • Non-renewable resource. • Environmental impact. ♦PE + KE = Internal Energy (E or U) ♦Int. U of a chemical system depends on • number of particles • type of particles • temperature SP 07 LU Slide 26 of 53 SP 07 LU Slide 25 of 53 +, + ♦The higher the T the The higher the internal energy energy ♦So, use changes in T So, ( T) to monitor changes in U ( U). changes U). Slide 27 of 53 SP 07 LU $ heat energy transferred U=q + w energy change work done by the by system system Slide 28 of 53 SP 07 LU heat transfer in (endothermic), +q heat transfer out heat (exothermic), -q (exothermic), +Most chemical reactions occur at constant P, so Heat transferred at constant P = qp Heat p qp = p SYSTEM U=q+w w transfer in (+w) Slide 29 of 53 H where H = enthalpy enthalpy E and so E = H + w (and w is usually small) H = heat transferred at constant P H = change in heat content of the system Slide 30 of 53 w transfer out (-w) SP 07 LU H = Hfinal - Hinitial SP 07 LU 5 Chemistry 140 Fall 2002 !"# Consider the formation of water Consider H2(g) + 1/2 O2(g) --> H2O(g) + 241.8 kJ (g) -O(g) 241.8 ' ' #./ Exothermic reaction — heat is a “product” Exothermic heat Slide 31 of 53 SP 07 LU and H = – 241.8 kJ and Most H values are labeled Ho Measured under '' ' Measured P = 1 bar Concentration = 1 mol/L Concentration T = usually 25 oC usually with all species in standard states e.g., C = graphite and O2 = gas Slide 32 of 53 SP 07 LU #./ Depend on how the reaction is written and on Depend on how the reaction is written and on phases of reactants and products phases of reactants and products phases ' H˚ = -242 kJ H˚ = -484 kJ H˚ = +242 kJ H˚ = -286 kJ ' #./ H2(g) + 1/2 O2(g) --> H2O(g) (g) -2 H2(g) + O2(g) --> 2 H2O(g) (g) -H2O(g) ---> H2(g) + 1/2 O2(g) O(g) --(g) H2(g) + 1/2 O2(g) --> H2O(l) (g) -Slide 33 of 53 NIST (Nat’l Institute for Standards and NIST Institute Technology) gives values of Technology) * ( ' ') # ) — the enthalpy change when 1 mol of the compound is formed from elements under standard conditions. standard ♦ The standard enthalpy of formation of a pure element in its reference state is 0. Slide 34 of 53 SP 07 LU SP 07 LU Standard States * 0 ' ') ) # ♦ C (graphite) + 2 H2 (g) ===> CH4 (g) H of = -74.8 kJ/mol ♦ KEY: elements in standard states elements forming one mole of product forming Hfo (CH4, g)= - 74.8 kJ/mol g)= By definition, By Hfo = 0 for elements in their standard for states. states. Slide 35 of 53 SP 07 LU Slide 36 of 53 SP 07 LU 6 Chemistry 140 Fall 2002 Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions Standard Enthalpies of Formation Slide 37 of 53 SP 07 LU Slide 38 of 53 SP 07 LU & Calculate H of reaction? reaction? ' ' #./ & ' ' #./ In general, when ALL ALL enthalpies of formation enthalpies are known, o Horxn = rxn o (products) Σ Hffo (products) Σ o - Σ Hffo (reactants) Σ Calculate the heat of combustion of methanol, i.e., Horxn for for CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 CH (g) -(g) H2O(g) Horxn = Σ Hfo (prod) - Σ (prod) Slide 40 of 53 Hfo (react) SP 07 LU Remember that Slide 39 of 53 always = final – initial SP 07 LU Enthalpy Diagrams Enthalpy of Reaction ∆Hrxn = Σ∆Hf°products- Σ∆H f°reactants Slide 41 of 53 SP 07 LU Slide 42 of 53 SP 07 LU 7 Chemistry 140 Fall 2002 1 1 $ $ & & & %$ & %$ Note that T is constant Note as ice melts as & $/ 2 & $/ 2 , , Changes of state involve energy " Ice + 333 J/g (heat of fusion) -----> Liquid water Ice ----!( / ") " Slide 43 of 53 SP 07 LU Slide 44 of 53 SP 07 LU What quantity of heat is required to vaporize 125 g of benzene, C6H6, at its boiling point, 80.1ºC? The heat of vaporization of benzene is 30.8 kJ/mol. Bomb Calorimeter qrxn = -qcal qcal = q bomb + q water + q wires +… Define the heat capacity of the calorimeter: qcal = Σmici∆T = C∆T all i heat Slide 45 of 53 SP 07 LU Slide 46 of 53 SP 07 LU $ $ ' !( & $ ' !( & $ !( %& %& &# &# Coffee Cup Calorimeter ♦A simple calorimeter. • Well insulated and therefore isolated. • Measure temperature change. Calculate heat of combustion of octane. C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O CO • Burn 1.00 g of octane ♦Temp rises from 25.00 to 33.20 oC Temp ♦Calorimeter contains 1200 g water ♦Heat capacity of bomb = 837 J/K Slide 47 of 53 SP 07 LU qrxn = -qcal • In lab we shall see that the calorimeter and solution will need to be separated out. Slide 48 of 53 SP 07 LU 8 Chemistry 140 Fall 2002 Indirect Determination of ∆H: Hess’s Law Making H2O from H2 involves two steps. Making H2(g) + 1/2 O2(g) ---> H2O(g) + 242 kJ (g) --H2O(g) ---> H2O(liq) + 44 kJ O(g) --O(liq) ----------------------------------------------------------------------- 3 4 & & ) 1 2 H2(g) + 1/2 O2(g) --> H2O(liq) + 286 kJ (g) -Example of 3 +, — Example If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. Slide 49 of 53 SP 07 LU Slide 50 of 53 Active Figure 6.18 Forming H2O can occur in a single step or in a two steps. Htotal is the same no matter which path is followed. SP 07 LU 3 4 & & ) 1 2 Σ H along one path = along Σ H along one path = Σ H along another path along Σ H along another path ♦ This equation is valid because H is a + $ ♦ These depend only on the state of the system and not how it got there. ♦ V, T, P, energy — and your bank account! ♦ Unlike V, T, and P, one cannot measure absolute H. Can only measure H. Slide 52 of 53 SP 07 LU Slide 51 of 53 Active Figure 6.18 Forming CO2 can occur in a single step or in a two steps. Htotal is the same no matter which path is followed. SP 07 LU Applying Hess’s Law ♦You wish to know the enthalpy change for the formation of liquid PCl3 from the elements. P4 (s) + 6 Cl2 (g) 4 PCl3 (l) Hº = ? The enthalpy change for the formation of PCl5 from the elements can be determined experimentally, as can the enthalpy change for the reaction of PCl3 (l) with more chlorine to give PCl5 (s): 1. P4 (s) + 10 Cl2 (g) 4 PCl5 (s) Hº = -1774.0 kJ 2. PCl3 (l) + Cl2 (g) PCl5 (s) Hº = -123.8 kJ Use these data to calculate the enthalpy change for the formation of 1 mol of PCl3 (l) from phosphorus and chlorine. Slide 53 of 53 SP 07 LU 9 ...
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