We are completely dependent on energy We are completely ...

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We are completely dependent on energy We are completely ...

2013/04/02ong>Weong> ong>areong> ong>completelyong> ong>dependentong> on energy12ong>Weong> ong>areong> ong>completelyong> ong>dependentong> on energyong>Weong> ong>areong> ong>completelyong> ong>dependentong> on energy Drive our machinery andappliances Power our transportvehicles Necessary for life :- Photosynthesis- Food from which wederive energy341


2013/04/02Nearly all our energy is derived from chemical reactions Combustion: Eskom power stations , taxis, braai’s, etcong>Weong> need to understand what energy changesaccompany chemical reactions Electrochemical: Cell phones, laptops, iPods, etc Nuclear reactions: Koeberg power station,submarines, most superheroes Biological: Photosynthesis, Metabolism of food5Why is it that some reactions release energy,whereas others require energy in order to occur ?Can we determine or estimate the amount ofenergy required or released for a given reaction/s ?6The energy used to causean object with mass tomove against a force…..or…Examining the ways in which matter can possessenergy and how that energy can be transferredform one piece of matter to another.The energy used tocause the temperatureof an object to rise782


2013/04/02 Energy of motion. Stored energy. Anything that has mass and is motion,possess kinetic energy. eg. Atoms, molecules, airplanes, etc.1 kg The energy possessed by an object byvirtue of its position, or its condition, orits composition.E k of an object depends on its mass (m) and speed (v). Potential energy arises when a forceoperates on an object.910Due to relative position : Arises when a force operates on anobject (push or pull) – most familiarGRAVITYHeightE p is given by the mass of the object (m), it’s relative height(h), and gravitational constant (g = 9.8 m/s 2 )11Gravity is important for large bodies but what about atomsand molecules ? ELECTROSTATIC potential energy, arisesfrom the interaction of charged particles.E el = K Q 1 Q 2dE el is given by the charges of the particles (Q 1 , Q 2 ), the distancebetween (d) them and K (8.99x109 J.m/C 2 ) Q1 and Q2 the same → repulsion : E el > 0 Q1 and Q2 opposite → aEracFon : E el < 0Q 1dQ 2+ -+ -+ +123


2013/04/02The SI unit for energy : Joule (J)E k ====1 J = 1 kg.m 2 .s -2Due to its motion :Due to its position :J is quite small so we often use kJTraditionally : energy changes ong>areong> expressed as calories (cal). The amount ofenergy needed to raise the temperature of 1g of water from 14.5 °C to 15.5 °C.1 cal = 4.184 J13As temperature changes :Moving an object :Unit for energy :Joule (J)14If everything has some sort of energy, how then do wemonitor changes in energy without going insane ?There ong>areong> essentially three types of systems that we considerThe portion we single out for study :SYSTEMEverything else :SURROUNDINGSMatter and energy can beexchanged with the surroundings.Can exchange energy but notmatter with the surroundings.15Neither energy or matter can beexchanged with the surroundings.164


2013/04/02There ong>areong> two ways in which we experience energy changes inour daily lives :WORKThere ong>areong> two ways in which we experience energy changes inour daily lives :HEATEnergy used to move anobject over some distancew = F x d(where w is work, F is force used,and d the distance over which theforce is exerted).Fd+ +17The energy transferred from ahotter object to a colder one..Heat can be viewed as theenergy transferred between asystem and its surroundings.18Eg. A guy weighing 70 kg is skipping, jumping up anddown on a spot. Each time he jumps he raises hisbody 40 cm of the ground.a. What happens to his potential energyeach time he jumps ?Eg. A guy weighing 70 kg is skipping, jumping up anddown on a spot. Each time he jumps he raises hisbody 40 cm of the ground.b. What quantity of work, in J, must he useto lift himself to this height19w = F x dd = 40 cm or 0.4 mF = ?F =w ====205


2013/04/02Eg. A guy weighing 70 kg is skipping, jumping up anddown on a spot. Each time he jumps he raises hisbody 40 cm of the ground.c. Assume the potential energy is completedconverted to kinetic energy when hits theground, at what speed will he hit the ground ?Potential Kinetic Work HeatOn coming back to the ground E p isconverted to E k , do we know E p ?E p = 274 JE k = ½ mv 2v 2 =v =2122Remember : In order to simplify our investigation of the universe (surroundings) wefocus on defined region of it (system).In general energy is converted from one form to another, andtransferred from one place to another.ENERGY OF THE SYSTEM WILL REMAIN CONSTANT UNLESS : HEAT IS ADDED TO OR TAKEN AWAY FROM THE SYSTEM. SOME WORK TAKES PLACE EITHER ON OR BY THE SYSTEM.The sum of all the kinetic and potential energies of all the components of the system.23246


2013/04/02It is nearly impossible to sum the contributions of all thecomponents of the system.ΔU = U FINAL – U INITIALHOWEVER MEASURING THECHANGE IN INTERNAL ENERGY ISPOSSIBLE (ΔU)ΔU =Subtract the initial energy of thesystem from the final energy of thesystemInternal energy (U)FinalstateInitialstateFinalstateΔU = U Final – U initialΔU > 0Energylost gainedΔU = U Final – U initialΔU < 025What info do we get from ΔU ?1. NUMBER AND UNIT – MAGNITUDE.2. SIGN – DIRECTION IN WHICH ENERGY IS TRANSFERRED.ΔU > 0 (some positive value, +)ΔU < 0 (some negative value, -)26How do we use the sign to determine the direction inwhich energy is transferred ?For : ΔU > 0 (+)How do we use the sign to determine the direction inwhich energy is transferred ?For : ΔU < 0 (-)U Final > U InitialFinalstateU Final < U InitialInitialstateEnergy lost fromthe systemThe system gains energy from thesurroundings.The internal energy of the systemtherefore increases, U Final is largerthan U Initial , and the change inenergy is positive.Internal energy (U)InitialstateΔU = U Final – U initialΔU > 0Energy gained fromthe surroundings27The system loses energy to thesurroundings.The internal energy of the systemtherefore decreases, U Final issmaller than U Initial , and the changein energy is negative.Internal energy (U)FinalstateΔU = U Final – U initialΔU < 0287


2013/04/02ANY CHANGE IN THE ENERGY OF THE SYSTEM ISACCOMPANIED BY AN OPPOSITE CHANGE IN THE ENERGY OFTHE SURROUNDINGS If q < 0 (negative) then heat is being transferredfrom the system to the surroundings. If q > 0 (positive) then heat is being transferredfrom the surroundings to the system.w > 0q < 029 If w < 0 (negative) then work is being done byw < 0q > 0system on the surroundings. If w > 0 (positive) then work is being done by thesurroundings on the system. 30ΔU ==ΔU < 0, therefore system istransferring energy to thesurroundings31 328


2013/04/02A(g) + B (g)Internal energy (U)ΔU < 0 ΔU > 0C (g)3334Internal energy of a certain mass ofwater at 25 °C is the same whether : the water is warmed from a lowertemperature to 25 °C or cooled from a higher temperatureto 25 °CBECAUSE U IS A STATE FUNCTION, ΔUDEPENDS ONLY ON THE FINAL AND INITIALSTATES OF THE SYSTEM, NOT ON HOW THECHANGE OCCURS.25 °C0 °C 100 °C35369


2013/04/02However some thermodynamic quantities such as heat (q) andwork (w) ong>areong> not state functions.Although ΔU (ΔU = q + w) does not depend on how the change occurs, thespecific amounts of q and w produced depends on the way in which the changeoccurs.Due to its motion :Due to its position :ΔUqwAs temperature changes :Moving an object :Unit for energy :Joule (J)3738ENERGY CAN BE NEITHER CREATED NOR DESTROYEDENERGY OF THE SYSTEM WILL REMAIN CONSTANT UNLESS : HEAT IS ADDED TO OR TAKEN AWAY FROM THESYSTEM. SOME WORK TAKES PLACE EITHER ON OR BY THESYSTEM.ΔU = U FINAL – U INITIALThe chemical and physical changesthat occur around us result from:w > 0q < 0w < 0q > 0Although ΔU (ΔU = q + w) does not depend on howthe change occurs, the specific amounts of q and wproduced depends on the way in which the changeoccurs. 39As gas is produced it expands against theatmosphere, system does work on the surroundingsUsually the only work is mechanicalwork associated with a change involume of the system(open to the atmosphere).4010


2013/04/02Let us consider a reaction taking place (atconstant pressure) in a vessel that is equippedwith a piston that is free to move up or down.At constant pressure : w = -PΔVSurroundingsSystem4142Enthalpy (H) accounts for the flow of heat in processes occurring atconstant pressure where only P-V work is performedong>Weong> cannot measure enthalpy (H), however we can measure thechange in enthalpy (ΔH).434411


2013/04/02ong>Weong> can further simplify :45 46---47-4812


2013/04/0249Note: Equations for chemical reactions recordedin this fashion ong>areong> known as thermochemicalequations 502 H 2 (g) + O 2 (g)The magnitude of ΔH is directlyproportional to the amount of reactantconsumed.CH 4 (g) +2O 2 (g)2CH 4 (g) +4O 2 (g)Reaction is exothermic :System releases energyProducts must have lowerenthalpy than reactantsEnthalpyΔH < 0Exothermic2 H 2 O(g)51Enthalpy5213


2013/04/02CH 4 (g) + 2O 2 (g)CH 4 (g) + 2O 2 (g)EnthalpyΔH =-890 kJΔH =890 kJEnthalpyΔH =-890 kJΔH =-802 kJ5354Eg. How much heat is released when 4.50 g of methane gas is burned in aconstant pressure system ?First write down the balanced chemical reaction equation :CH 4 (g) + 2O 2 (g)But remember that this is for 1 mole of CH 4 (g), we need for 4.50gEnthalpyΔH =-890 kJΔH =-802 kJong>Weong> therefore need to work out number of moles, n = m/M wM w (CH 4 ) = 16.0 g.mol -1n = m/M w = 4.50 g / 16.0 g.mol -1= 0.281 mols (CH 4 )ΔH = -890 kJ (CH 4 ) : 1mole of CH4ΔH = ? kJ : 0.281 molsΔH = -250 kJ55Is the reaction exothermic or endothermic ? 5614


2013/04/02Eg. Calculate the heat needed to convert 25 ml water to steam at atmosphericpressure (density = 1.0 g/ml)First write down the balanced chemical reaction equation :But remember that this is for 1 mole of H 2 O, we need for 25 mlDensity (ρ) = m / Vm =n ==ΔH = 44 kJ : 1moleΔH = ? kJ :ΔH =Recall :57 5859 6015


2013/04/02Calculate mass of water : m = ρ x V = 1 L x 1 kg.L -1 = 1.00 x 10 3 gΔT = 98.0 ⁰C – 23.0 ⁰C = 75.0 ⁰C or 75 K6162But we need moles not grams : n = m/M :63 6416


2013/04/02ΔT = 16.9 ⁰C – 22.0 ⁰C = -5.1 ⁰C or -5.1 Km = total mass of solution = 60.0 g + 4.25 g = 64.3 g (assumption)But we want kJmol -1 : n=m/Mn = m/M= 4.25 g / 80.052 g.mol -1= 0.0531 mol656667 6817


2013/04/02A19∆H overall reaction∆HAB∆HBD∆HAB∆HCD∆HBCBBC73Identical to the one step reaction7475 76


2013/04/0277 7879 8020


2013/04/02Enthalpy of formationO C O∆H fStandard State ∆H°CCarbonC graphiteOOxygenO 281 8283 8421


2013/04/02∆H f ⁰ values have been experimentally determined for a vast number ofcompounds. ong>Weong> use these values to calculate the enthalpies of many reactions(∆H⁰ rxn ).the ∆H f ⁰ values of reactant and products.∆H⁰ rxnfrom∆H⁰ rxn = ∑ n∆H f ⁰ (products) - ∑ m∆H f ⁰ (reactants)∆H f⁰ [C 3 H 8 (g)] x –∆H f⁰ [CO 2 (g)] x 3∆H f⁰ [H 2 O (g)] x 48586∆H⁰ rxn = ∑ n∆H f ⁰ (products) - ∑ m∆H f ⁰ (reactants)Foods∆H⁰ rxn for the combustion of propane ?values for the reactants and products∆H f ⁰∆H f ⁰ (kJ/mol)∆H⁰ rxn = ∑ n∆H f ⁰ (products) - ∑ m∆H f ⁰ (reactants)= [ 3(-393.5) + 4(-285.8) ] – [(-103.85) + 5(0) ]= 2220 kJ/mol878822


2013/04/02Nutritional InformationTypical values per 100 gramsNutritional InformationTypical values per 100 gramsEnergy245 kJ/ 58 kcalEnergy1050 kJ/ 251 kcalCarbohydratesProteinFats7.2 g4.6 g1.2 gCarbohydratesProteinFats5.2 g2.6 g5.2 g89 90Fuels919223


2013/04/02AlternativeEnergy SourcesClimate change concerns, coupled with highoil prices, and increasing governmentsupport, ong>areong> driving increasing renewableenergy legislation, incentives andcommercialization9324

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