2013/04/02**on** **energy**12**on** **energy****on** **energy** Drive our machinery andappliances Power our transportvehicles Necessary for life :- Photosynthesis- Food from which wederive **energy**341

2013/04/02Nearly all our **energy** is derived from chemical reacti**on**s Combusti**on**: Eskom power stati**on**s , taxis, braai’s, etc**energy** changesaccompany chemical reacti**on**s Electrochemical: Cell ph**on**es, laptops, iPods, etc Nuclear reacti**on**s: Koeberg power stati**on**,submarines, most superheroes Biological: Photosynthesis, Metabolism of food5Why is it that some reacti**on**s release **energy**,whereas others require **energy** in order to occur ?Can we determine or estimate the amount of**energy** required or released for a given reacti**on**/s ?6The **energy** used to causean object with mass tomove against a force…..or…Examining the ways in which matter can possess**energy** and how that **energy** can be transferredform **on**e piece of matter to another.The **energy** used tocause the temperatureof an object to rise782

2013/04/02 Energy of moti**on**. Stored **energy**. Anything that has mass and is moti**on**,possess kinetic **energy**. eg. Atoms, molecules, airplanes, etc.1 kg The **energy** possessed by an object byvirtue of its positi**on**, or its c**on**diti**on**, orits compositi**on**.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 positi**on** : 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 gravitati**on**al c**on**stant (g = 9.8 m/s 2 )11Gravity is important for large bodies but what about atomsand molecules ? ELECTROSTATIC potential **energy**, arisesfrom the interacti**on** 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 → repulsi**on** : E el > 0 Q1 and Q2 opposite → aEracF**on** : 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 moti**on** :Due to its positi**on** :J is quite small so we often use kJTraditi**on**ally : **energy** changes **energy** 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 wem**on**itor changes in **energy** without going insane ?There **on**siderThe porti**on** 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 **energy** changes inour daily lives :WORKThere **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 **on**e..Heat can be viewed as the**energy** 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 **energy**each 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 completedc**on**verted 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 isc**on**verted to E k , do we know E p ?E p = 274 JE k = ½ mv 2v 2 =v =2122Remember : In order to simplify our investigati**on** of the universe (surroundings) wefocus **on** defined regi**on** of it (system).In general **energy** is c**on**verted from **on**e form to another, andtransferred from **on**e 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 comp**on**ents of the system.23246

2013/04/02It is nearly impossible to sum the c**on**tributi**on**s of all thecomp**on**ents 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 directi**on** inwhich **energy** is transferred ?For : ΔU > 0 (+)How do we use the sign to determine the directi**on** 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 in**energy** 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 d**on**e byw < 0q > 0system **on** the surroundings. If w > 0 (positive) then work is being d**on**e 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) **on**s.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 moti**on** :Due to its positi**on** :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 **on**ly work is mechanicalwork associated with a change involume of the system(open to the atmosphere).4010

2013/04/02Let us c**on**sider a reacti**on** taking place (atc**on**stant pressure) in a vessel that is equippedwith a pist**on** that is free to move up or down.At c**on**stant pressure : w = -PΔVSurroundingsSystem4142Enthalpy (H) accounts for the flow of heat in processes occurring atc**on**stant pressure where **on**ly P-V work is performed

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2013/04/0249Note: Equati**on**s for chemical reacti**on**s recordedin this fashi**on** **on**s 502 H 2 (g) + O 2 (g)The magnitude of ΔH is directlyproporti**on**al to the amount of reactantc**on**sumed.CH 4 (g) +2O 2 (g)2CH 4 (g) +4O 2 (g)Reacti**on** is exothermic :System releases **energy**Products 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 ac**on**stant pressure system ?First write down the balanced chemical reacti**on** equati**on** :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 kJ**on** exothermic or endothermic ? 5614

2013/04/02Eg. Calculate the heat needed to c**on**vert 25 ml water to steam at atmosphericpressure (density = 1.0 g/ml)First write down the balanced chemical reacti**on** equati**on** :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 soluti**on** = 60.0 g + 4.25 g = 64.3 g (assumpti**on**)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 reacti**on**∆HAB∆HBD∆HAB∆HCD∆HBCBBC73Identical to the **on**e step reacti**on**7475 76

2013/04/0277 7879 8020

2013/04/02Enthalpy of formati**on**O C O∆H fStandard State ∆H°CCarb**on**C graphiteOOxygenO 281 8283 8421

2013/04/02∆H f ⁰ values have been experimentally determined for a vast number ofcompounds. **on**s(∆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 combusti**on** 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

2013/04/02Nutriti**on**al Informati**on**Typical values per 100 gramsNutriti**on**al Informati**on**Typical 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 c**on**cerns, coupled with highoil prices, and increasing governmentsupport, **energy** legislati**on**, incentives andcommercializati**on**9324