We are completely dependent on energy

We are completely dependent on energy• Drive our machinery andappliances• Power our transportvehicles• Necessary for life :- Photosynthesis- Food from which wederive energy

We are completely dependent on energy

Nearly all our energy is derived from chemical reactions• Combustion: Eskom power stations , taxis, braai’s, etc• Electrochemical: Cell phones, laptops, iPods, etc• Nuclear reactions: Kuberg power station, submarines,most superheroes• Biological: Photosynthesis, Metabolism of food

Energy**Chemistry** We need to understand what energy changesaccompany chemical reactions• Why 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 ?

Capacity todo WorkThe energy used to causean object with mass tomove against a force…..or…Transfer ofHeatThe energy used tocause the temperatureof an object to rise

Examining the ways in which matter can possessenergy and how that energy can be transferredform one piece of matter to another.Kinetic EnergyPotential Energy

• Energy of motion.• Anything that has mass and is motion,possess kinetic energy.1 kg• eg. Atoms, molecules, airplanes, etc.E k = ½ mv 2E k of an object depends on its mass (m) and speed (v).

• Stored energy.• The energy possessed by an object byvirtue of its position, or its condition, orits composition.• Potential energy arises when a forceoperates on an object.

Gravity 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 → attraction : E el < 0Q 1dQ 2+ -+ -+ +

The SI unit for energy : Joule (J)1 J = 1 kg.m 2 .s -2E k = ½ mv 2= ½ . (2 kg). (1m.s -1 ) 2= 1 kg.m 2 .s -2= 1 JJ is quite small so we often use kJTraditionally : energy changes are 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 J

Due to its motion :Due to its position :As temperature changes :Moving an object :Unit for energy :Joule (J)

If everything has some sort of energy, how then do wemonitor changes in energy without going insane ?The portion we single out for study :SYSTEMEverything else :SURROUNDINGS

There are essentially three types of systems that we considerMatter and energy can beexchanged with the surroundings.Can exchange energy but notmatter with the surroundings.Neither energy or matter can beexchanged with the surroundings.

There are two ways in which we experience energy changes inour daily lives :HEATThe energy transferred from ahotter object to a colder one..Heat can be viewed as theenergy transferred between asystem and its surroundings.

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 heightw = F x dd = 40 cm or 0.4 mF = ?F = m x gw = F x d= m x g x d= 70 kg x 9.8 m.s -2 x 0.40 m= 274 J

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.c. Assume the potential energy is completedconverted to kinetic energy when hits theground, at what speed will he hit the ground ?On 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 = 2E k /m = 2 x 274 kg.m 2 .s -2 / 70 kg = 7.8 m 2 .s -2v = 2.8 m.s -1

Potential Kinetic Work Heat

In general energy is converted from one form to another, andtransferred from one place to another.

Remember : In order to simplify our investigation of the universe (surroundings) wefocus on defined region of it (system).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.

Internal energy (U)It is nearly impossible to sum the contributions of all thecomponents of the system.HOWEVER MEASURING THECHANGE IN INTERNAL ENERGY ISPOSSIBLE (ΔU)ΔU = U FINAL – U INITIALFinalstateInitialstateΔU = U Final – U initialΔU > 0Energylost gainedSubtract the initial energy of thesystem from the final energy of thesystemFinalstateΔU = U Final – U initialΔU < 0

What info do we get from ΔU ?ΔU = U FINAL – U INITIAL1. NUMBER AND UNIT – MAGNITUDE.2. SIGN – DIRECTION IN WHICH ENERGY IS TRANSFERRED.ΔU > 0 (some positive value, +)ΔU < 0 (some negative value, -)

Internal energy (U)How do we use the sign to determine the direction inwhich energy is transferred ?For : ΔU > 0 (+)U Final > U InitialFinalstateThe system gains energy from thesurroundings.The internal energy of the systemtherefore increases, U Final is largerthan U Initial , and the change inenergy is positive.InitialstateΔU = U Final – U initialΔU > 0Energy gained fromthe surroundings

Internal energy (U)How do we use the sign to determine the direction inwhich energy is transferred ?For : ΔU < 0 (-)U Final < U InitialInitialstateEnergy lost fromthe systemThe system loses energy to thesurroundings.The internal energy of the systemtherefore decreases, U Final issmaller than U Initial , and the changein energy is negative.FinalstateΔU = U Final – U initialΔU < 0

ANY 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.• If w < 0 (negative) then work is being done bysystem on the surroundings.• If w > 0 (positive) then work is being done by thesurroundings on the system.w > 0w < 0q < 0q > 0

• Work is being on the system : w > 0 (positive)• Heat is being given off by the system : q < 0 (negative)ΔU = q + w= (-1150 J) + 480 J = -670 JΔU < 0, therefore system istransferring energy to thesurroundings

Exothermic

Endothermic

Internal energy (U)A(g) + B (g)ΔU < 0 ΔU > 0C (g)

Internal energy (U) isa STATE FUNCTION

Internal 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 °C

However some thermodynamic quantities such as heat (q) andwork (w) are 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.If changing the path changes theamount of heat , that path willalso change the amount of work.ΔUqw

Due to its motion :Due to its position :As temperature changes :Moving an object :Unit for energy :Joule (J)

ENERGY 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 INITIALw > 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.

The chemical and physical changesthat occur around us result from:HeatWorkAs 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).

Let 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 = -PΔVSurroundingsSystem

Enthalpy (H) accounts for the flow of heat in processes occurring atconstant pressure where only P-V work is performed

We cannot measure enthalpy (H), however we can measure thechange in enthalpy (ΔH).

We can further simplify :

Heat is being absorbed by the system q p > 0, and ΔH > 0Process is endothermic.Heat is being given off by the system q p < 0, and ΔH < 0Process is exothermic.

Note : Equations for chemical reactions recordedin this fashion are known as thermochemicalequations

Enthalpy2 H 2 (g) + O 2 (g)Reaction is exothermic :• System releases energy• Products must have lowerenthalpy than reactantsΔH < 0Exothermic2 H 2 O(g)

EnthalpyThe magnitude of ΔH is directlyproportional to the amount of reactantconsumed.CH 4 (g) +2O 2 (g)2CH 4 (g) +4O 2 (g)

EnthalpyCH 4 (g) + 2O 2 (g)ΔH =-890 kJΔH =890 kJ

EnthalpyCH 4 (g) + 2O 2 (g)ΔH =-890 kJΔH =-802 kJ

EnthalpyCH 4 (g) + 2O 2 (g)ΔH =-890 kJΔH =-802 kJ

Eg. 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 :But remember that this is for 1 mole of CH 4 (g), we need for 4.50gWe therefore need to work out number of moles, n = m/M wM w (CH 4 ) = 16.0 g.mol -1n = m/Mw = 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 kJIs the reaction exothermic or endothermic ?

Eg. 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 = ρ x V = 1.0 g.ml -1 x 25 ml = 25 gn = m/Mw = 25 g / 18.016 g.mol -1= 1.4 mol (H 2 O)ΔH = 44 kJ : 1moleΔH = ? kJ : 1.4 molsΔH = 62 kJRecall : ΔH = q p = 62 kJ

Calculate mass of water : m = ρ x V = 1 L x 1 kg.L -1 = 1.00 x 10 3 gΔT = 98.0 :C – 23.0 :C = 75.0 :C or 75 K

But we need moles not grams : n = m/M :

ΔT = 16.9 :C – 22.0 :C = -5.1 :C or -5.1 Km = total mass of solution = 60.0 g + 4.25 g = 64.3 g (assumption)

But we want kJ.mol -1 : n=m/Mn = m/M= 4.25 g / 80.052 g.mol -1= 0.0531 mol

ΔT = 30.57 :C – 23.44 :C = 7.13 :Cq rxn = -56.0 kJ : 2.200 g? kJ : 1 gq rxn = -25.5 kJ.g -1

q rxn = -56.0 kJ : 0.02035 mol? kJ : 1 molq rxn = -2.75 kJ.mol -1kJ.g -1 x

∆H overall reactionA∆H AB∆H BD∆H ABBB∆H∆H BCCDC

Identical to the one step reaction

Enthalpy of formationO C O∆H fStandard State ∆H°CCarbonOOxygenC graphiteO 2

∆H f : values have been experimentally determined for a vast number ofcompounds. We use these values to calculate the enthalpies of many reactions(∆H: rxn ).∆H f:[C 3 H 8 (g)]∆H f:[CO 2 (g)]∆H f:[H 2 O (g)]x –x 3x 4

the ∆H f : values of reactant and products.∆H: rxn from∆H: rxn = ∑ n∆H f : (products) - ∑ m∆H f : (reactants)

∆H: rxn = ∑ n∆H f : (products) - ∑ m∆H f : (reactants)∆H: rxn for the combustion of propane ?values for the reactants and products∆H f :∆H f : (kJ/mol)∆H: rxn = ∑ n∆H f : (products) - ∑ m∆H f : (reactants)= [ 3(-393.5) + 4(-285.8) ] – [(-103.85) + 5(0) ]= 2220 kJ/mol

Foods

Nutritional 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 g

Fuels

AlternativeEnergy SourcesClimate change concerns, coupled with highoil prices, and increasing governmentsupport, are driving increasing renewableenergy legislation, incentives andcommercialization