Chemical Energetics

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stpm term 2 modular lecture notes

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CHEMICAL ENERGETICSLearning Outcomes :Students able to :(a) explain that most chemical reactions are accompanied by enthalpy changes (exothermic or endothermic);(b) define enthalpy change of reaction, H, and state the standard conditions;(c) define enthalpy change of formation, combustion, hydration, solution, neutralisation, atomisation, bond energy, ionisation energy and electron affinity;

Heat is the transfer of thermal energy between two bodies that are at different temperatures.Energy Changes in Chemical ReactionsTemperature is a measure of the thermal energy.Temperature = Thermal Energy


400Cgreater thermal energy

Thermochemistry is the study of heat change in chemical reactions.The system is the specific part of the universe that is of interest in the study.

openmass & energyExchange:closedenergyisolatednothingSYSTEMSURROUNDINGSExothermic process is any process that release heat from the system to the surroundings. Endothermic process is any process that absorb heat from the surroundings to the system. 2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energyenergy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)State FunctionsUsually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

6State FunctionsHowever, we do know that the internal energy of a system is independent of the path by which the system achieved that state.In the system below, the water could have reached room temperature from either direction.

7State FunctionsTherefore, internal energy is a state function.It depends only on the present state of the system, not on the path by which the system arrived at that state.And so, E depends only on Einitial and Efinal.

8Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

DH = H (products) H (reactants)H = heat given off or absorbed during a reaction at constant pressureHproducts < HreactantsH < 0Hproducts > HreactantsH > 0Thermochemical Equations

H2O (s) H2O (l)H = +6.01 kJIs H negative or positive?System absorbs heatEndothermicH > 06.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.Thermochemical EquationsCH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890.4 kJIs H negative or positive?System releases heatExothermicH < 0890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

Energy Diagrams50 kJ/mol300 kJ/mol150 kJ/mol100 kJ/mol-100 kJ/mol+200 kJ/mol

ExothermicEndothermicActivation energy (Ea) for the forward reactionActivation energy (Ea) for the reverse reaction (c) Delta H Change in heat energy always accompany a chemical rx.Enthaply change of reaction, H is defined as heat change (released or absorbd) in a chemical rx for the no of moles of rxts (or pdts) shown in a balanced eqn.Enthaply change of reaction, HH2O (s) H2O (l)H = +6.01 kJThe stoichiometric coefficients always refer to the number of moles of a substanceThermochemical EquationsIf you reverse a reaction, the sign of H changesH2O (l) H2O (s)H = -6.01 kJIf you multiply both sides of the equation by a factor n, then H must change by the same factor n.2H2O (s) 2H2O (l) H = 2 x +6.01 = +12.0 kJH2O (s) H2O (l)H = +6.01 kJThe physical states of all reactants and products must be specified in thermochemical equations.Thermochemical EquationsH2O (l) H2O (g)H = +44.0 kJTo standardize enthalpy measurement,following std condts are adopted :T @ 25 C / 298 KP @ 1 atm / 101 kPaAq soltn @ [1 moldm-3]Most stable allotrope @ 25 C & 1 atmEnthalpy change measured under such condtns = std enthalpy change, H Std enthalpy change of formation, HfStd enthalpy change of combustion, Hc hydration, Hhyd solution, Hsoln neutralisation, Hn atomisation, HaBond energyIonisation energy, IEElectron affinity, EATypes of std enthalpies :Std enthalpy change of formationEnthalpy change when 1 mole of substance is formed from its constituent elements under standard conditions (298K, 1atm)

Std enthalpy change of combustionHeat evolved when 1 mole of substance is completely burnt in excess oxygen under standard conditions (298K, 1atm). negative.

Standard enthalpy change of hydrationHeat evolved when 1 mole of gaseous ions is surrounded by water molecules under standard conditions (298K, 1atm). negative.

Standard enthalpy change of solutionEnthalpy change when 1 mole of the substance is completely dissolved in a solvent to form an infinitely dilute solution under standard conditions (298K, 1atm)

Hsol = +2258 - 1650 + 2(-364)Hsol = -120 kJ mol-1

The Solution Process for NaClDHsoln = Step 1 + Step 2 = 788 784 = 4 kJ/molStandard enthalpy change of neutralizationHeat evolved when 1 mole of water is formed in the neutralization between an acid and a base under standard conditions (298K, 1atm). Standard enthalpy change of atomization Enthaply change when 1 mole of gaseous atoms is formed from its element under standard conditions (298K, 1atm).

Mg (s) Mg (g) H = +147.7 kJ/mol1 Cl2 (g) Cl (g) H = +121.7 kJ/mol2...?Ionisation energy (1st IE)Min energy required to remove 1 mole of electrons from one mole of gaseous atoms to form one mole of the gaseous positive ions. positive.Bond energyAverage energy required to break 1 mole of covalent bond in the gaseous state. positive.Cl2 (g) 2Cl (g) H = + kJ/molNa (g) Na+ (g) + e H = +496 kJ/molElectron affinity (1st EA)Enthalpy change when 1 mole of the gaseous atom acquired 1 mole of electrons to form one mole of gaseous negative ions. O (g) + e O- (g) H = -141.1 kJ/molLearning Outcomes :Students able to :(d) calculate the heat energy change from experimental measurements using the relationship: heat change, q = mcT or q = mc ;(e) calculate enthalpy changes from experimental results.

CalorimetryBomb Calorimeter

Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.27Bomb CalorimeterReactions can be carried out in a sealed bomb, such as this one, and measure the heat absorbed by the water.

28Bomb CalorimeterBecause the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.For most reactions, the difference is very small, E, consider as H

29 Constant Pressure Calorimetry

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.30 Constant Pressure Calorimetry

Because the specific heat for water is well known (4.184 J/molK), we can measure H for the reaction with this equation:q = m c T @ q = m c 31q Enthalpy change/heat change in JoulemMass in gramSpecific heat capacity, c The heat needed to the temperature of 1 g of a substance by 1 C. Symbol: c, units: J/(gC).TTemperature change in C

Sample problem: (must know water = 4.18 J/gC)Q1. When 12 g of a food was burned in a calorimeter, the 100 mL of water in the calorimeter changed from 20C to 33C. Calculate the heat released.Soltn :

= 100 gx 13C= 5.4 kJq=mcTx 4.18 J/(gC)Q25 g of copper was heated from 20C to 80C. How much energy was used to heat the Cu? [c Cu = 0.38 J/(gC)]Q3If a 3.1 g ring is heated using 10.0 J, its temp. rises by 17.9C. Calculate the specific heat capacity of the ring.q=cmT= 0.38 J/(gC)x 5 gx 60 C= 114 JQ25 g of copper was heated from 20C to 80C. How much energy was used to heat the Cu? [c Cu = 0.38 J/(gC)]

Soltn :

q=cmTQ3If a 3.1 g ring is heated using 10.0 J, its temp. rises by 17.9C. Calculate the specific heat capacity of the ring.

Soltn :

= 0.18 J/(gC)c = q/mT10.0 J3.1 gx 17.9C=q = mcTQ4:3.00 g of octane was burned in a calorimeter with excess oxygen, the 1000 mL of water in the calorimeter rose from 23.0C to 57.6C. Write the thermochemical equation for octane, representing the enthalpy change of combustion.Soltn :S1 => = 1000 x 4.18 x 34.6= 144 628 JS2 =>Octane (C8H18) has a molar mass of 114.26 g/moln = 3.00

S3 => = 5508 kJ/molJmol144 628 J0.0263 mol=H = = 0.0263 mol114.26C8H18(l) + 25O2 (g) 8CO2 (g) + 9H2O(l) H0C = -5508 kJ2Learning Outcomes :Students able to :(a) state Hess law, and its use to find enthalpy changes that cannot be determined directly, e.g. an enthalpy change of formation from enthalpy changes of combustion;(b) construct energy level diagrams relating the enthalpy to reaction path and activation energy;(c) calculate enthalpy changes from energy cycles.

Hess LawHesss Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.(Enthalpy is a state function. It doesnt matter how you get there, only where you start and end.)Hesss LawBecause H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

41Calculate the standard enthalpy of formation of CS2 (l) given that:

C(s) + O2 (g) CO2 (g) H0 = -393.5 kJS(s) + O2 (g) SO2 (g) H0 = -296.1 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJEnergy cycle :C(s) + 2S(s) CS2(l) + O2 (g) CO2 (g) + 2O2 (g) SO2 (g) +3O2Calculate the standard enthalpy of formation of CS2 (l) given that:

C(s) + O2 (g) CO2 (g) H0 = -393.5 kJS(s) + O2 (g) SO2 (g) H0 = -296.1 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJ1. Write the enthalpy of formation reaction for CS2C(s) + 2S(s) CS2 (l)2. Add the given rxns so that the result is the desired rxn.C(s) + O2 (g) CO2 (g) H0 = -393.5 kJ2S(s) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJ+C(s) + 2S(s) CS2 (l)H0 = -393.5 + (2x-296.1) + 1072 = +86.3 kJBenzene (C6H6) burns in air to produce carbon dioxide and liquid water. Calculate the std enthalpy change of combustion of benzene. The standard enthalpies of formation of benzene is -49.04 kJ/mol, water is -187.90 kJ/mol & carbon dioxide is -393.50 kJ/mol.

C6H6 (l) + 15/2O2 (g) 6CO2 (g) + 3H2O (l)= - 2875.66 kJ/molLearning Outcomes :Students able to :(a) define lattice energy for simple ionic crystals in terms of the change from gaseous ions to solid lattice;(b) explain qualitatively the effects of ionic charge and ionic radius on the numerical magnitude of lattice energy values;(c) construct Born-Haber cycle for the formation of simple ionic crystals.Born Haber CycleLattice Energy, Hlatt: heat change when 1 mole of solid ionic compound is formed from its constituent gaseous ions.Na+(g) + Cl-(g) NaCl(s) Hlatt = -771 kJ/molLattice dissociation energy, - Hlattenergy needed to break down lattice of 1 mole ionic crystalline solid to forms its gaseous ionsNaCl(s) Na+(g) + Cl-(g) -Hlatt = +771 kJ/mol Magnitude of lattice energy :Increases when ionic charges increasesIncreases when ionic radii decreases# strong attraction between small ions & highly charged ions => Hlatt > negativeHlatt MgO is more tive than Hlatt Na2O => Hlatt MgO > Hlatt Na2O because ?

Mg2+ is smaller in size & has bigger charge than Na+Thermochemical cycle used to determine Hlattice

eg : the lattice energy of NaCl cannot be determine directly from the reaction : Na+(g) + Cl-(g) NaCl(s)Born Haber CycleBorn-Haber Cycles for NaClenthalpy HNaCl (s)Na+ (g) + Cl- (g)H latticeenthalpyNa (s) + Cl2 (g)H formationH atomisationNa (g) + Cl2 (g)H atomisationNa (g) + Cl (g)Na+ (g) + e- + Cl (g)H first ionisation energyH first electron affinityBorn-Haber Cyclesenthalpy HNaCl (s)Na+ (g) + Cl- (g)H latticeNa (s) + Cl2 (g)H formationH atomisation Na (g) + Cl2 (g)H atomisationNa (g) + Cl (g)Na+ (g) + e- + Cl (g)H first ionisation energyH first electron affinityHatmNa + HatmCl + H1st IE + H1st EA + Hlattice = Hformation Apply Hesss Law:: applying Hesss LawThere are two routes from elements to ionic compound Born-Haber Cycles: applying Hesss LawHatmNa + HatmCl + H1st IE + H1st EA + Hlattice = Hformation Rearrange to find the lattice energy: Hlattice = Hformation - (HatmNa + HatmCl + H1st IE + H1st EA) Constructing Born Haber cycle to calculate lattice energy of LiF

DH1 = DHatomizationDH3 = BE / DHatomizationDH2 = IE1DH5 = Lattice EnergyDH4 = EA1Calculating Lattice EnergyStep 1: Convert elements to atoms in the gas statee.g. for Li, Li (s) Li (g)DH1 = DHatomizationfor F, 1/2 F2 (g) F (g)DH2 = DHatomizationStep 2: Electron transfer to form (isolated) ionsLi (g) Li+ (g) + eDH3 = IE1F (g) + e F (g)DH4 = EA1Step 3: Ions come together to form solidLi+ (g) + F (g) LiF (s)DH5 = Lattice EnergyOverall: Li (s) + 1/2 F2 (g) LiF (s) DH = DHf = S(DH15)Lattice Energy = DHf (DH1 + DH2 + DH3 + DH4)54Start here 11/9Born-Haber cycle for the formation of magnesium chlorideMg(s) + Cl2(g)Mg(g) + Cl2(g)DH1 = +147 kJenthalpy of atomisation of magnesiumMg+(g) +e- +2Cl(g)first ionisation energy of magnesiumDH3= +744kJsecond ionisation energy of magnesiumDH4 = +1460 kJMg2+(g) +2e- + 2Cl(g)Mg2+(g) + 2Cl-(g)2 x electron affinity of chlorineDH5 = -697.4 kJDH6 enthalpy of formation of magnesium chlorideDHformation= DH1 + DH2 + DH3 + DH4 + DH5 + DH6DH6 = DHformation - (DH1 + DH2 + DH3 + DH4 + DH5) Mg(g) + 2Cl(g)bond energy of chlorineDH2= +243 kJlattice enthalpy of magnesium chlorideMgCl2(s)DHformation = -429.4 kJEDHformation= DH1 + DH2 + DH3 + DH4 + DH5 + DH6DH6 = DHformation - (DH1 + DH2 + DH3 + DH4 + DH5) = - 429.4 (147 + 243 + 744 +1460 + (-697.4))DHlattice= - 2326.0 kJmol-156

Learning Outcomes :Students able to :(a) construct energy cycles for the formation of aqueous solutions of ionic compounds;(b) explain qualitatively the influence on solubility of the relationship between enthalpy change of solution, lattice energy of solid and enthalpy change of hydration or other solvent-solute interaction.Solubility of Solid in LiquidsEnthalpy of Solution is the enthalpy change that takes place when 1 mole of a solute dissolves in a solvent to forman infinitely dilute solution.

NaCl(s) + aq NaCl(aq) Hsol = +5 kJ mol-1

Enthalpy of SolutionThe sum of two imaginary stepsreverse of the Lattice Enthalpy

the Hydration Enthalpy(of the cations and anions)HLattice- Lattice EnthalpyHhyd- Hydration EnthalpyEnergy cycle for solutionIonic lattice + solventsolutionHsolutionGaseous ions + solvent- HLattHhyd (cation) + Hhyd (anion)

Hsolution = Hhyd (cation) + Hhyd (anion) - HLattice

Enthalpy Diagram of enthalpy of solution of NaClNaCl(s) + aqNa+(g) + Cl-(g)-HLattice =+776 kJ mol-1Na+(aq) + Cl-(aq)Hhyd =-771 kJ mol-1Hsol = +5 kJ mol-1Hsol = -HLatt + (Hhyd(cation) + Hhyd(anion))Hsol = - (-776) + -771 = + 5 kJ mol-1What decides solubility?Energy is absorbed to separate the ions in the lattice [ - HLattice]

Energy is released by hydration of the ions

If more energy is released than absorbed, Hsolution will be negative and the solute is more likely to dissolve.The more negative the value for Hsolutionthe more likely the solute is to dissolve.

If Hsolution is large and positive, the solute will not dissolve.enthalpySolute + solventGaseous ionsSolutionHsolution-HLattHhyd(cat)Hhyd(an)Is HsolutionEndothermic or exothermic ?-HLE+ Hhyd(cat)+ Hhyd(an)Hsolution =enthalpySolute + solventGaseous ionsSolutionHsolution-HLattHhyd(cat)Hhyd(an)-HLE+ Hhyd(cat)+ Hhyd(an)Hsolution =Is Hsolution exothermic or endothermic?Which solute / substance dissolve in water?Ans :

AgCl(s)Hsoln = +81kJ/molAg+(g)Cl-(g)Ag+(aq)Cl-(aq)AgCl.NaOHEND OF CHAPTER!!!