*ThermochemistryChapter 6

*Energy and changeThe study of the energy involved in a change is THERMODYNAMICSIn thermodynamics, the universe is divided intoSystem (what you are studying)andSurroundings (the rest of the universe!)

*SystemsA system may be OPENboth matter and energy can be exchanged with the surroundings

*SystemsA system may be CLOSEDonly energy can be exchanged with the surroundings

*SystemsA system may be ISOLATEDneither matter nor energy can be exchanged with the surroundings

*Systems may beopen, closed, or isolated

*Internal energyThe internal energy E of a system is all the energy (both kinetic & potential) contained in the systemChemists are especially interested inthermal energy (energy of random molecular motion)chemical energy (energy stored in chemical bonds and intermolecular forces)

*First Law of ThermodynamicsThe energy a system contains is its internal energy, EEnergy can move in or out of a system as heat (q) and/or work (w).Energy is conserved, so all energy lost or gained by a system must be accounted for as heat and/or work:

*E is a state functionA state function (aka function of state) is a property whose value depends only on the state of the system, not how it achieved that stateThe state of the system is specified by the pressure, temperature, and composition of the system.P, V, & T are state functions. E is a state function.q and w are NOT state functions.

*The elevation gain is the same by either path: a state functionHow E is distributed between q and w depends on which path you take: q and w are not state functions

*fully chargedbatteryfully dischargedbatteryPath 1: short circuit the battery with a wrench.Lots of heat, maybe even sparks and a fire, but no work!Path 2: connect the battery to a motor.Some heat, but also some work.E the samebyeither path

*HeatHeat is energy transferred because of a temperature differenceA system does not contain heat;it contains ENERGYHeat is just a form by which energy is transferredThe other form by which energy is transferred is workWhether q or w, energy IN is positive, energy OUT is negative!

*HeatThe amount of energy transferred asheat, q, is related tothe amount of matter gaining/losing energy (m)the specific heat of matter gaining/losing energy (c) the amount of the temperature change (T)James Prescott Joule

*Heat capacity and specific heatSpecific heat capacity = cIntrinsic capacity to gain/lose energy as heatunit = J/g K or J/g CMolar specific heat capacity = J/mol K or J/mol CHeat capacity of a system = CQuantity of energy to change temperature by 1 Cunit = J/K or J/C

*Significance of specific heat valuesTemperature change change in molecular motionLow specific heat less energy to increase motionHigh specific heat more energy to increase motion specific heat substance (J/g C)air1.00aluminum0.895copper0.387lead0.128ethanol2.45water4.18

*ExamplesHow much heat energy (in kJ) is required to raise the temperature of 237 g ice water from 4.0 C to 37.0 C? cwater = 4.18 J/g CHow much heat energy (in kJ) is required to raise the temperature of 2.50 kg Hg from 20.0 C to 6.0 C? cHg = 28.0 J/mol C

*Determination of specific heatHeat sample of metal totemperature of boilingwaterMeasure temperature of measured amount ofwater in insulated beakerPut hot metal in cold water and measure final temperatureAll energy is assumed to stay in the insulated containerqmetal + qwater = 0 or qmetal = qwater

*ExamplesWhen 1.00 kg Pb (specific heat = 0.13 J/g C) at 100.0 C is added to some water at 28.5 C, the final temperature is 35.2 C. What is the mass of the water?100.0 g Cu (specific heat 0.385 J/g C) at 100.0 C is added to 50.0 g water at 26.5 C. What is the final temperature of the mixture?

*WorkWork is done when a force acts for a distance:w = F x dEnergy is the capacity to do workkinetic energy is the energy of motionthe random motion of molecules (thermal energy) is kineticpotential energy is stored energy, that can do work when it is releasedthe energy in chemical bonds is potential energy

*Pressure-volume workImagine a gas trapped in a cylinder with a moveable lid (a piston)If the piston is not moving, how does the gas pressure inside compare to the external pressure?If the piston is not moving, Pgas = Pexternal

*Pressure-volume workWhat if Pgas increases, so Pgas > Pexternal?

*Pressure-volume workThe piston rises and the gas expands, until once again Pgas = Pexternal The gas has moved the piston some distance against the opposing Pext The gas has done work!

*Pressure-volume workHow much work did the gas do?h

*Pressure-volume workHow much work did the gas do?

When the gas did work by expanding, it lost energyh

*Pressure-volume workNow what if we increase the external pressure, so Pexternal > Pgas ?

*Pressure-volume workThe gas is compressed until once again Pgas = Pexternal The piston has been moved some distance against the opposing Pgas This time work was done on the gasHow much work?PgasPexternalh

*Pressure-volume (PV) workWhen the gas expands, it expends energy to do work = PextVV is positive energy leaves the system (the gas)When the gas is compressed, it gains the energy used to compress it = PextVV is negativeenergy enters the system (the gas)

*Sign conventions for workWhen energy goes into the system as work, w is positiveWhen energy leaves the system as work, w is negativew = PVwhen a gas expands V is + and w is (energy leaves the system)when a gas is compressed V is and w is + (energy enters the system)

*Unitsw = PVwork is in joules (J)P is in atm and V is in L, so PV is in atm LThe relationship between J and atm L is1 atm L = 101.325 J

*ExampleWhat is the work done on a gas (in J) when the gas is compressed from an initial volume of 35.0 L to a final volume of 23.5 L under a constant pressure of 0.987 atm?

*CalorimetryThe process for measuring the amount of heat energy exchanged by system and surroundings is CALORIMETRYAssume total energy is constant: heat lost by system is gained by surroundingsqsystem + qsurroundings = 0qsystem = qsurroundings

*Heat of reaction, qrxnIf the reaction releases energy, qrxn is negative (the reaction loses energy)The calorimeter gains that energy and gets hotterThe reaction is EXOTHERMICqrxn = qcalorimeter

If the reaction absorbs energy, qrxn is positive (the reaction gains energy)The calorimeter loses that energy and gets colderThe reaction is ENDOTHERMICqrxn = qcalorimeter

*Bomb calorimeterCombustion reaction occurs in sample compartmentReaction releases energy to water in calorimeterqrxn = qcalorimeter = CTC = heat capacity of calorimeter

*ExamplesThe combustion of 1.013 g vanillin (C8H8O3) in a bomb calorimeter with heat capacity = 4.90 kJ/C causes the temperature to rise from 24.89 C to 30.09 C. What is the heat of combustion of vanillin, in kJ/mol?Combustion of 1.176 g benzoic acid (HC7H5O2, heat of combustion 26.42 kJ/g) causes the temperature in a bomb calorimeter to increase by 4.96 C. What is C for that calorimeter?

*Coffee cup calorimeterReactants (usually in aqueous solution) mix and react in the insulated cupReaction releases energy to (or absorbs energy from) liquid in which it is dissolvedqrxn = qcal = (mcT)calCalorimeter = liquid in the cup (cup assembly is usually ignored)

*Example100.0 mL 1.00 M AgNO3 (aq) and 100.0 mL 1.00 M NaCl (aq), both initially at 22.4 C, are mixed in a coffee cup calorimeter. The temperature rises to 30.2 C. What is the heat of reaction, in kJ per mol AgCl, for the reaction:Ag1+ (aq) + Cl1 (aq) AgCl (s)

*Example100.0 mL 1.020 M HCl and 50.0 mL 1.988 M NaOH, both initially at 24.52 C, are mixed in a coffee cup calorimeter. If qneutralization = 56 kJ/mol H2O, what will be the final temperature in the mixture?Hint: this is a limiting reactant problem

*Energy and EnthalpyE = q + ww = PVAt constant volume, V = 0 so w = 0and E = qV We define H = E + PVH is called enthalpyAt constant pressure, H = E + PVBut E = qP PV, so H = qP PV + PVH = qP

*E and HAt constant volume, measured heat = EAt constant pressure, measured heat = HConstant pressure is the more common lab conditionH is the heat of reaction for a chemical change carried out at constant pressure

*How different are E and H?C (s) + 1/2 O2 (g) CO (g) H = 110.5 kJ H = E + (PV) or E = H (PV)(PV) = (nRT) = ngasRT (solids are not significant)ngas = 1/2 molAt 298 K,

E is 111.7 kJ, only 1% difference from H

*EnthalpyLike E, absolute values of H cannot be measured, but changes in H can be measured: H = qPH is part of the chemical reaction stoichiometryH is extensive (depends on amount of material)H is directional (sign reverses if direction of reaction reverses)H is a state functionValue of H same whether reaction occurs in a single step or a series of steps, if final result is the same

*Reactions add togetherLook at these reactions: C (s) + O2 (g) CO (g) + 1/2 O2 (g) CO (g) + 1/2 O2 (g) CO2 (g)The reactions add to give C (s) + O2 (g) CO2 (g)The CO and 1/2 O2 terms on both sides cancel

*Heats of reaction add, tooDH values add just like reactions: H (kJ/mol) C (s) + O2 (g) CO (g) + 1/2 O2 (g) 111 kJ/molCO (g) + 1/2 O2 (g) CO2 (g) 283 kJ/mol C (s) + O2 (g) CO2 (g) 394 kJ/mol

*Hess Law Germain Hess discovered that heats of reaction add together in 1840The additivity of H values is called Hess LawHess Law establishes enthalpy as a state functionState functions depend only on the state of the systemIt doesnt matter how the system achieved that state

*Hess LawHess Law allows us to calculate H for a reaction from measured H values for other reactionsThe rules of the gameMultiply reaction by a factor to get desired coefficientsMultiply H by the same factorReverse a reaction to get a substance on the desired sideIf you reverse the reaction, reverse the sign of H

*ExampleGiven these heats of reaction,H2 (g) + 1/2 O2 (g) H2O (l) H = 286 kJ2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (l) H = 3123 kJ2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (l) H = 2602 kJFind H for this reaction:C2H2 (g) + 2 H2 (g) > C2H6 (g)

*ExampleFirst look at the target reaction: C2H2 (g) + 2 H2 (g) C2H6 (g)You want 1 C2H2 on the left. We need to multiply the reaction with C2H2 by 1/2:1/2 [2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (l)]C2H2 (g) + 5/2 O2 (g) 2 CO2 (g) + H2O (l)Multiply H by 1/2: H = 1/2[2602 kJ] = 1301 kJ

*ExampleAnother look at the target reaction: C2H2 (g) + 2 H2 (g) C2H6 (g)You need 2 H2 on the left. We need to multiply the reaction with H2 by 2:2 [H2 (g) + 1/2 O2 (g) H2O (l)] 2 H2 (g) + O2 (g) 2 H2O (l)Multiply H by 2: H = 2[286 kJ] = 572 kJ

*ExampleThe target reaction: C2H2 (g) + 2 H2 (g) C2H6 (g)Finally, you want 1 C2H6 on the right. Multiply the reaction with C2H6 by 1/2, and reverse it:1/2 [2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (l)]reverse C2H6 (g) + 7/2 O2 (g) 2 CO2 (g) + 3 H2O (l)2 CO2 (g) + 3 H2O (l) C2H6 (g) + 7/2 O2 (g) Multiply H by 1/2, and reverse its sign:H = 1/2[+3123 kJ] = +1561 kJ

*ExampleNow you have C2H2 (g) + 5/2 O2 (g) 2 CO2 (g) + H2O (l) H = 1301 kJ 2 H2 (g) + O2 (g) 2 H2O (l) H = 572 kJ2 CO2 (g) + 3 H2O (l) C2H6 (g) + 7/2 O2 (g) H = +1561 kJNow add the reactions and cancel items identical on both sidesThe total isC2H2 (g) + 2 H2 (g) C2H6 (g) H = 312 kJ

*Formation reactionsTo organize collections of H values for reactions, chemists defined a formation reactionA formation reaction is the equation to form one mole of compound from its elements in their most stable stateThe formation reaction for H2O isH2 (g) + 1/2 O2 (g) H2O (l)Notice that the reactants are elements in their most common, stable formNotice that only one mole of product forms

*Standard enthalpy of formationThe enthalpy of a formation reaction is called the standard heat of formation (symbol Hf).The little f means a formation reactionThe means standard conditions, 1 atm and 1 M solutionsHf values are collected in reference books (see Appendix II in the back of your text)Hf for an element in its most stable state is zeroThe state (liquid vs. gas or solid) is importantEquations are not given (you are supposed to know how the write the appropriate formation reaction)

*Heats of formation combineYou can combine formation equations and their Hf values just like any other reactionsSame rules of the game:If you reverse the reaction, reverse the sign of HfIf you multiply the reaction by some factor, multiply Hf by the same factor

*But, of course, theres a SHORTCUTH = nHf products nHf reactants n is the coefficient for the substance, in molesC2H2 (g) + 2 H2 (g) C2H6 (g)Hf values: C2H2 = +227 kJ/mol, H2 = 0 kJ/mol (element), C2H6 = 85 kJ/molH = [1(Hf C2H6)] [1(Hf C2H2) + 2(Hf H2)]H = [1 mol (85 kJ/mol)] [1 mol (+227 kJ/mol) + 2 mol (0 kJ/mol)]H = 312 kJ

*A shortcut for the shortcutAn easy way to set up the shortcut isWrite the target equationBeneath each substance, write its Hf (watch signs!)Beneath Hf, write the coefficient for that substanceMultiply each Hf by its coefficientAdd the answers for the products, add the answers for the reactantsH = products reactants

*The short shortcut C2H2 (g) + 2 H2 (g) C2H6 (g)+227 kJ 0 kJ 85 kJ x 1 x 2 x 1+227 kJ 0 kJ 85 kJ+227 kJ 85 kJ

H = (85 kJ) (+227 kJ) = 312 kJThe same result we got earlier by adding reactions

*Reactions in aqueous solutionAdditivity also applies to reactions in solutionHf values for ions are given in the AppendixThe baseline for aqueous ions is H1+ (aq) = 0 kJ/mol instead of the element formAll rules apply as before