Thermochemistry

ThermochemistryEnergy Changes In Chemical Reactions

Learning Catalytics: 1Chapter 6Coverage:6.2 The Nature of Energy: Key Definitions6.3 The First Law of Thermodynamics: There is No Free Lunch6.4 Quantifying Heat and Work6.5 Measuring U for Chemical Reactions: Constant-Volume Calorimetry6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure6.7 Constant-Pressure Calorimetry: Measuring rH6.8 Relationships involving rH6.9 Determining Enthalpies of Reaction from Standard Enthalpies of FormationOmitted6.1 Chemical Hand Warmers6.10 Energy Use and the Environment2Energy basicsInternal energy, heat, work and all that3TerminologyThermochemistry: Study of changes in energy functions that accompany chemical reactionsE.g. 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) DrH = -1648.4 kJ/molSystem: Part of the universe under studyE.g. Known masses of Fe(s) and O2(g) in a vesselSurroundings: Everything elseSystem + Surroundings = Universe

4Energy from a chemists POVEnergy: Capacity of chemical system to do work or transfer heatKinetic energy: Energy of a system due to the motion of its particlesA.k.a. thermal energyPotential energy: Energy a system possesses due to the positions of the particles in a fieldE.g. molecules in a gas vs same molecules in a liquid5Units of EnergySI: joule, J (nJ, J, mJ, kJ, MJ, GJ,)Small j for joule, capital for symbolImportant: case of prefix matters (m = 10-3, M = 106)No period after symbolOther units in common usage:Calorie (cal) = 4.184 JBig calorie (Cal) = 4,184 J (a kilocalorie, or kcal)Kilowatt-hour (kWh)1000 W for one hour = 1000 (J/s) 3600 s = 3.6 MJ6QuestionsDoes water at 20C possess energy? Explain.What measurable value contains information about the kinetic energy of a chemical system?Which has higher energy: liquid water at 100C or steam at 100C? Explain.

7EnergyInternal energy (U): sum of energy of all particles in system (kinetic + potential)Heat (q): energy exchanged with surroundings by atomic-scale collisions, due to a temperature differenceWork (w): energy exchanged with surroundings by macroscopic motion of surroundingsHeat and work are mechanisms by which energy is exchanged between system and surroundingsU = q + w (by conservation of energy, U = -Usurr )8QuestionsWhich system has more energy, a litre of water at 100C or a swimming pool at 20C?Which can transfer more heat (to, say, melt ice)?A litre of water at 20C is put into contact with a litre of water at 10C. Which way will heat flow?Heat flows from high temp to low (compare diffusion)Does a litre of water at 20C possess more heat than a litre of water at 10C?No: Systems do not possess heat, they transfer heat9Heat/Work AnalogyDoes your bank account contain the cash and cheques you deposited?Cash & cheques are mechanisms for transferring valueBUT: Bank accounts do not contain cash or chequesWhat does a bank account contain? A balance.Does a chemical system contain heat and work?Heat & work are mechanisms for moving energyBUT: Chemical systems do not contain heat or workWhat does a chemical system contain?10

Sign Convention for q and wEnergy entering the system is positiveHeat absorbed by or work done on the systemEnergy leaving the system is negativeHeat released from or work done by systemU = q + w < 0 means more energy left the system than entered it during a changeU = q + w > 0 means more energy entered the system than left it during the change11Heat/Work AnalogyWhen money enters your bank account (by cheques or cash) your account balance (B) increasesB > 0When money leaves your bank account (by cheques or cash) your account balance (B) decreasesB < 012

Temperature questionThe temperature of a system decreases during a thermodynamic change. Does that mean that the system lost heat to the surroundings?NoWhat does temperature measure?Kinetic internal energy (thermal energy)A T decrease means the system lost internal energyIt might have lost energy as heat or as work13LC: Sign ConventionDuring a reaction, some of the energy released by the system is used to move a mass up a hill, while the air temperature of the surroundings near the system increases by 50 K. What are the relative values for q, w and DU, respectively?q > 0 ; w > 0 ; DU > 0q > 0 ; w < 0 ; DU = 0q < 0 ; w > 0 ; DU = 0q < 0 ; w < 0 ; DU < 014SolutionDuring a reaction, some of the energy released by the system is used to move a mass up a hill, while the air temperature of the surroundings near the system increases by 50 K. What are the relative values for q, w and DU, respectively? ANS = DIs w positive, negative, or zero?NEGATIVE: work done by system to raise the massIs q positive, negative, or zero?NEGATIVE: heat released raises T of surroundingsIs U positive, negative, or zero?NEGATIVE: q and w are both < 0 15LC: Sign ConventionA gas is contained in a piston-and-cylinder system, oriented vertically. On top of the cylinder there is a mass. When the gas is heated it expands, and the mass is lifted. The energy used to lift the mass is exactly the same as the energy absorbed from the heater. What are the signs for q, w and DU, respectively?q = +; w = + ; DU = +q = +; w = - ; DU = 0q = -; w = + ; DU = 0q = -; w = - ; DU = -16SolutionA gas is contained in a piston-and-cylinder system, oriented vertically. On top of the cylinder there is a mass. When the gas is heated it expands, and the mass is lifted. The energy used to lift the mass is exactly the same as the energy absorbed from the heater. What are the signs for q, w and DU, respectively? ANS = BIs q > 0, < 0, or = 0?ANS: q > 0. gas is heatedIs w > 0, < 0, or = 0?ANS: w < 0. Work done by system to lift massIs U > 0, < 0, or =0?ANS: U = q + w = 0. Work (-) and heat (+) balance

17State functions and path functions18

State & State FunctionThermodynamic state defined by specifying the TD variables; e.g. T, P, , compositionThermodynamic state functionTD quantity of the systemDepends solely on the TD state U is a state functionChange in a state function (e.g. U) does not depend on the mechanism to go from one state to the other19

Analogy: Altitude on a Mountain20LC: State FunctionsSample A: 50 g of H2O (g) at 400 K and 1 atm is compressed until it condenses, then the liquid is cooled to 300 K. Pressure is reduced to 1 atm while the temperature is maintained at 300 K

Sample B: 50 g of H2O (s) at 200 K and 1 atm is heated to the melting point, and then the liquid water is further heated to 300 K under 1 atm pressure

Which water sample has the higher internal energy in the final state?Sample ASample BNeither21SolutionAnswer is CInternal energy (U) is a state functionState functions have one value per state of the systemFinal state of A is same as final state of BTherefore, A and B have same internal energy22What isnt a state function?Heat (q) and work (w) are not state functionsIn going from state i (initial) to state f (final), U must change byU = Uf - Ui = q + w

Infinite pairs of q + w that add to Uq and w called path functions (depend on path)BUT: Uf and Ui do not depend on the path23Heat and heat transferMeasuring and calculating heat exchange between bodies24Specific Heat Capacity2510 g water 25.00 C100 J heat10 g water 27.39 C10 g Cu 25.00 C100 J heat10 g Cu 50.97 C= 4.18 Jg-1C-1Cs = 0.385 Jg-1C-1Specific Heat CapacityValues are specific to substance or mixture4.184 is for water, nothing else!Available in tables (e.g. Table 6.4) for pure substancesAllow calculation of temperature changesE.g. How much heat is required to raise the temperature of a glass of water (250 g) from 20C to 100CANS: q = mCsT = (250 g)(4.184 J/g/C)(100 20)C = 83.7 kJ

26Heat exchangeHeat entering the system is equal and opposite heat leaving the surroundings q = -qsurrmCsT = - msurrCs,surrTsurrE.g. A block (25 g) of copper (the system) at 100C is submerged in 100 g of water (the surroundings) at 20C. What is the final temperature of the water when equilibrium is reached?Cs,H2O = 4.184 J/g/C; Cs,Cu = 0.385 J/g/C27Heat exchange28Notes on specific heat capacity1C change in temp = 1 K change in tempBUT: temp in K temp in C!

Molar heat capacityFor pure substances onlyE.g. for water(4.184 J/g/C)(18.0 g/mol) = 75.38 J/mol/CMore generally: Cm = CsM

29Pressure-Volume Work30Pressure-Volume Work31PextPi > PextPextPf = PextGas expands against Pext until pressures equaliseViVfWork done:w = - Pext(Vf Vi)Initial StateFinal StatePressure-Volume Workw = FdF is the force due to the external pressured is the distance the piston travelsw = (PextA)dA is the cross-sectional area of the pistonw = Pext(Ad) = PextVSign convention: gas does work on surroundings when it expands; thereforew = -PextVIf Pext and P are in equilibrium during changew = -P V

32PV work example11.7 L of gas in a cylinder-and-piston system is initially at equilibrium with its surroundings (Pext = 1.03 bar). Heat is added slowly and the gas expands in equilibrium with the surroundings until the piston reaches the end of the cylinder and stops. Heat continues to be added while the piston is stopped. The final volume of the gas is 24.1 L and the total heat added is 1500 J. Calculate the work and U for the system.33PV work exampleWork: System and surroundings are in equilibrium sow = -P V

Remember to change units: L to m3 and bar to Paw = - (1.03105 Pa)(24.1 11.7)10-3 m3 = - 1.28103 J

Heat is +1500 J soU = q + w = 1500 J 1280 J = +220 J34Constant-Volume CalorimetryMeasuring heats of reaction35CalorimetryPurpose: measure heat of combustionV = 0 so w = 0Process:Burn substance completely in bombCarefully monitor T of calorimeterCalculate heat transferred36

Heat CapacityFirst step: determine heat capacity of calorimeterCcal = qcal/TcalBurn substance that gives off known amt of heatOnce Ccal is known, can use bomb calorimeter to measure heats of other combustion reactionsHeat of reaction (qr) equal and opposite to heat gained by calorimeterqr = -qcal = - Ccal Tcalqr = - Ccal Tcal37Note on constant V reactionsThe energy change associated with a reaction is denotedrU Since w = 0 when V = 0,rU = q + w = qr,vwhere the subscript v denotes constant volume38LC: CalorimetryWhen 1.730 g of benzoic acid is burned in oxygen in a calorimeter, the temperature of the calorimeter increases by 2.95 K. If the heat released from combustion of 1.000 g of benzoic acid is 26.38 kJ, what is the heat capacity of the calorimeter?13.0 kJ/K45.64 kJ/K15.5 kJ/K135 kJ/K39AnswerC. 15.5 kJ/KHeat absorbed by the calorimeter isqcal = (1.730 g)(26.38 kJ/g) = 45.64 kJAnd Tcal = 2.95 KCcal = qcal/Tcal = (45.64 kJ)/(2.95 K) = 15.5 kJ/K15.5 kJ of heat required to raise temp by 1 K40LC: CalorimetryA calorimeter (Ccal = 13.0 kJ/K) is used to measure the energy of reaction for combustion of a substance with molar mass 205.9 g/mol. When 1.151 g of the substance is burned in O2, the temperature of the calorimeter increases by 1.62 K. What is the energy of reaction (rU) in kJ/mol?-0.118 -18.3-1440-3770

41AnswerD. -3770 kJ/molqr = -Ccal Tcal = -(13.0 kJ/K)(1.62 K) = -21.06 kJIt took 1.151 g to generate that much heat, so qr,m = - 18.30 kJ/gwhere the subscript m represents per unit massThe molar mass is 205.9 g/mol, sorU = (-18.30 kJ/g)(205.9 g/mol) = -3770 kJ/mol42EnthalpyOur second energy state function43EnthalpySymbol: HChange in enthalpy, HChange in enthalpy on reaction, rHRelationship to UH = U + PVRelationship to qU = qv (heat for constant volume process)H = qP (heat for constant pressure process)Since chemical reactions are often performed at constant pressure, H tends to be used more than U44Enthalpy and heatStarting with the definition,H = U + PVIf a change happens at constant pressureH = U + PVFor a constant-pressure process occurring at equilibrium with the surroundings,U = qP + w = q PVSubstitutingH = q PV + PV = qP

45Endothermic vs ExothermicEndothermic process:H > 0Heat is taken in from the surroundingsExothermic processH < 0Heat is released to surroundings46Source of ConfusionExothermic reactions release heatCombustion is exothermicTemp of combusting system rises as reaction proceedsIf system releases heat, shouldnt temp decrease?47

Endothermic reactionBa(OH)2.8H2O(s) + 2 NH4SCN(s) Ba(SCN)2(s) + 10 H2O(l) + 2 NH3(g) 48

Source of ConfusionTables of H values for processes at 1 atm and 298KInitial state: reactants at 1 atm and 298KFinal state: products at 1 atm and 298KFor exothermic reaction, heat has to be removed to get products down to 298KFor endothermic reaction, heat has to be added to warm products up to 298K49Endothermic exampleInitial state: reactants at 25CMix and reaction cools (allowing heat to enter) Final state: products warm to 25C (endothermic, H>0)50NH4SCN(s)Ba(OH)2(s)25CSurr = 25C25CSurr = 25C