Chapter 6§6.1 Thermochemistry§6.1 Some Definitions§6.1 System vs Surroundings§6.1 Units of Energy§6.2 The First Law of Thermodynamics§6.2 What Type of Energy Transfer?§6.3 State Functions§6.3 Internal Energy§6.3 Components of Energy§6.3 Calorimetry§6.4 Heat§6.4 Pressure-Volume Work§6.4 Work, HeatThe Thermite ReactionSlide 16§6.4 Thermal Heat TransferAn Endothermic ReactionSome Endothermic Processes§6.5 Constant Volume CalorimetrySlide 22§6.6 Enthalpy (H) vs Internal Energy (E)§6.6 EnthalpySlide 25Slide 26§6.6 Stoichiometry and DHrxn§6.7 Constant Pressure Calorimetry§6.7 Two Types of Calorimetry§6.8 Calculating DHreaction§6.8 Calculating DHreactionSlide 32§6.9 Standard States§6.9 Standard Enthalpy of FormationSlide 35§6.9 Standard Enthalpy of Reaction,§6.9 Standard Enthapy of ReactionEthanol is used as an additive in many fuels today. What is ΔHºrxn in kJ for the combustion of ethanol?Slide 39§6.9 Fraction Reaction CoefficientsSlide 41Keeping all the DHs sortedTHERMOCHEMISTRYChapter 61skip §6.10§6.1 ThermochemistryThermochemistry – the relationship between matter and energyEnergy – capacity to do workWork – force acting over a distance2§6.1 Some DefinitionsPotential Energy (PE) – energy stored due to position or compositionKinetic Energy (KE) – energy due to motion of an object3§6.1 System vs SurroundingsSystem – any defined environmentSurroundings – everything elseSystem + surroundings = the universe4§6.1 Units of EnergyJoule (J) – the energy needed to move a 1 kg mass 1 m.o1 J = 1 N∙m = 1 kg∙m2/s2Calorie (cal) – the energy needed to raise 1 g of water 1 °C.okcal = energy needed to raise 1 kg of water 1 °Cofood calories = kcalsKilowatt-hour (kWh) – used to measure electrical energy5§6.2 The First Law of ThermodynamicsThe total energy of the universe is constant. Or, energy is neither created nor destroyed.Energy can be transferred between objects.Energy can be transformed from one form to another.6§6.2 What Type of Energy Transfer?electrical to thermal 7light to electricalnuclear to thermal to mechanical to electricalchemical to light and thermalchemical to light§6.3 State FunctionsState functions only depend on initial and final conditions, not the pathway between them.Internal energy, enthalpy, entropy, mass, altitude and density are state functions.8§6.3 Internal EnergyInternal energy (E) – a system’s total amount of KE and PEE = Efinal – EinitialEreaction = Eproducts - EreactantsEsys = -Esurround-Esys = EsurroundEnergy is neither created nor destroyed.9§6.3 Components of EnergyHeat (q) – transfer of energy due to a temperature difference. Work (w) – force acting over a distance.q and w are not state functions.E = q + wpositive value negative valueqsystem gains heat system releases heatwsystem gains energy from work done on it system releases energy by doing workEsystem gains energy system releases energy10§6.3 CalorimetryCalorimetry is the science of measuring heat.Specific heat capacity (Cs) – the energy required to raise 1 g of a substance by 1 °C (or 1 K).Molar heat capacity – the energy required to raise 1 mol of a substance by 1 °C (or 1 K).H2O has an extremely high specific heat.11§6.4 HeatWhen heat flows into a system, the temperature increases.q = C · T q = m · Cs · Tq = heatCs = system’s specific heatm = mass of systemT = temperature increase12§6.4 Pressure-Volume Workw = -PV Fluids do work via volume change according to this formula. The system is often a fluid-filled piston or balloon. When a fluid expands (+V), work is negative. When a fluid compresses (-V), work is positive. Units of PV work are often L · atm. 1 L · atm = 101.3 J 13§6.4 Work, HeatA 100. watt light bulb inside a 1.00 L rubber balloon is lit for 0.020 h, causing the balloon to expand to 5.00 L at a constant pressure of 1.0 atm. Calculate E , q and w assuming the light bulb+balloon is the system. 1 watt · hour (W · h) = 3.60 x 103 J101.3 J = 1 L · atmw = -PV = - (1.0 atm)(5.00 – 1.00 L) = -4.0 L · atm = -405.2 J E = (100 watt)(0.020 h) = 2.0 W · h = 7.2 x 103 Jq = E - w = 7.2 x 103 J – (-405.2 J) = 7.6 x 103 J E = q + wThe Thermite ReactionThis is one of the most exothermic chemical reactions known (H = -850 kJ).Iron (III) oxide (rust) is reduced at such a high temperature that the iron metal is in the liquid state. The melting point of iron is 1538J°C; skin burns at 54 °C.thermite2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(l)The Thermite ReactionThe thermite reaction is used in railway construction. Liquid iron drops into a mold between two rails, welding them together.thermite2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(l)17qwater = -qFeCs, water · mwater · Twater = -(Cs, Fe · mFe · TFe )4.18 J/g·°C · 1000 g · Tfinal - 25 °C = -[0.449 J/g·°C · 111.78 g · Tfinal - 1538 °C)]Tfinal = 43 °CTinit = 1538 °CCs = 0.449 J/g·°C In the reaction you saw, if 2 mol of Fe forms and falls into 1 kg of water at 25 °C, what is the final temperature of the water?2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(l)Cs, water = 4.18 J/g·°C §6.4 Thermal Heat TransferAn Endothermic ReactionBa(OH)2.8H2O(s) + 2NH4SCN(s)  Ba(SCN)2(s) + 10H2O(l) + 2NH3(g)Most reactions are exothermic. Endothermic reactions occur less often because they constantly absorb heat from their surroundings.The temperature of this reaction after completion is about -20 °C:18Some Endothermic ProcessesH2O (s)  H2O (l)H2O (l)  H2O (g) below 1oo °CNH4NO3 (s)  NH4+ (aq) + NO3- (aq) xylitol (s)  xylitol (aq)19xylitol gumNH4NO3 cold pack melting ice§6.5 Constant Volume CalorimetryThe reaction whose q is to be measured occurs in an internal steel chamber (called a bomb) surrounded by water. The chamber’s volume is constant, therefore no work is done. Energy is only released as heat.E = q + wE = q at constant volumeE = qcal = Ccal · T qcal = heat absorbed by calorimeter Ccal = heat capacity of calorimeterT = temperature changeqcal = - qrxn 21§6.5 Constant Volume CalorimetryIn a bomb calorimeter with a heat capacity of 10.9 kJ/°C, 249.4 g AgF2 reacted