2 Energy is... The ability to do work. Conserved. made of heat and work. a state function.independent of the path, or how you get from point A to B.Work is a force acting over a distance.Heat is energy transferred between objects because of temperature difference.
3 The Universe is divided into two halves. the system and the surroundings.The system is the part you are concerned with.The surroundings are the rest.Exothermic reactions release energy to the surroundings.Endothermic reactions absorb energy from the surroundings.
10 First Law of Thermodynamics The energy of the universe is constant.Law of conservation of energy.q = heatTake the systems point of view to decide signs.q is negative when the system loses heat energyq is positive when the system absorbs heat energy
11 Enthalpy Heat content of a substance Depends on many things Symbol is “H”Can’t be directly measuredChanges in enthalpy can be calculated, DH = Hfinal - HinitialChanges in heat (q) can be used to find DH
12 Same rules for Enthalpy Heat energy given off by the system is negative, DH < 0Heat energy absorbed by the system is positive, DH > 0
13 q and DHq is the “heat flow for a system” It can be for any amount of mass or moles of substance DH is “change in heat” It is for a specific chemical or physical change and it refers to that reaction or process.
14 Heat CapacityThe quantity of heat required to change the temperature of a system by one degree.Three “types” with different “systems:Specific Heat CapacityMolar Heat CapacityHeat Capacity
15 Heat Capacity Types q = (m)(C)(T) Specific heat capacity System is one gram of substanceUnits are J/goCq = heat changem = massC = Specific Heat CapacityDT = final temperature – initial temperatureq = (m)(C)(T)
16 Heat Capacity Types q= nCT Molar heat capacity. System is one mole of substance.Units are J/mol oCq = heat changen = molesC = molar heat capacityDT = final temperature – initial temperatureq= nCT
17 Heat Capacity Types q = CT Heat capacity of an object Specific for a given amount or unitso it is not mass dependentUnits are J/oCq = heatC = heat capacityDT = final temperature – initial temperatureq =CT
18 Heat Capacity Types So how do you know which one to use when? Look at the known informationSelect the value with appropriate units to cancel
19 ExampleThe specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.
20 Calorimetry Measuring temperature changes to calculate heat changes. Use a calorimeter.Two kindsConstant pressure calorimeter (called a coffee cup calorimeter)Constant volume calorimeter (called a bomb calorimeter)
21 CalorimetryA coffee cup calorimeter measures temperature and calculates q.An insulated cup, full of water (constant pressure, variable volume)Water is the surroundings in which the system changesCalculate the heat change of water.The specific heat of water is J /gºCHeat of water qH2O = CH2O x mH2O x DT
22 qsystem + qsurroundings = 0 qH2O + qrxn = 0qH2O = - qrxnIn interactions between a system and itssurroundings the total energy remains constant—energy is neither created nor destroyed.Law of Conservation of Energy
23 Coffee Cup Calorimeter A simple calorimeter.Well insulated and therefore isolated.Measure temperature change.qrxn = -qcal
25 Example qlead = -qwater Determining Specific Heat from Experimental Data.Use the data presented on the last slide to calculate the specific heat of lead.qlead = -qwaterqwater = mcT = (50.0 g)(4.184 J/g °C)( )°Cqwater = 1.4x103 Jqlead = -1.4x103 J = mcT = (150.0 g)(c)( )°Cclead = 0.13 Jg-1°C-1
26 ExampleWhen a piece of copper (5.0 g) is heated for 2.0 seconds, and 100 J of heat energy is transferred to the copper, the temperature increases from 20.0 ˚C to 71.9 ˚C.What is the specific heat of the copper?
27 ExamplesIf 10.0 g of Cu is heated for 2.0 seconds from 20.0 ˚C and 200 J of heat are absorbed, what is the final temperature of the block? a ˚C d ˚C b. 100 ˚C e. 719 ˚C c. 46 ˚C
28 ExampleEqual masses of liquid A, initially at 100 ˚C, and liquid B, initially at 50 ˚C, are combined in an insulated container. The final temperature of the mixture is 80 ˚C. Which has the larger specific heat capacity, A or B?ABA and B have the same specific heat capacity.
29 ExampleWhen 86.7 grams of water at a temperature of 73.0 ˚C is mixed with an unknown mass of water at a temperature of 22.3 ˚C the final temperature of the resulting mixture is 61.7 ˚C. What was the mass of the second sample of water?a g c gb. 302 g d.419 g
32 Changes of State of Water qphase change = DHphase change (mass or moles)Molar enthalpy of vaporization:H2O (l) → H2O(g)Hv = KHv = (40.65 kJ/mol)(1 mol/18.02 g) = 2256 J/gMolar enthalpy of fusion (melting):H2O (s) → H2O(l)Hf = KHf = (6.01 kJ/mol)(1 mol/18.02 g) = 333 J/g
33 Heating Curve for Water Total heat absorbed by the water as it is warmed can be determined by finding the heat involved in each “step” of the process.q1+ q2 + q3 …. = qtotalNote:Cice= Csteam = 2.09 J/goCq =(m)(C)(T)qphase change = DHphase change (mass or moles)
34 ExampleCalculate q for the process in which 50.0 g of water is converted from liquid at 10.0°C to vapor at 125.0°C.
35 ExampleWhat is the heat of fusion of lead in J/g if 6.30 kilojoules of heat are required to convert 255 grams of solid lead at its melting point into a liquid?a J/g c J/gb J/g d J/g
36 ExampleWhat quantity of heat is required to heat 1.00 g of lead from 25 ˚C to the melting point (327 ˚C) and melt all of it? (The specific heat capacity of lead is J/g • K and it requires 24.7 J/g to convert lead from the solid to the liquid state.)a J c Jb J d J
37 Calorimetry Constant volume calorimeter is called a bomb calorimeter. Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water.The heat capacity of the calorimeter is known and tested.
38 Propertiesintensive properties not related to the amount of substance.density, specific heat, temperature.Extensive property - does depend on the amount of stuff.Heat capacity, mass, heat from a reaction.
39 Bomb Calorimeter thermometer stirrer full of water ignition wire Steel bombsample
40 Bomb Calorimeter qrxn = -qcal qcal = qbomb + qwater + qwires +… Define the heat capacity of the calorimeter:qcal = miciT = CcalTheatSystem includes everything inside the double walled container. All the water, wires, stirrer, thermometer, and the reaction chamber.Bomb is filled with sample and assembled. Then pressurized witih O2. A short pulse of electric current heats the sample and ignites it. The final temperature of the calorimeter assembly is determined after the combustion reaction.Because the reaction is carried out at a fixed volume we say that this is at constant volume.
41 q and DH revisitedIf you know “q” for a process for a known amount of sampleAnd you have a balanced chemical equation“q” can be converted to DH
42 q to DHIf you know the heat per amount of substance, find it per gram then use the molar mass to find the heat per mole.
43 Thermochemical Equations Balanced chemical equationIncludes a term which indicates the change in enthalpy, DHEnthalpy term can be embedded in the reaction or written after the reaction with DH notation.
44 ExamplesThe combustion of g sucrose, in a bomb calorimeter, causes the temperature to rise from to 28.33°C. The heat capacity of the calorimeter assembly is 4.90 kJ/°C.What is the heat of combustion of sucrose, expressed in kJ/mol C12H22O11Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 calories.
46 Exothermic Thermochemical Equations H2(g) + ½ O2(g) H2O(l) kJSame as:H2(g) + ½ O2(g) H2O(l)DHrxn = kJ/mol rxn
47 Stoichiometry and Thermochemical Equations Just as the mole ratios can be used to determine reacting and produced amounts of chemicalsThe ratios can be used to relate amounts of chemicals with energy
48 Example DHrxn = 1,676 kJ/mol rxn Al2O3(s) 2Al(s) + 3/2 O2(g)DHrxn = 1,676 kJ/mol rxnHow many grams of aluminum would be produced if the aluminum absorbed 3500 kJ of energy?
49 DHrxn DHrxn is for ANY general chemical or physical reaction There are specific types of reactions which have a special designation for their DH.
50 Some Important Types of Enthalpy Changes Heat of combustion (DHcomb)C4H10(l) /2O2(g) CO2(g) + 5H2O(g)Heat of formation (DHf)K(s) + 1/2Br2(l) KBr(s)Heat of fusion (DHfus)NaCl(s) NaCl(l)Heat of vaporization (DHvap)C6H6(l) C6H6(g)
51 Standard States and Standard Enthalpy Changes Define a particular state as a standard state.Standard enthalpy of reaction, H°The enthalpy change of a reaction in which all reactants and products are in their standard states.Standard StateThe pure element or compound at a pressure of 1 bar and at the temperature of interest.Temperature must be specified because H varies with temperature.
52 Indirect Determination of H: Hess’s Law H is an extensive property.Enthalpy change is directly proportional to the amount of substance in a system.N2(g) + O2(g) → 2 NO(g) H = kJ½N2(g) + ½O2(g) → NO(g) H = kJH changes sign when a process is reversedNO(g) → ½N2(g) + ½O2(g) H = kJ
53 Hess’s law of constant heat summation If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.½N2(g) + ½O2(g) → NO(g) H = kJNO(g) + ½O2(g) → NO2(g) H = kJ½N2(g) + O2(g) → NO2(g) H = kJ
56 Standard Enthalpies of Formation fH°The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states.The standard enthalpy of formation of a pure element in its reference state is 0.Absolute enthalpy cannot be determined.H is a state function so changes in enthalpy, H, have unique values.Reference forms of the elements in their standard states are the most stable form of the element at one bar and the given temperature.The superscript degree symbol denotes that the enthalpy change is a standard enthalpy change andThe subscript “f” signifies that the reaction is one in which a substance is formed from its elements.
59 Writing Formation Equations Remember that DfHo refers to forming ONE mole of a pure substance FROM ITS ELEMENTS in their standard stateWrite thermochemical formation equations for the formation of:Al2O3(s)NaHCO3(s)
60 Standard Enthalpies of Reaction H°rxnH°overall = -2fH°NaHCO3+ Hf°Na2CO3+ Hf°CO2 + Hf°H2O
61 Enthalpy of ReactionH°rxn = SHf°products- SHf°reactants
62 Enthalpies of Formation of Ions in Aqueous Solutions