2 Energy is the capacity to do work Radiant energy comes from the sun and is earth’s primary energy sourceThermal energy is the energy associated with the random motion of atoms and moleculesChemical energy is the energy stored within the bonds of chemical substancesNuclear energy is the energy stored within the collection of neutrons and protons in the atomPotential energy is the energy available by virtue of an object’s positionKinetic energy is energy associated with the motion of an object.6.1
3 Energy cannot be created or destroyed. Energy can be converted from one form of energy to another.Potential EnergyKinetic EnergyEnergy cannot be created or destroyed.=> Law of conservation of energy = total quantity of energy in the universe is constant
4 Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures.Temperature is a measure of the thermal energy.Temperature = Thermal Energy°CK°F1 joule = 1 kg m /sec1 calorie = joule (exactly)400C900Cgreater thermal energy
5 Study the interconversion of energy from one form to another. Thermochemistry is the study of heat change in chemical & physical processes.=> chemical reactions=> phase changesStudy the interconversion of energy from one form to another.
6 The system is the specific part of the universe that is of interest in the study. The surroundings is the rest of the universe.SURROUNDINGSSYSTEMSystem Types:openclosedisolatedExchange:mass & energyenergynothing6.2
7 State – specific characteristics of a system (physical or chemical) Examples: PressureTemperatureCompositionState Function – a property of a system that characterizes a state of a system and is independent of how the system got to that state.Examples: PressureVolumeComposition
8 Energy it took each hiker to get to the top is different !! ThermodynamicsState functions are properties that are determined by the state of the system, regardless of how that condition was achieved.energy, pressure, volume, temperatureDE = Efinal - EinitialDP = Pfinal - PinitialDV = Vfinal - VinitialDT = Tfinal - TinitialPotential energy of hiker 1 and hiker 2 is the same even though they took different paths.Energy it took each hiker to get to the top is different !!
9 2H2 (g) + O2 (g) 2H2O (l) + energy Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.2H2 (g) + O2 (g) H2O (l) + energyH2O (g) H2O (l) + energyEndothermic process is any process in which heat has to be supplied to the system from the surroundings.energy + H2O (s) H2O (l)energy + 2HgO (s) Hg (l) + O2 (g)6.2
11 First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.DEsystem + DEsurroundings = 0orDEsystem = -DEsurroundingsE is a state function.C3H8 + 5O CO2 + 4H2OExothermic chemical reaction!Chemical energy lost by combustion = Energy gained by the surroundingssystemsurroundings6.3
12 Another form of the first law for DEsystem DE = q + wEnergy is expressed in units of joules of kJ.DE is the change in internal energy of a systemq is the heat exchange between the system and the surroundingsw is the work done on (or by) the systemw = -PDV when a gas expands against a constant external pressureNote the conditions that define each process !!
13 Work is done by the surroundings on the system Work Done On The Systemw = FdWork is done by the surroundings on the systemDV < 0-PDV > 0wsys > 0w = -P DVP x V = x d3 = Fd = wFd2initialfinalWork is not a state function!Dw = wfinal - winitial
14 Work is done by the system on the surroundings Work Done By The Systemw = FdWork is done by the system on the surroundingsDV > 0-PDV < 0wsys < 0w = -P DVP x V = x d3 = Fd = wFd2Work is not a state function!Dw = wfinal - winitialinitialfinal6.3
15 W = -0 atm x 3.8 L = 0 L•atm = 0 joules A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm?w = -P DV(a)DV = 5.4 L – 1.6 L = 3.8 LP = 0 atmW = -0 atm x 3.8 L = 0 L•atm = 0 joules(b)DV = 5.4 L – 1.6 L = 3.8 LP = 3.7 atmw = -3.7 atm x 3.8 L = L•atmEnergy needs to be converted to joules. Use 1 L . atm = Jw = L•atm x101.3 J1L•atm= JThe change in volume was the same in both cases, but the work done was not the same => work is not a state function.
16 Chemistry in Action: Making Snow DE = q + wInside machine = air + water vapor @ 20 atmOutside machine = 1 atmAir w/ water vapor rapidly expandsq = adiabatic processDE = ww < 0DE < 0Kinetic Energy (gas) a T (K)DE = CDTDT < 0, SNOW !!
17 Note the conditions that define each process !! Enthalpy and the First Law of ThermodynamicsDE = q + wNote the conditions that define each process !!q = DH and w = -PDVAt constant pressure:DE = DH - PDVDH = DE + PDV6.4
18 Enthalpy is a state function. Enthalpy (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)Enthalpy is a state function.DH = heat given off or absorbed during a reaction at constant pressureHproducts < HreactantsHproducts > HreactantsDH < 0DH > 06.4
19 Thermochemical Equations Is DH negative or positive?System absorbs heatEndothermicDH > 06.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.H2O (s) H2O (l)DH = 6.01 kJ6.4
20 Thermochemical Equations Is DH negative or positive?System gives off heatExothermicDH < 0890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DH = kJ6.4
21 Enthalpy is a state function. We only care about where we start and where we finish, not the route.DHnet = DH1 + DH2 + DH3 + DH4DH2DH3EnergyDH1endDHnetDH4start=> We can construct a convenient path.
22 Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substanceH2O (s) H2O (l)DH = 6.01 kJIf you reverse a reaction, the sign of DH changesH2O (l) H2O (s)DH = kJIf you multiply both sides of the equation by a factor n, then DH must change by the same factor n.2H2O (s) H2O (l)DH = 2 x 6.01 = 12.0 kJYou must pay attention to the physical state and the quantity of all reactants and products !!!
23 Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations.H2O (s) H2O (l)DH = 6.01 kJH2O (l) H2O (g)DH = 44.0 kJHow much heat is evolved when 266 g of white phosphorus (P4) burn in air?P4 (s) + 5O2 (g) P4O10 (s) DH = kJ1 mol P4123.9 g P4x3013 kJ1 mol P4x266 g P4= 6470 kJ6.4
24 Flameless Ration Heaters “Feeding the Army”Flameless Ration HeatersChemical Equation:Mg (s) + 2H2O (l) Mg(OH)2 (s) + H2 (g)+ 353kJSolid magnesium and water yields solid magnesium hydroxide, hydrogen gas & kilojoules of heat energySource- Tocci, Viehland Holt Chemistry Visualizing Matter. Austin, Texas: Holt, Rinehart and Winston, Inc. pg. 271, 303and
25 It takes about 353kJ of heat energy is needed to heat the MRE to 60°C It takes about 353kJ of heat energy is needed to heat the MRE to 60°C . Calculate the amount of magnesium needed.353 kJ = 1 mol Mg = g MgCalculate the volume of water needed.353 kJ = 2 mol H g / mol H20 H20 density = 1.00 g/mL(2 mol H20) x (18.01 g / mol H20) x (1 mL H20 / 1.00 g H20) = mL H20Calculate the mass of magnesium hydroxide Mg(OH)2 produced.24.0g Mg = 1 mol Mg = 1 mol Mg(OH) g / mol Mg(OH)258.0 g Mg(OH)2Calculate the number of moles of hydrogen gas H2 produced.353 kJ = 1 mol Mg = 1.00 mol H2Calculate the volume of hydrogen gas H2 produced at 1.00 atm and 25°C.PV = nRT => V = nRT/PV = (1.00 mol)( L atm/K mol)(298 K)/1.00 atm = L
26 What would happen if 25 mL of water is added? What would happen is the soldier were smoking while heating dinner?Is the material in a used flameless ration heater safe to touch?Can the heater be reused?
27 A Comparison of DH and DE 2Na (s) + 2H2O (l) NaOH (aq) + H2 (g) DH = kJ/molDE = DH - PDVAt 25 0C, 1 mole H2 = 24.5 L at 1.0 atmPDV = 1.0 atm x 24.5 L = 2.5 kJDE = kJ/mol – 2.5 kJ/mol = kJ/mol6.4
29 Heat (q) absorbed or released: The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.C = msHeat (q) absorbed or released:q = msDtq = CDtDt = tfinal - tinitial6.5
30 Dt = tfinal – tinitial = 50C – 940C = -890C How much heat is given off when an 869 g iron bar cools from 940C to 50C?s of Fe = J/g • 0CDt = tfinal – tinitial = 50C – 940C = -890Cq = msDt= 869 g x J/g • 0C x –890C= -34,000 J6.5
31 DHnet = DH1 + DH2 + DH3 + DH4 + DH5 How much heat is required to take exactly one mole of water from a solid at -25.0°C to steam at 125.0°C?DHnet = DH1 + DH2 + DH3 + DH4 + DH5DH1 = q1 = msDt1 m = (1 mol ) (18.01 g/mol) = gs = J / g °C Dt1 = (0°C - -25°C) = 25°CDH1 = (18.01 g) (2.030 J / g °C) (25°C) = 914 J = kJDH2 = DHfus = (6.01 kJ/mol ) (1 mol) = 6.01 kJH2O (s) H2O (l)DH = 6.01 kJDH3 = q3 = msDt s = J / g °CDt3 = (100.°C -0°C) = 100.0°CDH3 = (18.01 g)(4.184 J / g °C)(100.0°C) = 7540 J = 7.54 kJDH4 = DHvap = (44.0 kJ/mol ) (1 mol) = 44.0 kJH2O (l) H2O (g)DH = 44.0 kJDH5 = q5 = msDt s = J / g °CDt5 = (125.0°C – 100.0°C) = 25.0°CDH5 = (18.01 g) (2.043 J / g °C) (25.0°C) = 920 J = kJDHnet = DH1 + DH2 + DH3 + DH4 + DH5= kJ kJ kJ kJ kJ = 56.1 kJ
34 What happens is we mix two liquids at different temperatures? Example: Mix 100 g water at 25°C with 100 g water at 75°C What is the final temperature?Since both liquids are the same (water) and both have the same weight (100 g) => take the average75°CLoss of thermal energy50°CEnergyGain of thermal energy25°CWhat happens if the substances are different or the amounts are different?
35 | Heat lost by copper | = | Heat gained by water | If g of Copper at °C is added to g water at 25.0°C What is the final temperature?| Heat lost by copper | = | Heat gained by water |qsys = qwater + qcal + qCu = 0 assume qcal is very very small- qCu = qH2O-qCu = - mCusCuDtCu = mH2OsH2ODtH2ODt = (tfinal – tinitial)sCu = J / g °C sH2O = J / g °C-(60.56 g)(0.385 J/g°C)(tfinal °C) = (150. g)(4.184 J/g°C)(tinitial -25.0°C)(-23.3 J/g°C) tfinal J = (628 J/g°C) tfinal – J15700 J J = (( ) J/g°C) tfinaltfinal = (18000/651.3)°Ctfinal = 27.7°C
37 Chemistry in Action: Fuel Values of Foods and Other Substances C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) DH = kJ/mol1 cal = J1 Cal = 1000 cal = 4184 J
38 (24.76 kJ / g Mg)(24.31 g Mg / mol Mg) = 601.9 kJ / mol Mg A g sample of solid magnesium is burned in a constant volume bomb calorimeter that has a heat capacity of 3027 J/°C. The temperature increases by °C. Calculate the heat change by the burning Mg in kJ/g and in kJ/mol.| Heat given off by reaction | = | Heat gained by calorimeter |qCal = CpDt = (3024 J/°C)(1.126 °C) = x103 JqRxn = (3.405 x103 J)(1 kJ / 1000 J) = kJ / g Mgg Mg(24.76 kJ / g Mg)(24.31 g Mg / mol Mg) = kJ / mol MgIs the reaction exothermic or endothermic?What would happen to the temperature of the water if it were the opposite type reaction?
39 Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions.fStandard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.fThe standard enthalpy of formation of any element in its most stable form is zero.DH0 (O2) = 0fDH0 (C, graphite) = 0fDH0 (O3) = 142 kJ/molfDH0 (C, diamond) = 1.90 kJ/molf
41 The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxnaA + bB cC + dDDH0rxndDH0 (D)fcDH0 (C)=[+]-bDH0 (B)aDH0 (A)DH0rxnnDH0 (products)f=SmDH0 (reactants)-Hess’s 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 doesn’t matter how you get there, only where you start and end.)
42 Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0 = kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJrxn1. Write the enthalpy of formation reaction for CS2C(graphite) + 2S(rhombic) CS2 (l)2. Add the given rxns so that the result is the desired rxn.rxnC(graphite) + O2 (g) CO2 (g) DH0 = kJ2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJrxn+CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJrxnC(graphite) + 2S(rhombic) CS2 (l)DH0 = (2x-296.1) = 86.3 kJrxn
43 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)DH0rxnnDH0 (products)f=SmDH0 (reactants)-DH0rxn6DH0 (H2O)f12DH0 (CO2)=[+]-2DH0 (C6H6)DH0rxn=[ 12x– x–187.6 ] – [ 2x49.04 ] = kJ-5946 kJ2 mol= kJ/mol C6H66.6
44 DHsoln = Hsoln - Hcomponents The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.DHsoln = Hsoln - HcomponentsWhich substance(s) could be used for melting ice?Which substance(s) could be used for a cold pack?6.7
45 The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol6.7