Presentation on theme: "Chapter 2 Elements of Chemical Thermodynamics Atkins: Ch6 & 7"— Presentation transcript:
1Chapter 2 Elements of Chemical Thermodynamics Atkins: Ch6 & 7 Silberberg: Ch 6 & 20
2Thermodynamics is the study of heat and its transformations. Thermochemistry is a branch of thermodynamics that deals withthe heat involved with chemical and physical changes.Fundamental premiseWhen energy is transferred from one object to another,it appears as work and/or as heat.For our work we must define a system to study; everything elsethen becomes the surroundings.The system is composed of particles with their own internal energies (U or E).Therefore the system has an internal energy. When a change occurs, theinternal energy changes.
3A chemical system and its surroundings. the surroundingsthe system
4System & SurroundingsSystem - the specific part of the universe that is of interest in the study. often the reacting vessel in the chemistry labA beaker of NaOH and HClSurroundings - the rest of the universeWhen heat is given off from the beaker, the energy is transferred out of the system
7Exothermic & Endothermic Reactions Exothermic - transfer of heat from the system to the surroundings2H2(g) + O2(g) 2H2O(l) + energyEndothermic - the transfer of heat from the surroundings to the systemenergy + 2HgO(s) 2Hg(l) + O2(g)
9Copyright McGraw-Hill 2009 Internal EnergyInternal energy: 2 componentsKinetic energy - molecular motionPotential energy - attractive/repulsive interactionsCopyright McGraw-Hill 2009
10Copyright McGraw-Hill 2009 Internal EnergyWe cannot calculate the changes in internal energy with any certaintyWe can calculate the changes in energy of the system experimentallyU = U(products) - U(reactants)Copyright McGraw-Hill 2009
11DU = Ufinal - Uinitial = Uproducts - Ureactants Energy diagrams for the transfer of internal energy (U) between a system and its surroundings.DU = Ufinal - Uinitial = Uproducts - Ureactants
12DUuniverse = DUsystem + DUsurroundings Units of EnergyJoule (J)1 J = 1 kg*m2/s2Calorie (cal)1 cal = 4.18JBritish Thermal Unit1 Btu = 1055 J
13First Law of Thermodynamics Energy can be converted from one form to another but cannot be created nor destroyed(the Law of Energy Conservation)Copyright McGraw-Hill 2009
14Energy and the Universe Usystem + Usurroundings = 0When a system releases heat, some of the chemical energy is released as thermal energy to the surroundings but this does not change the total energy of the universe.Copyright McGraw-Hill 2009
15Energy and the Universe Usystem = - UsurroundingsWhen a system undergoes a change in energy, the surroundings must undergo a change in energy of equal magnitude but opposite in sign.Copyright McGraw-Hill 2009
16Copyright McGraw-Hill 2009 Work and HeatUsys = q + w“q” = heat“w” = workCopyright McGraw-Hill 2009
18+ = + + + + - - + - - - q w DU The Sign Conventions* for q, w and DU depends on sizes of q and w-+depends on sizes of q and w---* For q: + means system gains heat; - means system loses heat.* For w: + means work done on system; - means work done by system.
19Copyright McGraw-Hill 2009 Work and HeatCalculate the overall change in internal energy for a system that absorbs 125 J of heat and does 141 J of work on the system.“q” = + (heat absorbed)“w” = - (work done)Usys = q + w = (+125 J) + (-141 J)= -16 JCopyright McGraw-Hill 2009
20State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.energy, pressure, volume, temperatureDU = Ufinal - UinitialDP = Pfinal - PinitialDV = Vfinal - VinitialDT = Tfinal - TinitialPotential energy of hiker 1 and hiker 2 is the same even though they took different paths.
21Copyright McGraw-Hill 2009 Constant VolumeConstant volume: (V = 0)U = q - PVqv = U(not easy to achieve in the chem lab)Copyright McGraw-Hill 2009
23Enthalpy and Enthalpy Changes Enthalpy (H) is a state functionEnthalpy measures heat contentConstant pressure process: (most are)qp = HEnthalpy of reaction:H = H(products) - H(reactants)H = - (exothermic) H = + (endothermic)Copyright McGraw-Hill 2009
24Thermochemical Equations Equations that represent mass and enthalpy relationshipsH2O(s) H2O(l) rH = kJ/molThis is an endothermic process. It requires 6.01 kJ to melt one mole of ice.(the “per mole” means per mole of reaction as written - respect stoichiometry)Copyright McGraw-Hill 2009
25Thermochemical Equation Guidelines Always specify state of reactants and products.When multiplying an equation by a factor (n), multiply the H value by same factor.Reversing an equation changes the sign but not the magnitude of the H.Copyright McGraw-Hill 2009
26Enthalpy and the First Law of Thermodynamics DU = q + wq = DH and w = -PDVAt constant pressure:DU = DH - PDVDH = DU + PDV
27The Meaning of Enthalpy w = - PDVDH ≈ DU inH = U + PV1. Reactions that do not involve gases.where H is enthalpy2. Reactions in which the number of moles of gas does not change.DH = DU + PDV3. Reactions in which the number of moles of gas does change but q is >>> PDV.qp = DU + PDV = DH
28Enthalpy diagrams for exothermic and endothermic processes. CH4(g) + 2O2(g) CO2(g) + 2H2O(g)H2O(l) H2O(g)CH4 + 2O2H2O(g)HfinalEnthalpy, HHinitialEnthalpy, Hheat outheat inDH < 0DH > 0CO2 + 2H2OH2O(l)HfinalHinitialA Exothermic processB Endothermic process
30The standard enthalpy of reaction (DrH0 ) is the enthalpy of a reaction carried out at 1 atm. aA + bB cC + dDDrH0dDfH0 (D)cDfH0 (C)=[+]-bDfH0 (B)aDfH0 (A)DrH0nDfH0 (products)=SmDfH0 (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.)
31Sample Problem 2.1Using Hess’s Law to Calculate an Unknown DHPROBLEM:Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:CO(g) + NO(g) CO2(g) + 1/2N2(g) DH = ?Given the following information, calculate the unknown DH:Equation A: CO(g) + 1/2O2(g) CO2(g) DHA = kJ/molEquation B: N2(g) + O2(g) NO(g) DHB = kJ/molPLAN:Equations A and B have to be manipulated by reversal and/or multiplication by factors in order to sum to the first, or target, equation.SOLUTION:Multiply Equation B by 1/2 and reverse it.CO(g) + 1/2O2(g) CO2(g) DHA = kJ/molNO(g) /2N2(g) + 1/2O2(g)DHB = kJ/molCO(g) + NO(g) CO2(g) + 1/2N2(g)DHrxn = kJ/mol
32Because 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 (DfH0) as a reference point for all enthalpy expressions.Standard enthalpy of formation (DfHm) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.The standard enthalpy of formation of any element in its most stable form is zero.Df H m (O2) = 0DfH0 (C, graphite) = 0DfH0 (O3) = 142 kJ/molDfH0 (C, diamond) = 1.90 kJ/mol
33Selected Standard Heats of Formation at 250C(298K) FormulaDfHm (kJ/mol)calciumCa(s)CaO(s)CaCO3(s)carbonC(graphite)C(diamond)CO(g)CO2(g)CH4(g)CH3OH(l)HCN(g)CSs(l)chlorineCl(g)-635.11.9-110.5-393.5-74.9-238.613587.9121.0hydrogennitrogenoxygenFormulaDfHm (kJ/mol)H(g)H2(g)N2(g)NH3(g)NO(g)O2(g)O3(g)H2O(g)H2O(l)Cl2(g)HCl(g)-92.3218-45.990.3143-241.8-285.8107.8FormulaDfHm (kJ/mol)silverAg(s)AgCl(s)sodiumNa(s)Na(g)NaCl(s)sulfurS8(rhombic)S8(monoclinic)SO2(g)SO3(g)-127.0-411.12-296.8-396.0
34The general process for determining DrH0 from DfH0 values. Elementsdecompositionformation-DfH0DfH0Enthalpy, HReactantsHinitialDrH0ProductsHfinalDrH0 = S mDfH0 (products) - S nDfH0 (reactants)
35Sample Problem 2.2Calculating the Heat of Reaction from Heats of FormationPROBLEM:Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:Calculate DH0rxn from DH0f values.4NH3(g) + 5O2(g) NO(g) + 6H2O(g)PLAN:Look up the DH0f values and use Hess’s Law to find DHrxn.SOLUTION:DHrxn = S mDfH0 (products) - S nDfH0 (reactants)DrH = [4(DfH0 NO(g) + 6(DfH0 H2O(g)]- [4(DfH0 NH3(g) + 5(DfH0O2(g)]= (4)(90.3 kJ/mol) + (6)( kJ/mol) -[(4)(-45.9 kJ/mol) + (5)(0 kJ/mol)]DrH = -906 kJ/mol
36Some Important Types of Enthalpy Change heat of combustion (DcombH)C4H10(l) /2O2(g) CO2(g) + 5H2O(g)heat of formation (DfH)K(s) + 1/2Br2(l) KBr(s)heat of fusion (DfusH)NaCl(s) NaCl(l)heat of vaporization (DvapH)C6H6(l) C6H6(g)
37Limitations of the First Law of Thermodynamics DU = q + wUuniverse = Usystem + UsurroundingsDUsystem = -DUsurroundingsDUsystem + DUsurroundings = 0 = DUuniverseThe total energy-mass of the universe is constant.However, this does not tell us anything about the direction of change in the universe.