Chapter 6: Energy and Chemical Reactions Chem 103: Chapter 6 Chapter 6: Energy and Chemical Reactions Copyright 2006, David R. Anderson

Energy Types of energy Energy units Specific heat and energy transfer Changes of state Enthalpy: Thermochemical Equations Definitions Enthalpies of reaction Hess's law Enthalpy of formation standard state; standard molar enthalpies of formation enthalpies of reaction from enthalpies of formation Calorimetry Coffee-cup calorimetry Bomb calorimetry

= capacity to do work or transfer heat Energy = capacity to do work or transfer heat chemical energy in gasoline & O2 move car & generate heat (w = F•d)  A. Types of energy Kinetic energy: energy of motion mechanical: motion of objects (KE = ½mv2) thermal: motion of atoms and molecules electrical: motion of electrons Potential energy: stored energy gravitational energy (PE = mgh) chemical energy: energy stored in chemical bonds electrostatic energy: attraction or repulsion of charged particles  Total energy = KE + PE conserved! (law of conservation of energy)

SI: define 1 joule (J) = KE of a 2-kg object traveling at 1 m/s Energy Energy units SI: define 1 joule (J) = KE of a 2-kg object traveling at 1 m/s KE = ½mv2 = ½(2 kg)(1 m/s)2 = 1 kg·m2/s2 = 1 J 1 kJ = 1000 J English: 1 cal = heat required to raise the temperature of 1 g of water by 1ºC 1 kcal = 1000 cal 1 cal = 4.184 J 1 kcal = 4.184 kJ 1 Calorie (Cal) = 1000 cal = 1 kcal (food calorie)

heat capacity: heat required to raise the temperature of something Energy Specific heat heat capacity: heat required to raise the temperature of something (extensive) by 1ºC (1 K) specific heat: heat required to raise the temperature of 1 g of something (intensive) by 1ºC for water, specific heat, C = = or C = 1 cal/g ºC or 4.184 J/g ºC or rearranging for heat: q = C  m  DT How many kJ of heat are required to raise the temperature of 5.00 x 102 g of water by 24.0 ºC? q m•T cal g ·ºC J g·K

qfusion = heat required to melt a substance vaporization (boiling): Energy Changes of state fusion (melting): solid + heat  liquid qfusion = heat required to melt a substance vaporization (boiling): liquid + heat  gas qvaporization = heat required to vaporize a substance

Heating/cooling curves: Energy Changes of state Heating/cooling curves: gas (e) liquid + gas b.p. (d) T liquid (c) solid + liquid m.p. (b) solid (a) heat added  (heating)  heat removed (cooling) e.g. H2O (a) C(s) = 2.1 J/g·ºC (d) qvap = 2256 J/g or 40.7 kJ/mol (b) qfus = 333.5 J/g or 6.01 kJ/mol (e) C(g) = 2.0 J /g·ºC (c) C(l) = 4.2 J /g·ºC

Energy Changes of state How much heat is required to convert 10.0 g of solid water at -10ºC to gas at 100 ºC?

Internal energy, heat, and work: For a system: DEsys = qsys + wsys Enthalpy: Thermochemical Equations Definitions universe: surroundings system (energy can be transferred) Internal energy, heat, and work: For a system: DEsys = qsys + wsys change in energy for a system heat absorbed or released by the system work done on or by the system

Esys = energy of a system = sum of all energies (kinetic, bond, etc.) Enthalpy: Thermochemical Equations Definitions DEsys = qsys + wsys Esys = energy of a system = sum of all energies (kinetic, bond, etc.) DEsys = Efinal - Einitial system gains energy: DEsys > 0 (DEsurr < 0) system loses energy: DEsys < 0 (DEsurr > 0) qsys = heat heat added to system: qsys > 0 endothermic (qsurr < 0) heat lost from system: qsys < 0 exothermic (qsurr > 0) wsys = work work done on system: wsys > 0 (wsurr < 0) work done by system: wsys < 0 (wsurr > 0) can’t measure can measure

A balloon is heated by adding 240 J of heat. It expands, doing 135 J Enthalpy: Thermochemical Equations Definitions A balloon is heated by adding 240 J of heat. It expands, doing 135 J of work on the atmosphere. What is DE for the system?

taken to get from one to the other Enthalpy: Thermochemical Equations Definitions DE is a state function -depends only on the initial and final states of a system, not the path taken to get from one to the other 1 g H2O, 10 ºC 1 g H2O, 65 ºC 1 g H2O, 25 ºC 1 g H2O, 50 ºC DE = 25 cal First Law of Thermodynamics: energy is neither created nor destroyed; it can only change forms (conservation of energy). For a system: DEsys = qsys + wsys

Enthalpy change, DH = heat gained or lost at constant pressure Enthalpy: Thermochemical Equations Enthalpies of reaction Enthalpy change, DH = heat gained or lost at constant pressure DH = Hfinal - Hinitial = qP (does not include work) -state function Enthalpy of reaction Reactants  Products DH = Hproducts - Hreactants DH > 0: endothermic (products higher energy than reactants) DH < 0: exothermic (products lower energy than reactants)

1. DH is an extensive property What is DH per mole of Fe? Enthalpy: Thermochemical Equations Enthalpies of reaction e.g., 2Al + Fe2O3  2Fe + Al2O3 DH = -851.5 kJ (thermochemical equation) 1. DH is an extensive property What is DH per mole of Fe? 2. reverse equation: change sign of DH 2Fe + Al2O3  2Al + Fe2O3 DH = ? 3. enthalpy diagrams: 2Al + Fe2O3 DH = –851.5 kJ DH = +851.5 kJ H 2Fe + Al2O3

DHoverall process = SDHindividual steps (by any path; 1st law) Enthalpy: Thermochemical Equations Hess’s law DHoverall process = SDHindividual steps (by any path; 1st law) 3Mg + N2  Mg3N2 DH = ? given: Mg3N2 + 3H2  3Mg + 2NH3 DH = +371 kJ and: 1/2N2 + 3/2H2  NH3 DH = –46 kJ

S + O2  SO2 DH = ? given: 2SO2 + O2  2SO3 DH = –196 kJ Enthalpy: Thermochemical Equations Hess’s law S + O2  SO2 DH = ? given: 2SO2 + O2  2SO3 DH = –196 kJ and: 2S + 3O2  2SO3 DH = –790 kJ

1. standard state = most stable form of a substance at 1 bar pressure Enthalpy: Thermochemical Equations Enthalpy of formation 1. standard state = most stable form of a substance at 1 bar pressure and a given temperature, usually 25ºC (298 K). noble gases: monatomic gases diatomic gases: brinclhof metals: Mº carbon: graphite standard molar enthalpy of formation, DHfº = enthalpy of reaction for formation of one mole of a substance from its elements, all in their standard states (DHfº(element) = 0) (Table 6.2, Appendix L) e.g., DHfº(CO(g)) = –110.5 kJ/mol means: C(graphite) + ½O2(g)  CO(g) DHrxnº = –110.5 kJ/mol = DHfº

DHrxnº =  nDHfº(products) –  nDHfº(reactants) Enthalpy: Thermochemical Equations Enthalpy of formation enthalpies of reaction from enthalpies of formation DHrxnº =  nDHfº(products) –  nDHfº(reactants) C2H2(g) + 2H2(g)  C2H6(g) DHrxnº = ? C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) DHrxnº = ? DHfº (kJ/mol) C2H2(g) 226.7 C2H6(g) -84.68 C2H5OH(l) -277.7 CO2(g) -393.5 H2O(l) -285.8

Enthalpy: Thermochemical Equations Enthalpy of formation enthalpies of reaction from enthalpies of formation The heat of combustion of glucose is –2802.5 kJ. What is DHfº for glucose? C6H12O6(s) +

two styrofoam cups (insulation) Calorimetry Coffee-cup calorimetry thermometer stopper or lid two styrofoam cups (insulation) -enclose the universe: system (reaction) and surroundings (solution) reaction in solution generates or consumes heat, raising or lowering temperature qsolution = C  m  T Hrxn = qrxn = –qsolution i.e., if T increases (T>0), rxn is exothermic (H<0) if T decreases (T<0), rxn is endothermic (H>0)

Calorimetry Coffee-cup calorimetry When 4.25 g of NH4NO3(s) dissolves in 60.0 g of H2O in a coffee cup calorimeter, the temperatures drops from 22.0ºC to 16.9ºC. Assuming the specific heat of the solution is 4.184 J/g·ºC, what is H (in kJ/mol) for the dissolution of NH4NO3? (fw NH4NO3 80.04)

thermometer heating element stirrer insulation heat capacity of Calorimetry Bomb calorimetry thermometer heating element stirrer insulation heat capacity of calorimeter = C qrxn = -qcal = –C  T bomb

Calorimetry Bomb calorimetry A 1.80-g sample of octane, C8H18, was combusted in a bomb calorimeter with a heat capacity of 11.66 kJ/ºC. The temperature increased from 21.36ºC to 28.78ºC. What is H of combustion per mole of octane? (mw C8H18 114.23)