Chapter 5ThermochemistryA. DefinitionsB. CalorimetryC. Enthalpies of FormationD. Hess’s LawE. Energy SourcesHW #P4#P4 will be covered on Quiz 2 and Exam 2Chapter 5ThermochemistryA. Definitions (pp 164-181)Chemical reactions are accompanied by energy changes.Conservation of Energy  Energy is neither created nor destroyed.Processes that liberate heat are exothermic.Processes that absorb heat are endothermic.Enthalpy  H  is the heat content of a substance.The enthalpy change for a reaction (at constant pressure) is our focus.Hrxn = Hproducts  HreactantsHrxn is negative for exothermic and positivefor endothermic processes.2H2(g) + O2(g)  2H2O(g) H = 483.6 kJGuidelines:1. Hrxn depends on amount. 2 moles of H2(g) reacting with 1 mole O2(g) releases 483.6 kJ. 4 moles H2(g) reacting with 2 moles O2(g) gives 967.2 kJ.2. For the reverse reaction the sign of Hrxn changes.2H2O(g)  2H2(g) + O2(g) H = 483.6 kJ3. The value of Hrxn depends on the state of the reactants and products.2H2(g) + O2(g)  2H2O(l) H = 571.0 kJKinetic Energy – energy of motionKE = ½ mv2Potential Energy – energy of objectelectrical potential energy is given byE  Q1Q2d Work – energy to move object against a forceE = q + wq is heatw is workB. CalorimetryCalorimetry measures heat flow.1. DefinitionsHeat Capacity is the amount of heatrequired to raise an object’s temperature 1K.Specific Heat is the heat capacity of 1 g of a substance.Specific heat =quantity of heat transferred(grams of substance) × (temperature change)= qm × TExample:209 J is required to raise the temperature of 50.0 g of water by 1.00 K. What is the specific heat of water?C. Enthalpies of FormationEnthalpies can’t be measured for a substance, but we can use H for a defined reaction.We use tables (Table 5.3 or Appendix C) of standard heats of formation,Hf°Hf° is the change in enthalpy for formation of 1 mole of substance from the elements in their standard states at 25C.Example: Ethanol, C2H5OH2C(graphite) + ½O2(g) + 3H2(g) C2H5OH(l) Hrxn = 277.7 kJSince the reaction is for formation of 1 mole of C2H5OH(�)from the elements in their standard statesHrxn = Hf°(C2H5OH(l)) = -277.7kJ/mole Hf° for any element in standard state is zero.We use the Hf° to represent the enthalpies of substances for calculations.Examples:1. What is the standard enthalpy change for the conversionCaO(s) + CO2(g)  CaCO3(s)2. The standard heat of formation of TNT is 35.4 kJ. What is its heat of decomposition?2C7H5(NO2)3(s)  7C(s) + 7CO(g) + 3N2(g) + 5H2O(g)Bonus Example:C. How much heat is evolved from combustion of1.00 g of C2H5OH?D. Hess’s LawIf a reaction can be carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes for each step.Examples:1. What is the Hrxn for making diamonds,CH4(g) + O2(g)  C(dia) + 2H2O(l)givenCH4(g) + 2O2(g)  CO2(g) + 2H2O(l)Hrxn = 890 kJCO2(g)  O2(g) + C(dia)Hrxn = 395 kJE. Energy SourcesEnergy Problem, Fig 5.26Foods, Table 5.5Energy content of food comes from combustion of carbohydrates, fats and proteins.Carbohydrate (C6H12O6  glucose)C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)Hrxn = 17 kJ/gFatsC57H110O6(s) + 163/2 O2(g) 77CO2(g) + 55H2O(l)Hrxn = 38 kJ/gProteins are very complex  average about 17 kJ/gFoods use Calories;1 Calorie = 4.18 kJExample:A granola bar which weighs 1.0 oz (28 g) contains 3.0 gof protein, 5.0 g of fat and 17 g of carbohydrate. Whatis the energy value in