2 Thermochemistry Deals with the energy changes Chemical reactions Determines whether a reaction is exothermic or endothermicPhysical ChangesEnergy changes
3 Vocabulary Energy - The ability to do work (E) in J Potential Energy – Energy of position (PE)Kinetic Energy – Energy of motion (KE)Temperature – Random motion of particlesHeat – Transfer of energy (q) in JWork – Force x Distance (W) in JEnthalpy – Energy Change for reactions(ΔH) in J
4 Vocabulary Pathway – How energy is transferred State Function – Properties that depend on current state. (Independent of pathway)Energy is a state functionTemperature, Volume, PressureHeat and Work are not state functions
15 Hess’s LawIn going from a particular set of reactants to a particular set of products, the change in enthalpy (H) is the same whether the reaction takes place in one step or in a series of stepsEnergy is a state function
17 Hess’s Law RulesStart with the end reaction in mind and work backward rearrange your equationsWork to cancel like termsWhen reversing the equation the sign of ΔH must be changedWhen multiplying an equation by a coefficient ΔH must be multiplied by the same coefficient
18 Example #54 p. 283NH3 ½ N2 + 3/2 H2 H = 46 kJ2H2 + O2 2H2O H = -486 kJ2N2 + 6H2O 3O2 + 4NH3 H = ?Based on enthalpy is this a useful synthesis?
21 Calculate how much energy (kJ) is required to heat 50 Calculate how much energy (kJ) is required to heat 50.0 L of tap water from 10.0 ºC to 45.0 ºC. Assume the density of water is 1.00 g/mL and specific heat of water is 4.18 J/g* ºC7320 kJA McDonalds hamburger contains Calories (that’s 2.50x105 calories). How many burgers would it take to supply the same amount of energy? calorie = 4.18 J7.00 burgers
22 Enthalpy of FormationThe change in enthalpy that accompanies the formation of one mole of a compound from its elements with all elements in their standard stateSymbol = ΔHfºf is formationDegree symbol means standard states
23 Standard States Standard States Temperature is 25 ºC Pressure = 1.00 atmConcentration = 1.00 MThe standard state is how something exists at these conditionsO2(g), Na(s), Br2(l)
24 Reactions 4C(s) + 2H2(g) CH4(g) ΔHfº = -75 kJ/mol H2(g) + ½ O2(g) H2O(l) ΔHfº = -286 kJ/molWrite the enthalpy of formation reaction for gaseous carbon dioxide. Don’t worry about the valueC(s) + O2(g) CO2(g)Values are in Appendix 4 (A21)ΔHfº = kJ/mol
25 Enthalpy of Formation Values Most values are negativeCompounds are more stable than elementAll elements are zeroEven diatomic elements!States matterLiquid water is -286Gaseous water is -242Use this information to calculate enthalpy of the reaction!
27 Standard Enthalpy of Reaction Standard Enthalpy of Reaction is the energy change at standard conditionsSymbol = ΔHºΔHº = Σ ΔHfº(Products) - Σ ΔHfº(Reactants)Remember to include coefficientsRemember elements are zero!
28 Al(s) + Fe2O3(s) Al2O3(s) + Fe(s) Calculate the standard enthalpy of reaction for the followingAl(s) + Fe2O3(s) Al2O3(s) + Fe(s)Al = 0, Fe = 0, Fe2O3 = -826, Al2O3 = -1676