Tro's "Introductory Chemistry", Chapter 3 Thermodynamics Temperature: the measurement of the average kinetic energy of the particles in an object. Energy: the ability (or capacity) of a system to do work or supply (or produce) heat. Heat: heat is the transfer of energy between two objects due to temperature differences. Tro's "Introductory Chemistry", Chapter 3

Tro's "Introductory Chemistry", Chapter 3 Temperature K = 0C + 273 Water boils at 100 0C or 212 0F or 373 K Water freezes at 0 0C or 32 0F or 273 K Absolute Zero (0 K = -2730C): theoretical temperature at which all atoms cease motion Tro's "Introductory Chemistry", Chapter 3

Law of Conservation of Energy “Energy can neither be created nor destroyed.” The total amount of energy in the universe is constant. There is no process that can increase or decrease that amount. However, we can transfer energy from one place (a system) in the universe to another (surroundings), and we can change its form. Energy is stored in chemical bonds of matter.

Units of Energy Calorie (cal) is the amount of energy needed to raise one gram of water by 1 °C. kcal = energy needed to raise 1000 g of water 1 °C. food calories = kcals. Energy Conversion Factors 1 calorie (cal) = 4.184 joules (J) 1 Calorie (Cal) 1000 calories (cal)

Burning of a Match (chemical change) System Surroundings D(PE) (Reactants) Potential energy Energy released to the surrounding as heat (Products) Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293

An Energy Diagram (chemical change) activated complex activation energy Ea reactants products Potential energy course of reaction

Exothermic Reaction (chemical change) Reactants  Products + Thermal Energy (heat) Energy of reactants Energy of products Potential Energy Reactants -DHrxt Exothermic Products Reaction Progress

An Exothermic Reaction Surroundings reaction Potential energy Reactants Products Amount of energy released

The Zeppelin LZ 129 Hindenburg catching fire on May 6, 1937 at Lakehurst Naval Air Station in New Jersey.

Endothermic Reaction (chemical change) Thermal Energy + Reactants  Products (heat) Activation Energy Energy of reactants Energy of products Products Potential Energy +DHrxt Endothermic Reactants Reaction progress

An Endothermic Reaction Surroundings reaction Potential energy Products Reactants Amount of energy absorbed

Effect of Catalyst on Reaction Rate What is a catalyst? What does it do during a chemical reaction? Catalyst lowers the activation energy for the reaction. No catalyst activation energy for catalyzed reaction reactants Potential Energy A catalyst lowers the activation energy for the reaction. This allows the reaction to occur at a much faster rate. The catalyst is not a reactant or product. It is not consumed during the chemical reaction. products Reaction Progress

Barbecue An LP gas tank in a home barbecue contains 11.8 X 103g of propane (C3H8). Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank. The heat of reaction is -2044 kJ/mol C3H8. __C3H8 + __O2(g)  __CO2(g) + __H2O(g) 1 mol C3H8 -2044 kJ kJ = 11.8 X 103 g C3H8 44 g C3H8 1 mol C3H8 kJ = -5.47 X 105 kJ

Water Molecules in Hot and Cold Water Hot water Cold Water 90 oC 10 oC Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

Heat Transfer (ex: physical change) Surroundings System ENDOthermic EXOthermic System H2O(s) + heat  H2O(l) melting H2O(l)  H2O(s) + heat freezing Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207

Heating Curve of Water Gas - KE  Boiling - PE  Liquid - KE  140 Gas - KE  120 100 Boiling - PE  80 60 40 Liquid - KE  Temperature (oC) 20 Melting - PE  -20 -40 Solid - KE  -60 -80 -100 Thermal Energy added from Surroundings

Heat Gain or Loss by an Object The amount of heat energy gained or lost by an object depends on 3 factors: 1. how much material there is 2. what the material is 3. how much the temperature changed. Amount of Heat = Mass x Specific Heat x Temperature Change q = m x C x DT Tro's "Introductory Chemistry", Chapter 3

Thermometer Styrofoam cover cups Stirrer A Coffee Cup Calorimeter Calorimetry: experimental technique for investigating heat transfer Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302

Heat Capacity Heat capacity is the amount of heat a substance must absorb to raise its temperature by 1 °C. cal/°C or J/°C. Metals have low heat capacities; insulators have high heat capacities. Specific heat = heat capacity of 1 gram of the substance. cal/g°C or J/g°C. Water’s specific heat = 4.184 J/g°C for liquid. Or 1.000 cal/g°C. It is less for ice and steam.

Specific Heat Capacities

Calorimetry q = m . C . DT ΔT = TF - TI m . C . DT = - [m . C . DT] Object gaining heat = Object losing heat +q = -q m . C . DT = - [m . C . DT] Material whose temperature increases gains heat while materials whose temperature decreases lose heat. Metals generally lose heat, while liquids (such as water) generally gain heat in a system. We must account for all of the heat energy in the system due to conservation of energy.

Enthalpy Changes H2(g) + ½ O2(g) H2O(g) H2O(l) DH = +242 kJ Endothermic -242 kJ Exothermic -286 kJ Endothermic DH = -286 kJ Exothermic H2O(g) Energy 44 kJ Exothermic +44 kJ Endothermic H2O(l) H2(g) + 1/2O2(g)  H2O(g) + 242 kJ DH = -242 kJ Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211

Enthalpy Changes DH3 = DH1 + DH2 (Hess’s Law) Change in enthalpy does not depend on path of reaction a) H2(g) + 1/2O2(g)  H2O(g) DH1 = -242 kJ b) H2O(g)  H2O(l) DH2 = -44 kJ c) H2(g) + 1/2O2(g)  H2O(l) DH3 = -286 kJ

Hess’s Law a) N2(g) + O2(g)  2NO(g) DH1 = +180 kJ b) 2NO(g) + O2(g)  2NO2(g) DH2 = -112 kJ c) N2(g) + 2O2(g)  2NO2(g) DH3 = +68 kJ

Hess’s Law Calculate the enthalpy of formation of carbon dioxide from its elements. C(g) + 2O(g)  CO2(g) Use the following data: 2O(g)  O2(g) DH = - 250 kJ C(s)  C(g) DH = +720 kJ CO2(g)  C(s) + O2(g) DH = +390 kJ 2O(g)  O2(g) DH = - 250 kJ C(g)  C(s) DH = - 720 kJ C(s) + O2(g)  CO2(g) DH = - 390 kJ C(g) + 2O(g)  CO2(g) DH = -1360 kJ

Hess’s Law Calculate DH for the synthesis of diborane from its elements. 2B(s) + 3H2(g)  B2H6(g) DH = ? a) 2B(s) + 3/2O2(g)  B2O3(s) DH = -1273 kJ b) B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g) DH = -2035 kJ c) H2(g) + 1/2O2(g)  H2O(l) DH = - 286 kJ d) H2O(l)  H2O(g) DH = + 44 kJ