Chapter 6 – Energy Types of Systems Energy Energy (E) The capacity to do work or transfer heat. Kinetic Energy The energy of motion. Potential Energy Energy due to condition, position, or composition. Internal Energy The sum of all kinetic & potential energy contained in a system First Law of thermodynamics: The energy of the universe is constant (i.e. energy is neither created nor destroyed) Energy Units Calorie (cal) The quantity of heat required to change the temperature of one gram of water by one degree Celsius. Joule (J) SI unit for heat 1 cal = 4.184 J (exactly) Internal Energy Changes ΔE = q + w q = Heat The transfer of energy due to temperature changes Heat and temperature are not the same w = Work Force acting through a distance. Energy can be transferred between a system and its surroundings as a result of a temperature difference (q) and/or work being done (w). Internal Energy Changes The sign of q and w are from the system’s point of view In an endothermic process, heat flows into a system - ΔE is positive In an exothermic process, heat flows out of a system - ΔE is negative Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. Pressure – Volume Work • The expansion/compression of the system against the external atmosphere is pressure–volume work. Calculating Energy Changes (ΔE) When a reaction is run, the internal energy (kinetic, potential) can be transferred as heat and/or pressure-volume work: ΔE = q + w = q – PΔV How much work is associated with the expansion of a gas from 46L to 64L at a constant pressure of 15 atm? Filling (and heating) a hot air balloon takes 1.3X108 J of heat. At the same time, the volume changes from 4.00X106 L to 4.5X106 L. Assuming a constant pressure of 1 atm, what is the energy change for the process? Heat Capacities Heat capacity, C • Amount of heat needed to raise the temperature of the system by one degree Kelvin Heat Capacities Heat capacity, C • Amount of heat needed to raise the temperature of the system by one degree Kelvin Molar heat capacity, C q = CΔT • System = one mole of substance. q = mcΔT Specific heat capacity, c • System = one gram of substance If we add 500 J of heat to 25g of water initially at 25 °C, what is the final temperature of the water? (the specific heat of water = 4.184 J/g•°C) Heat Capacity and Conservation of Energy If substances at two different temperatures are mixed and allowed to come to a constant temperature: qa = –qb What is the final temperature when 125 g of iron at 92.3 °C is dropped into 50.0 g of water at 27.7 °C? The specific heat of iron is 0.444 J/g•°C and the specific heat of water is 4.184 J/g•°C. (Assume an isolated system) A 150.0 gram sample of metal at 75.0 °C is added to 150.0g of water at 15.0 °C . The temperature of the water rises to 23.0 °C. What is the specific heat of the metal? (Assume an isolated system) Reaction (Bomb) Calorimetry If we run a reaction in an isolated system (a calorimeter), we can very accurately measure the heat transferred as a result of the reaction. Note that this is a constant volume process. As long as we know the calorimeter (C): qcal = CΔT qrxn = –qcal = –CΔT heat Well insulated – considered isolated Reactions at Constant Volume ΔE = q – PΔV In a system at constant volume, no pressure-volume work is done: PΔV = P(0) = 0 Therefore, at constant volume, the internal energy change is equal to the heat of reaction: ΔE = q + 0 = qv A Coffee Cup (Simple) Calorimeter Note that this is a constant pressure process. qrxn = -qcal Well insulated – considered isolated 50.0 mL each of 1.0M HCl and 1.0M NaOH at 25 °C at mixed in a calorimeter. After reaction, the temperature of the calorimeter is 31.9 °C. What is the heat generated for the reaction? (We will estimate that the specific heat of the solution/calorimeter is about the same as that of water = 4.184 J/g•°C) State Functions Any property that has a unique value for a specified state of a system is said to be a State Function. • • • • • Water at 293 K and 1.00 atm is in a specified state. In this state, the density of water is 0.99820 g/mL This density is a unique function of the state. It does not matter how the state was established. Capitalized letters are used to identify State functions Internal Energy – A State Function ΔE has a unique value between two states Reactions at Constant Volume In a system at constant volume, no pressure-volume work is done: PΔV = P(0) = 0 Therefore, at constant volume, the internal energy change is equal to the heat of reaction: ΔE = qv Reactions at Constant Pressure Normally, reactions are run at constant pressure (and changing volume). At constant pressure, both heat and pressure-volume work results from energy changes: Enthalpy Change Because we are usually only interested in the heat of reaction at constant pressure, we will define a new state function: ΔE = qP - PΔV Let qP = ΔE + PΔV H = E + PV qP = ΔH = ΔE + PΔV ΔH, the enthalpy change, is the measurement we will generally use to describe thermal changes in a chemical system. Exothermic and Endothermic Reactions Negative ΔH = an exothermic reaction Positive ΔH = an endothermic reaction Note: Enthalpy change is an extensive property – it is directly proportional to the amount of substances in the system How Does a “Hand Warmer” Work? How much energy is needed to heat the water used in a 5 minute shower on Colby’s campus? (Assume Colby showers are set to 2 gal/min) Heat (Enthalpy) of Reaction All reactions will have an accompanying enthalpy change: ΔHrxn = Hfinal - Hinitial ΔHrxn = ΔHproducts- ΔHreactants For any reaction, the enthalpy change is the sum of the product enthalpies minus the sum of the starting material enthalpies. Why Do We Use These Energy Sources? How much natural gas must we burn to produce the heat (4435 kJ) needed for a single 5 minute hot shower on Colby’s campus? Changes in States of Matter Why doesn’t a pot of (boiling) water at 100 °C all become steam at once? Why do we have to continually apply heat? Changes in States of Matter Any change of state will cause an enthalpy change : Unless stated otherwise, ΔH values are assumed to be per mole Changes in States of Matter Any change of state will cause an enthalpy change : How much energy is required to convert 5 g of ice at 0 °C to water at 50 °C? To steam at 100 °C? Changes in States of Matter Any change of state will cause an enthalpy change : Which will cause a more damaging burn: skin exposed to 1 g of water at 100 °C or skin exposed to 1 g of steam at 100 °C? Manipulating Reaction Enthalpies The “reverse” of any reaction will have an equal enthalpy change of opposite sign. Standard States and Standard Enthalpy Changes • First we must define a particular state as a standard state • ΔH° is the standard enthalpy of reaction – The enthalpy change of a reaction in which all reactants and products are in their standard states • The Standard States are defined as: – The pure element or compound at a pressure of 1 bar (approximately 1 atm) and at the temperature “of interest” (usually 25 °C). Standard Enthalpies of Formation • ΔHf° , the standard enthalpy of formation, is the enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference (most common) forms of the elements in their standard states. • The standard enthalpy of formation of a pure element in its standard state is 0. Standard Enthalpies of Formation ΔHf° for CH2O = – 108.6 kJ/mol. ΔHf° for Al2O3 = – 1670 kJ/mol. ΔHf° for Fe2O3 = – 822 kJ/mol. What reactions do these heats of formation represent? Standard Enthalpies of Formation Standard Enthalpies of Formation Reaction Summation – Hess’s Law Hess’s law: If a process occurs in stages or steps (even hypothetically), then the enthalpy change for an overall process is the sum of the enthalpy changes for the individual steps. Manipulating ΔH – Hess’s Law • The enthalpy change of a chemical transformation is directly proportional to the amounts of substances: • The reverse of a chemical reaction has an equal but opposite DH: What is the standard enthalpy of reaction for the thermite reaction? How can we use a table of standard heats of formation to determine this? ΔHf° for Al2O3 = – 1670 kJ/mol. ΔHf° for Fe2O3 = – 822 kJ/mol. What is the standard enthalpy of reaction for the reaction below? How can we use a table of standard heats of formation to determine this? 2 NaHCO3 Na2CO3 + H2O + CO2 What is the standard enthalpy of reaction for the reaction below? How can we use a table of standard heats of formation to determine this? 2 NaHCO3 Na2CO3 + H2O + CO2 ΔHrxn = ΣΔHf°products- ΣΔHf°reactants What is the standard enthalpy of reaction for the formation of N2O5 as shown below? 2 N2 (g) + 5O2 (g) 1/2 N2 (g) + 3/2 O2 (g) + 1/2 H2 (g) N2O5 (g) + H2O (l) H2 (g) + 1/2 O2 (g) 2N2O5 HNO3 (l) (g) DH = -174.1kJ 2 HNO3 (l) DH = -76.6kJ H2O (l) DH = -285.8kJ