Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

Kinetic vs. potential energy Chemical energy is a form of potential energy. The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

total potential energy Kinetic vs. potential energy Kinetic energy: the energy of motion; dependent on mass and speed Ek increases as: 1) mass increases; 2) speed increases For example, think of different model cars traveling at different speeds. 2 volkswagen beetles drag racing – the faster has greater kinetic energy Beetle & SUV crashing head-on at identical speeds - SUV has more kinetic energy Potential Energy: stored energy… either physical or chemical Caused by the attractions & repulsions an object experiences in relation to other objects. Chemical potential energy? A gallon of gasoline Reagents which undergo spontaneous redox reactions zero kinetic energy, maximum potential energy kinetic energy, reduced potential energy total potential energy energy zero kinetic energy, zero potential energy p. 161

Energy within the atom Chemical potential energy: energy stored in the structure of the atom or in the chemical bonds of a molecule Results from electrostatic attractions between: - electrons & nucleus cations & anions covalently bonded atoms with EN differences + Chemical reactions may release potential energy. Example: combustion of gasoline (octane, C8H18) + heat 51 bonds created, so net energy release 33 bonds broken p. 161

Example: calculating kinetic energy What is the kinetic energy (J) of: An Ar atom moving at 650 m/s A mole of Ar atoms moving at 650 m/s Ar = 39.95 g/mol = 0.03995 kg 1 mol = 6.636 x 10-26 kg/atom 1 mol 6.02 x 1023 atoms Ek of one mole of Ar = (1/2)(0.03995 kg)(6502 m2/s2) = 8.4 x 103 J Ek of one atom of Ar = (1/2)(6.636 x 10-26 kg)(6502 m2/s2) = 1.4 x 10-20 J p. 164

Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

Thermodynamics is the study of energy tranfer Energy is transferred as pure energy, work or heat. The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

Thermodynamics & energy transfer Thermodynamics is the study of energy and its transformations. energy work heat Energy is the capacity to do work or to transfer heat. Energy is transferred in one of two ways: 1) causing motion of an object against a force  work 2) causing a change in temperature  heat Heat flow: heat always from hot regions to cold regions p. 163-4

Visualize energy transfer as work & heat It takes energy, in the form of work, to lift this ball of clay to the top of the brick wall. This work increases the potential energy of the ball. The ball may now roll off of the wall, and fall. Potential energy is transformed into kinetic energy. The ball hits the ground. Kinetic energy = zero. That energy is transferred in 2 ways: 1) work to squash the ball flat 2) into heat transferred to the ground p. 163-4

Thermal energy The energy a substance possesses because temperature (its thermal energy) is associated with the kinetic energy of the molecules in that substance. So rank the physical states of water in terms of levels of kinetic energy & temperature. steam > liquid water > ice Heat flows from regions of high temperature to regions of lower temperature. Our discussions of thermodynamics will focus on the transfer of chemical and thermal energy from reacting substances to the surrounding environment as heat. p. 163-4

Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

System vs. surroundings Closed systems exchange energy, but not matter, with their surroundings. Open systems exchange both energy & matter with their surroundings. The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

System vs. surroundings Science divides the world into two parts: The system; and Its surroundings System: the component parts of the experiment Surroundings: everything else in the universe So, imagine that we’re in the lab doing an experiment in a beaker on a lab bench. The beaker contains two ionic compounds, in water, that react with one another to form two products. What’s the system ? The ionic reactants & products only! Define the surroundings ? Everything but the ionic compounds. Everything includes the water, beaker, lab bench, lab, building… and the rest of the universe. Closed systems are the most easily studied: A closed system exchanges energy with surroundings, but does not exchange matter with surroundings. Open systems exchange both energy & matter with their surroundings. p. 162-3

Example: a closed system A mixture of octane (C8H18) and oxygen (O2) gases in a cylinder are compressed and then ignited. C8H18 (gas) + 15/2O2 (gas)  9H2O + 8CO2 + energy A chemical reaction occurs, breaking 33 bonds and forming 51 bonds to change changing octane and oxygen to water and carbon dioxide, and energy! No matter leaves the cylinder (the system). But the volume of gases expands pushes the piston up, doing work. And excess energy heats the apparatus and its surroundings work H2O C8H18 CO2 O2 p. 162-3

Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

First law of thermodynamics Energy is conserved. Energy is transferred & transformed. Internal energy is the sum of all energies of a system. The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

First law of thermodynamics The first law of thermodynamics governs interconversion of energies. Energy is conserved; energy is neither created nor destroyed. Thus, energy is only transferred or transformed. So where does energy go? What does it become? Energy lost by the system is gained by the surroundings. Energy gained by the system is lost from the surroundings. Kinetic energy is transformed into potential energy or vv. Work is transformed into heat or vv. Internal energy is the sum of all potential and kinetic energies of a system. This includes the all energy down to the movement of e- around the nucleus! Sign of E, q or w indicates whether heat is entering the system (+) or leaving the system (-). E = Ef - Ei Where f = final & i = initial E = q + w Where q = heat & w = work Most chemical reactions don’t “do work” and therefore internal energy is often equal changes in heat. Exception? Reactions that produce gas may result in changes in pressure & volume. These changes do work: explosives produce gas; NaN3….. p. 164-5

Internal energy of a system Addition, or loss, of energy as either heat or work changes the overall internal energy (ΔE) of a system. initial state final state E initial E final Energy lost Energy gained ΔE of system ΔE of system ΔE = E final – E initial ΔE = E final – E initial final state initial state E final E initial - ΔE, so system’s energy decreased + ΔE, so system’s energy increased p. 166

Examples: internal energy Gases A & B are confined in a cylinder-piston system and react to form the solid, C: A(g) + B(g)  C(s) As the reaction occurs, the system loses 1150 J of heat to the surroundings. As C is made, gas volume decreases, the piston drops, and the surroundings do 480 J of work on the system. Calculate the change in ΔE. ΔE = q + w = (-1150 J) + (+480 J) = -670 J So it’s all about knowing the signs of q & w. And that’s all about language and wording…. Subtle! p. 167

Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

Enthalpy Enthalpy is heat flow at constant pressure. Endothermic systems (reactions) require energy input (+ΔH). Exothermic systems (reactions) produce (release) energy (-ΔH). Thermochemical equations include energy as a product or reactant. Hess’s law allows enthalpies of reaction to be summed. The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

Enthalpy H = Hfinal - Hinitial = Hproducts - Hreactants Enthalpy (H): heat absorbed or released under constant pressure Enthalpy is a state function; only current state matters. As with internal energy, we cannot measure absolute enthalpy at any one point in time. Instead we measure change in enthalpy (H). H = Hfinal - Hinitial = Hproducts - Hreactants = qp Where qp is heat gained or lost by the system under constant pressure. ∆H = ∆(E + PV) = ∆E + P∆V Generally, enthalpy is more useful than internal energy, since most chemical rxns occur at constant pressure. For most reactions, PV is generally a very small factor, so there is little difference between enthalpy and internal energy. If H is positive, the reaction is endothermic; if negative then exothermic. Endothermic when Hproducts > Hreactants Exothermic when Hreactants > Hproducts p. 167-9

Exothermic vs. endothermic surroundings system system heat heat H > 0 (+) endothermic H < 0 (-) exothermic Endothermic: a process in which system absorbs heat from surroundings melting of ice (the container feels cool) Exothermic: a process in which system transfers heat to surroundings combustion of gasoline (the container feels hot) dilution of concentrated acid with water (steam is generated) thermite hand-warmers = Al + Fe2O3 p. 170-4

NH4Cl + H2O + heat  NH4+1 + Cl-1 + H2O Examples: exo- vs. endothermic Fe2O3 + 2Al  2Fe + Al2O3 + heat exothermic NH4Cl + H2O + heat  NH4+1 + Cl-1 + H2O endoothermic p. 167-8

Examples: exo- vs. endothermic For each example, indicate the sign of ΔH and describe the process as wither exo- or endothermic. An ice cube melts. One gram of butane is combusted completely. Energy input is required to melt ice, so ΔH is + & the process is endothermic. Heat is produced; ΔH is negative; process is exothermic. p. 171

Guidelines for thermochemical equations 1. Enthalpy is an extensive property. magnitude is directly proportional to the amount of reactant consumed CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ 2. Enthalpy change (H) for a reaction is equal in magnitude, but opposite in sign, to the H for the reverse reaction. CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) H = +890 kJ 3. Enthalpy change (H) for a reaction depends on the state of reactants and products. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ 2H2O(l) - 2H2O(g) H = 88 kJ Thermochemical equation = a balanced equation showing the H associated associated with the complete chemical reaction p. 173-4

Examples: thermochemical equations How much heat is released when 4.50 g of methane gas is burned at constant pressure? CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l) Hrxn = -890 kJ/mol MW CH4 = 16.042 g/mol 4.50 g 1 mol -890 kJ = -250 kJ 16.042 g 1 mol How many grams of ammonia can be decomposed with 1 MJ of energy? 2NH3 --> 3H2 + N2 Hrxn = +92 kJ 1 MJ = 1000 kJ (1000 kJ)(2 mol NH3/92 kJ)(17 g/mol NH3) = 370 g NH3 p. 174

Hess’s law Hess’s Law allows us to estimate enthalpy changes for complex chemical reactions without doing lab work. If a reaction occurs in a series of steps, the H for the whole reaction is equal to the sum of each step’s H. What property of enthalpy allows Hess’s Law to work? Because enthalpy is a state function (pathway-independent) it doesn’t matter how you get from reactants to products; you can use one step or ten, energy changes will be the same. N2 + 2O2 --> 2NO2 H = 68 kJ N2 + O2 --> 2NO H = 180 kJ 2NO + O2 --> 2NO2 H = -112 kJ N2 + 2O2 --> 2NO2 H = 68 kJ Guidelines for using Hess’s Law 1. If the reaction must be reversed, change the sine of H. (Enthalpy reverses with the direction of the reaction.) 2. If equations must be multiplied or divided by a fudge factor, do the same to H. (Enthlapy is quantitative.) p. 181-3

Examples: Hess’s law Carbon and CO can be combusted in O2 to form CO2. C(s) + O2(g) --> CO2(g) H = -393.5 kJ/mol C CO(g) + 1/2O2(g) --> CO2(g) H = -283.0 kJ/mol CO Use Hess’s Law to calculate H of: C(s) + 1/2O2(g) --> CO(g) C(s) + O2(g) --> CO2(g) H = -393.5 kJ/mol C CO2(g) --> CO(g) + 1/2O2(g) H = +283.0 kJ/mol CO (reversed) C(s) + 1/2O2(g) --> CO(g) H = - 110.5 kJ/mol C Carbon exists in two natural forms: graphite and diamond. Each form can be combusted: C(graphite) + O2(g) --> CO2(g) H = -393.5 kJ/mol C(diamond) + O2(g) --> CO2(g) H = -395.4 kJ/mol Use Hess’s Law to calculate H of the conversion of graphite to diamond: C(graphite) + O2(g) --> CO2(g) H = -393.5 kJ/mol CO2(g) --> C(diamond) + O2(g) H = +395.4 kJ/mol (reversed) C(graphite) --> C(diamond) H = +1.9 kJ/mol p. 181-3

Example: one more Hess’s Calculate H for this reaction: 2C(s) + H2(g) --> C2H2 (gas) Given the following thermochemical reactions: C2H2(g) + 5/2 O2(g) --> 2CO2 + H2O(l) H = -1229.6 kJ/mol C(s) + O2(g) --> CO2(g) H = -393.5 kJ/mol H2(g) + 1/2O2(g) --> H2O(l) H = -285.8 kJ/mol 2CO2 + H2O(l) --> C2H2(g) + 5/2 O2(g) H = +1229.6 kJ/mol (reversed) (2)(C(s) + O2(g) --> CO2(g)) H = (2)(-393.5 kJ/mol) (x2) H2(g) + 1/2O2(g) --> H2O(l) H = -285.8 kJ/mol 2C(s) + H2(g) --> C2H2 (gas) H = +156.8 kJ/mol p. 181-3

Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

Calorimetry is the lab study of heat flow. Constant pressure calorimetry for reactions that don’t change volume (do work). The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

Calorimetry: Is heat lost or gained? The amount of heat lost or gained by any material can be determined using calorimetry and specific heat values for that material. Calorimetry determination of enthalphy by experimental measurement of heat flow Specific heat the amount of energy required to raise the temp of 1 g of material by 1C J/C-g Heat capacity the amount of energy required to raise the temp of an object by 1C J/C Molar heat capacity the amount of energy required to raise the temp of 1 mole by 1C J/C-mol H = q = (specific heat)(grams)(T) = J How much heat is needed to warm 250 g of water from 22°C to 98°C? H = (4.18 J/g-K)(250 g)(98 - 22C) = 7.9x104 Sh of water = 4.184 J/g-K What is the molar heat capacity of water? Molar heat capacity = (4.18 J/g-K)(18.0g/1 mole) = 75.2 J/mol-K p. 175-6

Examples: calorimetry What is the specific heat of a substance that takes 547 J to raise 43 g from 25C to 53C? q = (sh)(mass)(T) --> sh = q/(mass)(T) sh = +547 J/(43 g)(53 - 25C) = 0.454 J/g-C How much heat is given off when 75.0 g of water is cooled from 88 to 25C? q = (sh)(mass)(T) q = (4.184 J/g-C)(75.0 g)(25 - 88C) = -19.7 kJ Some solar homes use large beds of rock to absorb heat. Calculate the amount of heat absorbed by 50.0 kg of rocks if their temperature increases by 12°C. Assume that rock has a specific heat of 0.82 J/g-C. q = (0.82 J/g-C)(5x104 g)(12°C) = 4.92x105 J

Constant-pressure calorimetry Remember that at constant pressure H = qp. So the most common form of calorimetry is constant pressure. The calorimeter serves to prevent loss of heat from the system to the surroundings. Thus, all heat generated by the system is absorbed by the system. System = water + calorimeter exothermic  heat produced causes the temperature within the calorimeter to increase endothermic  heat absorbed by the system causes the temperature within the calorimeter to decrease qrxn = - ((mass sol’n)(sh sol’n)(T) + (mass cal)(sh cal)(T)) T for both is change in water temp. p. 177-8

Measuring enthalpy with calorimeters Hsol’n = - Hrxn Why? An exothermic reaction produces heat (-), and that heat is absorbed (+) by the solution contained within the calorimeter. In a calorimeter, 50 mL of 0.65 M HCl is reacted with 50 mL of 1.00 M NaOH (excess base). Temperature of the solution increases from 25.0 to 38.4C. Calculate Hrxn. qrxn = -qsoln = -((4.184 J/g-C)(100 g)(13.4C) = -5.6 kJ Molar enthalpy of rxn? (0.050 L)(0.65 mol/L) = 0.033 mol HCl or H2O -5.6 kJ/0.033 mole = -170 kJ/mol p. 178-9

Example: calorimetry When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant pressure calorimeter, the temperature increases from 22.3C to 23.11C. The reaction is as follows: AgNO3 + HCl --> AgCl (insoluble) + HNO3 Calculate H of this reaction, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g-C. AgNO3  (0.050 L)(0.100 M) = 5x10-3 moles HCl --> (0.050 L)(0.100 M) = 5x10-3 moles T = 0.81C qrxn = - (4.18 J/g-C)(100.0 g)(0.81C) = - 3.38x102 J H= - 3.38x102 J/5x10-3 moles = - 6.77x104 J/moles p. 178-9

Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

Enthalpy of formation Enthalpy of formation: enthalpy to form 1 mole Enthalpies of formation can be used to calculate enthalpy of reaction. The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

Enthalpy of formation (ΔHf) There are many forms (or expressions) of enthalpy including Hvaporization, Hfusion, Hcombination & Hformation. The most useful is standard enthalpy of formation (H°f): the enthalpy change that accompanies the formation of 1 mole of a compound from elements in their standard states. The form of the element at 1 atm, 298 K/25°C gases - 1 atm solutions - 1 M standard states? Na? Hg? N2? solid liquid gas The enthalpy of formation of any substance in its most stable form is always zero: Carbon as graphite; oxygen as O2 gas; copper as an elemental solid (metal). Write the reactions for H°f for: CH3CH2OH FePO4 2C + 1/2O2 + 3H2  CH3CH2OH Fe + P + 2O2  FePO4 see table 5.3, p. 177 & appendix C p. 183-5

Example: ΔHf For which reaction (at 25°C) would enthalpy change represent standard enthalpy of formation? If not, how could you change the reaction conditions? a. 2Na(s) + 1/2O2(g)  Na2O(s) 2K(l) + Cl2(g)  2KCl(s) C6H12O6(s)  6C(diamond) + 6H2(g) + 3O2(g) Yes - all in standard states & 1 mole product is formed No - K should be (s) & 2 moles product are formed No - this is a decomposition, not a formation; reverse it p. 183-5

ΔHfs can be summed to calculate ΔHrxn So this concept combines Hess’s Law with Hf . If we know the Hf of all reactants & products, we can calculate Hrxn: H°rxn = nH°f(products) - nH°f(reactants) Propane (C3H8) is combusted to form CO2 and H2O under standard cond. C3H8 + 5O2  3CO2 + 4H2O (gas) (gas) (gas) (liquid) H°rxn = -Hf [C3H8(g)] + 3Hf [CO2(g)] + 4Hf [H2O(l)] = -(-103.85 kJ) + 3(-393.5 kJ) + 4(-285.8 kJ) = -2220 kJ (moles)(kJ/mole) Expressed via Hess’s Law: C3H8 --> 3C + 4H2 H1 = -Hf [C3H8(g)] reversed 3C + 3O2 --> 3CO2 H2 = 3Hf [CO2(g)] stoichiometry 4H2 + 2O2 --> 4H2O H3 = 4Hf [H2O(l)] state ! C3H8 + 5O2 --> 3CO2 + 4H2O H°rxn = H1 + H2 + H3 p. 185-7

Examples: ΔHf Calculate the H for combustion of 1 mole benzene (C6H6). C6H6 + 15/2O2  6CO2 + 3H2O (liquid) (gas) (gas) (liquid) Hrxn = [6Hf(CO2) + 3Hf(H2O)] - [Hf(C6H6) + 15/2 Hf(O2)]  products  reactants = [6(-393.5 kJ) + 3(-285.8 kJ)] - [(49.0 kJ) + 15/2(0 kJ)] = (-2361 - 857.4 - 49.0) kJ = - 3267 kJ Use enthalpies of formation to calculate the Hrxn for: CaCO3(s) --> CaO(s) + CO2(g) Hrxn = [Hf(CaO) + Hf(CO2)] - [Hf(CaCO3)]  products  reactants = [(-635.5 kJ) + (-393.5 kJ)] - [(-1207.1 kJ)] = 178.1 kJ p. 185-7

Enthalpies of combustion of fuels Carbohydrates? Glucose = C6H12O6 C6H12O6 + 6O2  6CO2 + 6H2O Hrxn = - 2803 kJ Average fuel value for carbohydrates (& proteins) is ~17 kJ/g. Fats? Triacylglycerol (body fat) = C57H110O6 2C57H110O6 + 163O2 --> 114CO2 + 110H2O Hrxn = - 75,520 kJ Average fuel value for carbohydrates (& proteins) is ~38 kJ/g. What about industrial fuels? fuel value (kJ/g) Coal 31 - 32 Oil 45 Natural gas 49 Gasoline 48 H2 142 p. 188-9

Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

Lecture 8: Terms to know Kinetic energy Potential energy Electrostatic force Energy Heat Work Thermal energy Heat flow System vs. surroundings Closed vs. open system First Law of Thermodynamics Internal energy (ΔE) Transfer vs. transformation Final vs. initial Enthalpy (ΔH) Exothermic vs. endothermic Thermochemical equation Hess’s Law Calorimetry Specific heat Heat capacity Molar heat capacity Enthalpy of formation (ΔHf) Standard states Enthalpy of combustion PLEASE note that the formula calculates a weighted average, so there’s not need to add and then divide the sum by the number of isotopes in the problem. This is an incredibly common student error so don’t get caught!