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THERMOCHEMISTRY: ENERGY FLOW AND CHEMICAL FLOWIn general…All matter contains energy, so whenever matter undergoes a change, the quantity of energy that the matter contains also changesWhen a burning candle melts a piece of iceHigher energy reactants, wax and oxygen, form lower energy products, carbon dioxide and waterThe difference is released as heat and light Some of the heat is absorbed when the lower energy ice becomes higher energy waterIn a thunderstormLower energy N2 and O2 absorb energy from lightning and form higher energy NO, and the higher energy water vapor releases energy as it condenses to lower energy water that falls as rainEveryday examplesFuels such as oil and wood release energy to heat our housesFertilizers help convert solar energy into foodMetal wires increase the flow of electrical energyPolymer fibers in winter clothing limit the flow of thermal energy away from our bodiesThermodynamics – the study of energy and its transformations (will be addressed in chapter 20 as well for students taking CHEM 102!)Thermochemistry – the branch of thermodynamics that deals with heat in chemical and physical changeForms of Energy and Their InterconversionsWhen energy is transferred from one object to another, it appears as work and/or heatDefining the System and SurroundingsSystem – part of the universe we are focusing onSurroundings – everything that is not the systemIn the above picture, the contents of the flask is the system and everything thing else (even the flask) is the surroundingsEnergy Transfer to and From a SystemInternal Energy, E, is the sum of the potential and kinetic energy found in each particle in a systemWhen the reactants in a chemical system change to products, the system’s internal energy changesThis is represented by ?E?E = Efinal - Einitial = Eproducts - EreactantsA change in the energy of the system must be accompanied by an equal and opposite change in the energy of the surroundingsA system can change its internal energy in one of two waysBy releasing some energy in a transfer to the surroundings (diagram A below)Efinal < Einitial so that ?E < 0By absorbing some energy in a transfer from the surroundings (diagram B below)Efinal > Einitial so that ?E > 0?E is a transfer of energy from system to surroundings, or vice versa…Heat and Work: Two Forms of Energy TransferEnergy transferred from system to surroundings or vice versa appears in two forms…Heat. Heat or thermal energy (symbolized as q) is the energy transferred as a result of a difference in temperature between the system and the surroundings. Movement from warmer system/surroundings to cooler system/surroundings.Work. All other forms of energy transfer involve some type of work (w), the energy transferred when an object is moved by a force.The total change in a system’s internal energy, E, is the sum of the energy transferred as heat and/or work.?E = q + wThe values of q and w can have either positive or negative signsWe define the sign (+ or -) of the energy change from the system’s perspectiveEnergy transferred into the system is positive because the system ends up with more energyEnergy transferred out from the system is negative because the system ends up with less energyEnergy Transferred as Heat OnlyHeat flowing out of a system… (example A)q is negative, so ?E is negativeHeat flowing into a system… (example B)q is positive, so ?E is positiveThermodynamics in the kitchen…The air is your refrigerator (surroundings) has a lower temperature than a newly added piece of food (system)Food releases energy to the surroundings, so q < 0 (system loses energy)The hot air in an oven (surroundings) has a higher temperature than a newly added piece of food (system)Food is absorbing energy from the surroundings, so q > 0 (system gained energy)Just to clarify… your favorite beverage in a cooler full of ice makes the ice warmer (the ice does not make your beverage cooler)The beverage is the system and the cooler is the surroundings. The beverage is releasing heat to the surroundings. So… q < 0 (system lost energy)Energy Transferred as Work OnlyWork done by a system… (in the above figure)The piston is pushed out by the system, so w is negative and ?E is negativeWork done on a system… (no picture) so… imagine the piston being pushed in by the surroundingsw is positive and ?E is positiveLaw of Conservation of EnergyEnergy changes form (from chemical to kinetic to potential for example) but does not simply appear or disappear – energy cannot be created or destroyed.Law of conservation of energy (first law of thermodynamics) – the total energy of the universe is constant?Euniverse = ?Esystem + ?EsurroundingsUnits of EnergyThe SI unit of energy is the joule (J)The calorie is an older term originally defined as the quantity of energy required to raise the temperature of 1 gram of water 1 oCNow defined in terms of the joule…1 calorie = 4.184 J1 J = 0.2390 calories1 kJ = 1000 J = 0.2390 kcal = 239.0 caloriesBritish thermal unit (Btu) is a unit used for energy output of appliancesQuantity of energy required to raise the temperature of 1 lb. of water 1oF1 Btu = 1055 JSample Problem 6.1 p. 234Determining the Change in Internal Energyq = -325 J (system releases 325 J to surroundings)w = -451 J (system does 451 J on the pistons)?E = q + w?E = -325 J + (- 451 J)?E = - 776 J- 776 J1 kJ= - 0.776 kJ1000 J- 0.776 kJ1 kcal= - 0.185 kcal4.184 kJFollow-Up Problem 6.1 p. 23426.0 kcal4.184 kJ= 108.784 kJ1 kcal15.0 Btu1055 J1 kJ= 15.825 kJ1 Btu1000 Jq = -108.784 kJ (absorbed by surroundings)w = 15.825 kJ (work done on the system)?E = q + w?E = -108.784 kJ + 15.825 kJ?E = - 92.959 kJ?E = - 93.0 kJState Functions & the Path Independence of the Energy ChangeThe internal energy (E) of a system is called a state functionA property dependent only on the current state of the system, not on the path the system takes to reach that state.?E is also a state function because it does not depend on how the change takes place, but only on the difference between the final and initial states.q and w are not state functions because their values do depend on the path the system takesIn the above example, q and w for the two paths are different, ?E is the samePressure (P), volume (V), and temperature are other state functionsPath independence means that changes in state functions - ?E, ?P, ?V, and ?T – depend only on the initial and final states.Enthalpy: Chemical Change at Constant PressureTwo most important types of chemical work…Electrical workWork done by moving charged particlesPressure-volume workThe mechanical work done when the volume of a system changes in the presence of an external pressure019050The quantity of PV work equals P times the change in V… w = - P?V (it is negative because the gas is doing work on the surroundingsAt constant pressure, a thermodynamic variable called enthalpy (H) eliminates the need to measure PV work. The enthalpy of a system is defined as the internal energy plus the product of pressure and volume…H = E + PVThe change in enthalpy (?H) is the change in internal energy plus the product of the pressure, which is constant, and the change in volume?H = ?E + P?VMost important point… the change in enthalpy (?H) equals the heated absorbed or released at constant pressure.?H is more relevant than ?E and easier to obtainComparing ?E and ?HKnowing the enthalpy change of a system tells us a lot about its energy change as wellMany reactions involve little (if any) PV workMost (or all) the energy change occurs as a transfer of heatFor most reactions, ?H equals or is very close to ?EExothermic and Endothermic ProcessesBecause H is a combination of the three state functions E, P, and V, it is also a state function?H = Hfinal – Hinitial = Hproducts - HreactantsExothermic processAn exothermic process releases heat and results in a decrease in the enthalpy of the systemHproducts < Hreactants so… ?H < OHeat is found on the product side of the equationCH4(g) + 2O2(g) CO2(g) + 2H2O(g) + heatDiagram A below is showing an exothermic reactionEndothermic processAn endothermic process absorbs heat and results in an increase in the enthalpy of the systemHproducts > Hreactants so… ?H > OHeat is found on the reactant side of the equationHeat + H2O(s) H2O(l)Sample Problem 6.2 p. 237Drawing Enthalpy Diagrams and Determining the Sign of ?HFollow-Up Problem 6.2 p. 237487683204212243074320294C3H5(NO3)3(l) (reactant)-767715372110Enthalpy00Enthalpy?H = - 5.72x103 kJ Exothermic3CO2(g) + 5/2 H2O(g) + ? O2(g) + 3/2 N2(g) (products)Calorimetry: Measuring the Heat of a Chemical or Physical ChangeTemperature change depends on the object…Every object has its own heat capacity, the quantity of heat required to change its temperature 1 KSpecific heat capacity (c), the quantity of heat required to change the temperature of 1 gram of an object by 1 KTo calculate the heat absorbed or released…q = c x mass x ?Tq = heat absorbed or releasedc = specific heat capacity (J/g K)mass in grams?T in KMolar heat capacity – the quantity of heat required to change the temperature of 1 mole of a substance 1 KThe combination of high specific heat of water and the amount of water that covers the Earth play an important role maintaining an environment that is favorable to life as we know it…025400If the Earth had no oceans, it would take one-sixth as much energy for the Sun to heat the rocky surface, and the surface would five off its heat six times as much heat after sundown. Water has a high specific heat which means it takes a lot of energy input to change the temperature and it takes a considerable loss for the temperature to go down. People that live in Chicago experience this with Lake Michigan as well. (The lake stays cold until well into the summer and stays warm well into the fall and early winter)Sample Problem 6.3 p. 239Finding the Quantity of Heat from a Temperature Change?T = Tfinal - Tinitial?T = 300. oC – 25oC = 275 oC = 275 Kq = c x mass x ?Tq = 0.387 J/g K x 125 g x 275 Kq = 1.33x104 JFollow-Up Problem 6.3 p. 239?T = Tfinal - Tinitial?T = 25.0 oC – 37.0oC = 12.0 oC = -12.0 K5.50 L1000 mL1.11 g= 6050 g1 L1 mLq = c x mass x ?Tq = 2.42 J/g K x 6050 g x -12.0 Kq = - 175,692 J = -176,000 J-176,000 J1 kJ= -176 kJ1000 JThe Two Major Types of CalorimetryCalorimeter – a device used to measure the heat released or absorbed by a physical or chemical processTwo types…One designed to measure the heat at constant pressureOne designed to measure the heat at constant volume304803987800Constant Pressure CalorimetryOften measured in a coffee-cup calorimeterCan be used to determine the specific heat capacity of a solid (as long as it does not react with or dissolve in the water)The solid (system) is weighed, heatto some known temperature, and added to a known mass and temperature of water (surroundings) in the calorimeter.After stirring, the final water temperature is measured, which is also the temperature of the solid. Assuming no heat escapes the calorimeter, the heat released by the system (-qsystem or –qsolid) is equal in magnitude but opposite in sign to the heat absorbed by the surroundings (+qsurroundings or +qH2O)So… -qsolid = qH2OSubstituting for q on both sides of the equation…-(csolid x masssolid x ?Tsolid) = cH2O x massH2O x ?TH2OSolving for csolidcsolid = - (cH2O x massH2O x ?TH2O / masssolid x ?Tsolid)Sample Problem 6.4 p. 240Determining the Specific Heat Capacity of a SolidFinding ?Tsolid and ?Twater?Twater = Tfinal - Tinitial?Twater = 28.49oC – 25.10oC = 3.39oC = 3.39 K?Tsolid = Tfinal - Tinitial?Tsolid = 28.49oC – 100.00oC = - 71.51oC = - 71.51KSolving for csolid-(csolid x masssolid x ?Tsolid) = cH2O x massH2O x ?TH2Ocsolid = - (cH2O x massH2O x ?TH2O / masssolid x ?Tsolid)csolid = - (4.184 J/g K x 50.00 g x 3.39 K / 22.05 g x – 71.51K)csolid = 0.450 J/g KFollow-Up Problem 6.4 p. 240Finding ?Tsolid and ?Twater?Twater = Tfinal - Tinitial?Twater = 27.25oC – 25.55oC = 1.70oC = 1.70 K?Tsolid = Tfinal - Tinitial?Tsolid = 27.25oC – 65.00oC = - 37.75oC = - 37.75KSolving for csolid-(csolid x masssolid x ?Tsolid) = cH2O x massH2O x ?TH2Ocsolid = - (cH2O x massH2O x ?TH2O / masssolid x ?Tsolid)csolid = - (4.184 J/g K x 25.00 g x 1.70 K / 12.18 g x – 37.75K)csolid = 0.387 J/g K [unknown is Copper, Cu]Sample Problem 6.5 p. 240Determining the Enthalpy(a)Finding masssolution and ?Tsolutionmasssolution = (25.0 mL + 75.0 mL) x 1.00 g/mL = 75.0 g?Tsolution = 27.21oC – 25.00oC = 2.21oC = 2.21Kqsolution = csolution x masssolution x ?Tsolutionqsolution = 4.184 J/g K x 75.0 g x 2.21Kqsolution = 693 J(b)HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)25.0 mL HCl1 L0.500 mol HCl= 0.0125 mol HCl1000 mL1L50.0 ml NaOH1L0.500 mol NaOH= 0.0250 mol NaOH1000 mL1L**HCl is the limiting reactant, so 0.0125 mol HCl is producedHeat absorbed by the solution was released by the reactionqsoln = -qrxn = 693 J so qrxn = -693 J-693 J1 kJ= -55.4 kJ / mol H2O0.0125 mol H2O1000 JFollow-Up Problem 6.5 p. 241Step 1Calculate moles of reactants50.0 mL1L0.500 mol Ba(OH)2= 0.025 mol Ba(OH)21000 mL1L50.0 mL1L0.500 mol HCl = 0.025 mol HCl1000 mL1LStep 2Determine the limiting reactant…0.025 mol Ba(OH)22 mol H2O= 0.500 mol H2O1 mol Ba(OH)20.025 mol HCl2 mol H2O= 0.025 mol H2O2 mol HCl**HCl is the limiting reactant (produces 0.025 mol H2O)- 1.386 kJ / 0.0250 mol H2O = - 55.4 kJ/molConstant-Volume Calorimetry0-1905Constant-volume Calorimetry is often carried out in a bomb calorimeter (a device commonly used to measure the heat of combustion reactions, such as for fuels and foods)With the much more precise bomb calorimeter, the heat capacity of the entire calorimeter is known (or can be determined)Sample Problem 6.6 p. 242Calculating the Heat of a Combustion ReactionStep 1Calculate ?T?T = ?Tfinal - ?Tinitial = 26.799oC – 21.862oC = 4.937oC = 4.937KStep 2Calculate the heat absorbed by the calorimeterWhen the dessert (system) burns, the heat released is absorbed by the calorimeter-qsystem = qcalorimeterqcalorimeter = heat capacity x ?Tqcalorimeter = 8.151 kJ/K x 4.937 Kqcalorimeter = 40.24 kJStep 3Compare answer to 10 Calories10 Calories1kcal4.184 kJ=41.84 kJ1 Calorie1 kcal40.24 kJ < 41.84 kJ The claim is correct…Follow-Up Problem 6.6 p. 242Step1Calculate the amount of heat released (qsample)0.8650 g C1 mol C393.5 kJ= 28.34 kJ12.01 g C1 mol CStep 2Calculate heat capacity of bomb calorimeter-qsample = qcalorimeterqcalorimeter = heat capacity x ?T28.34 kJ = heat capacity x 2.613 KHeat capacity = 10.85 kJ/KStoichiometry of Thermochemical EquationsThermochemical equation – a balanced equation that includes the enthalpy change of the reaction (?H)?H refers only to the amounts (mol) of substances and their states of matter in that equationThe sign of ?H depends on whether the reaction is exothermic (-) or endothermic (+)A forward reaction has the opposite sign of the reverse reaction2H2O(l) 2H2(g) + O2(g)?H = 572 kJBecause ?H is positive, this reaction is endothermic2H2(g) + O2(g) 2H2O(l)?H = -572 kJBecause ?H is negative, this reaction is exothermicThe magnitude of ?H is proportional to the amount of substance H2(g) + ? O2(g) H2O(l)?H = -286 kJNotice that this is ? the above reactionTwo key points to understand about thermochemical equations…When necessary, we use fractional coefficients to balance the equation, because we are specifying the magnitude of ?H for a particular amount, often 1 mol of substance? S8(s) + O2(g) SO2(g)?H = -296.8 kJFor a particular reaction, a certain amount of substance is thermochemically equivalent to a certain quantity of energy. For example…296.8 kJ is thermochemically equivalent to ? mol of S8286 kJ is thermochemically equivalent to ? mol of O2286 kJ is thermochemically equivalent to 1 mol H2OSample Problem 6.7 p. 243Using the Enthalpy Change of a Reaction to Find Amounts of Substance1.000x103 kJ2 mol Al26.98 Al= 32.20 g Al1676 kJ1 mol AlFollow-Up Problem 6.7 p. 243C2H4(g) + H2(g) C2H6(g)?H = -137 kJ15 kg C2H61000 g1 mol C2H6137 kJ= 6.83x104 kJ1 kg30.07 g C2H61 mol C2H6Hess’s Law: Finding ?H of Any ReactionIn some cases, a reaction is difficult, even impossible, to carry out…May be part of a complex biochemical processMay take place under extreme conditionsMay need a change of conditions to take placeEven if we can’t run a reaction in the lab, we can still find its enthalpy change…We can find the ?H for any reaction for which we can write an equationHess’s law – the enthalpy change of an overall process is the sum of the enthalpy changes of the individual steps?Hoverall = ?H1 + ?H2 + …. + ?HnWe apply Hess’s Law by…Imagining that an overall reaction occurs through a series of individual reaction steps, whether or not it actually does. Adding the steps must give the overall reaction.Choosing individual reaction steps that each have a known ?HAdding the known ?H values for the steps to get the unknown ?H of the overall reaction. We can also find the unknown ?H of any step by subtraction (if we know the ?H values for the overall reaction and all the other steps.Example of the use of Hess’s law – the oxidation of sulfur to sulfur trioxideEquation 1: S(s) + O2(g) SO2(g)?H1 = -296.8 kJEquation 2: 2SO2(g) + O2(g) 2SO3(g)?H2 = -198.4 kJEquation 3: S(s) + 3/2 O2(g) SO3(g)?H3 = ????Need to manipulate equations 1 and/or 2 so that they add up to equation 3We need to have the same coefficients for SO2 so that they will cancel (this will mean that we have to divide equation 2 by 2)Equation 2: SO2(g) + ? O2(g) SO3(g)?H2 = -99.2 kJAdd equations 1 and 2 together…Equation 1: S(s) + O2(g) SO2(g)?H1 = -296.8 kJEquation 2: SO2(g) + ? O2(g) SO3(g)?H2 = -99.2 kJ441896560960245618060960Equation 3: S(s) + O2(g) + SO2(g) + ? O2(g) SO2(g) + SO3(g)?H1 = -396.0 kJDelete SO2 because it is common to both sidesAdd O2 together on reaction sideFinal answer…S(s) + 3/2 O2(g) SO3(g)?H1 = -396.0 kJSummarize calculating an unknown ?HIdentify the target equation, the step whose ?H is unknown, and note the amount (mol) of each reactant and productManipulate each equation with known ?H values so that the target amount (mol) of each substance is on the correct side of the equation. Remember to:Change the sign of ?H when you reverse an equationMultiply amount (mol) and ?H by the same factorSample Problem 6.8Using Hess’s Law to Calculate an Unknown ?HEquation A:CO(g) + ? O(g) CO2(g)?H = -283.0 kJEquation B:N2(g) + O2(g) 2NO(g)?H = 180.6 kJEquation C:CO(g) + NO(g) CO2(g) + ? N2(g)?H = ?Reverse equation B so that N2 is on the correct side of equation and divide by 2 so that you have the correct number of mol of N2 , O2 and NOEquation B:NO(g) 1/2 N2(g) + ? O2(g) ?H = -90.3 kJAdd equations A & B together…Equation A:CO(g) + ? O(g) CO2(g)?H = -283.0 kJEquation B:NO(g) 1/2 N2(g) + ? O2(g) ?H = -90.3 kJ196291236576Equation C: CO(g) + ? O(g) + NO(g) 32918404013200CO2(g) + 1/2 N2(g) + ? O2(g) ?H = -373.3 kJEquation C: CO(g) + NO(g) CO2(g) + 1/2 N2(g) ?H = -373.3 kJFollow-Up Problem 6.8 p. 245Equation 1: N2O5(s) 2NO(g) + 3/2 O2(g)?H = 223.7 kJEquation 2: NO(g) + ? O2(g) NO2 (g)?H = -57.1 kJEquation 3: 2NO2(g) + ? O2(g) N2O5(s)?H = ???Reverse equation 1…Equation 1: 2NO(g) + 3/2 O2(g) N2O5(s) ?H = -223.7 kJMultiply equation 2 x 2 and reverse…Equation 2: 2NO2(g) 2NO (g) + O2(g)?H = 114.2 kJ201777636830106680036830Equation 3: 2NO(g) + 3/2 O2(g) + 2NO2(g) 2316480281941322832281942NO (g) + O2(g) + N2O5(s)?H = -109.5 kJEquation 3: 2NO2(g) + ? O2(g) N2O5(s)?H = -109.5 kJStandard Enthalpies of Reaction - ?H orxnWe can use Hess’s Law to determine the ?H values of an enormous number of reactions…?H varies somewhat with conditions… so chemist have established a set of specific conditions called standard statesFor a gas, it is 1 atm and ideal behaviorFor a substance that aqueous solution, it is 1 M concentrationFor a pure substance (element or compound), it is usually the most stable form of the substance at 1 atm and the temperature of interest (usually 25oC or 298K)The standard state symbol (shown as a degree sign) indicates that the variable has been measured with all the substances in their standard statesStandard enthalpy of formation, ?H of is the enthalpy change if measured at the standard stateAlso called the standard heat of formationC (graphite) + 2H2(g) CH4(g)?H of = -74.9 kJFractional coefficients are often used with reactants to obtain 1 mol of productNa(s) + ? Cl2(g) NaCl(s)?H of = -411.1 kJTable 6.3 p. 246 shows selected standard enthalpy of formations at 298KFor an element in its standard state, ?H of = 0The standard state for molecular elements, such as the halogens, is the molecular form, not separate atoms. An example would be… Cl2 = ?H of = 0Some elements exist in different forms, but only one is the standard state…Carbon ?H of = 0, not diamond ?H of = 1.9 kj/molOxygen ?H of = 0, not ozone ?H of = 143 kJ/mol Sulfur (S8) ?H of = 0, not S ?H of = 0.3 kJ/molMost compounds have a negative ?H ofMost compounds have exothermic formation reactions (under standard conditions, heat is released when most compounds form from their elements) 025400Sample Problem 6.9 p. 246Writing Formation Equations(a)Ag(s) + ? Cl2(g) AgCl(s)?H of = -127.0 kJ(b)Ca(s) + C(graphite) + 3/2 O2 CaCO3(s)?H of = -1206.9 kJ(c)? H2(g) + C(graphite) + ? O2(g) HCN(g) ?H of = 135.0 kJFollow-Up Problem 6.9 p. 247(a)C(graphite) + H2(g) + ? O2(g) CH3OH(l) ?H of = -238.6 kJ(b)Ca(s) + ? O2(g) CaO(s)?H of = -635.1 kJ(c)C(graphite) + 1/4S8(rhombic) CS2?H of = 87.9 kJDetermining ?H orxn from ?H of Values for Reactants & ProductsWe can use ?H of values to determine ?H orxn for any reactionStep 1Each reactant decomposes to its elements. This is the reverse of the formation reaction for the reactant, so the standard enthalpy change is -?H ofStep 2Each product forms from its elements. This step is the formation reaction for the product, so the standard enthalpy change is ?H ofText example… p. 247TiCl4(l) + 2H2O(g) TiO2(s) + 4HCl(g)TiCl4(l) Ti(s) + 2Cl2(g)-?H of [TiCl4(l)]2H2O(g) 2H2(g) + O2(g)-2?H of [H2O(g)]Ti(s) + O2(g) TiO2(s)?H of [TiO2(s)]2H2(g) + 2Cl2(g) 4HCl(g)4?H of [HCl(g)]426110433528337108833528259689633528196291233528TiCl4(l) + 2H2O(g) + Ti(s) + O2(g) + 2H2(g) + 2Cl2(g) 66446418796157276818796240792018796-3048018796Ti(s) + 2Cl2(g) + 2H2(g) + O2(g) + TiO2(s) + 4HCl(g)TiCl4(l) + 2H2O(g) TiO2(s) + 4HCl(g)It is important to realize that when titanium (IV) chloride and water react, the reactions don’t actually decompose to their elements, which then recombine to form the products. ?H orxn is the difference between two state functions, H oproducts minus H oreactants, so it doesn’t matter how the reaction actually occurs. We simply need to add the individual enthalpy changes to find ?H orxn?H orxn = ?H of [TiO2(s)] + 4?H of [HCl(g)] + {-?H of [TiCl4(l)] + {-2?H of [H2O(g)]}OR?H orxn = {?H of [TiO2(s)] + 4?H of [HCl(g)]} - {?H of [TiCl4(l)] + 2?H of [H2O(g)]}The standard enthalpy of reaction (?H orxn) is the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants?H orxn = ∑m?H of(products) - ∑n?H of(reactants)Sample Problem 6.10 p. 248Calculating ?H orxn from ?H of Values4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)?H orxn = ∑m?H of(products) - ∑n?H of(reactants)?H orxn = {4?H of [NO(g)] + 6?H of [H2O(g)]} –{4?H of [NH3(g)] + 5?H of [O2(g)]}?H orxn = {(4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol)} –{(4 mol)(-45.9 kJ/mol) + (6mol)(0 kJ/mol)?H orxn = [361 kJ + (-1451 kJ)] – [(-184 kJ) + 0 kJ]?H orxn = - 906 kJFollow-Up Problem 6.10 p. 248CH3OH(l) + 3/2O2(g) CO2(g) + 2H2O(g)?H orxn = -683.5 kJ?H orxn = ∑m?H of(products) - ∑n?H of(reactants)-683.5 kJ = {(1 mol)(-393.5 kJ) + (2 mol)(-241.8) – (3/2 mol)(0.0 kJ) + ?H of [CH3OH(l)]?H of [CH3OH(l)] = 638.5 kJ + (-393.5 kJ) + (-483.6)?H of [CH3OH(l)] = -238.6 kJ ................
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