# Chapter 6 Thermochemistry

Thermochemistry Copyright 2011 Pearson Education, Inc. Chapter Outline Overview of Energy in Chemical Reaction First Law of Thermodynamics Heat, Specific heat, Calorimetry Enthalpy of Reaction Standard Enthalpy of Formation Hesss law Tro: Chemistry: A Molecular Approach, 2/e 2 Copyright 2011 Pearson Education, Inc.

Chemical Hand Warmers and much more Most hand warmers uses the heat from rxn: 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) The temperature change in your hand depends on the size of the hand warmer the size of your glove, etc. mainly, the amount of heat from the reaction Most chemical rxn involves energy change Chemical energy provides nearly 70% electricity in the US. Nuclear 11%, Hydroelectric 17%. All batteries (car, portable electronic device) Tro: Chemistry: A Molecular Approach, 2/e 3 Copyright 2011 Pearson Education, Inc. About Energy Even though chemistry is the study of matter, energy affects matter

Energy is anything that has the capacity to do work. Example: Kinetic energy in moving water; Potential energy: meteorite near the Earth Chemical energy (a form of potential energy) in gasoline or battery Energy exchange through doing work or heat exchange. Tro: Chemistry: A Molecular Approach, 2/e 4 Copyright 2011 Pearson Education, Inc. Units of Energy Joule (J): the amount of energy for 1 Newton force for a distance of 1 meter 1 J = 1 Nm = 1 kgm2/s2 calorie (cal): the amount of energy needed to raise the temperature of one gram of water 1C kcal = energy needed to raise 1000 g of water 1C food Calories = kcals

Energy Conversion Factors 1 calorie (cal) = 4.184 joules (J) (exact) 1 Calorie (Cal) = 1000 cal = 1 kcal = 4184 J 1 kilowatt-hour (kWh) = 3.60 x 106 J Tro: Chemistry: A Molecular Approach, 2/e 5 Copyright 2011 Pearson Education, Inc. Classification of Energy Kinetic energy is energy of motion or energy that is being transferred Thermal energy is the energy associated with temperature thermal energy is a

form of kinetic energy Tro: Chemistry: A Molecular Approach, 2/e 6 Copyright 2011 Pearson Education, Inc. Some Forms of Energy Electrical kinetic energy associated with the flow of electrical charge Heat or thermal energy kinetic energy associated with molecular motion Light or radiant energy kinetic energy associated with energy transitions in an atom Nuclear potential energy in the nucleus of atoms Chemical potential energy due to the structure/bonding of the

atoms/molecules Tro: Chemistry: A Molecular Approach, 2/e 7 Copyright 2011 Pearson Education, Inc. Energy Change: Work vs. Heat Work = force acting over a distance Energy = Work = Force x Distance Example: Compressor convert electric energy to take away heat from room. Heat: Energy flow caused by a difference in temperature, from high temp to low temp. Heat transfer requires physical contact (molecular collision): Fastmoving particles pass the kinetic energy to slow moving particles Tro: Chemistry: A Molecular Approach, 2/e

8 Copyright 2011 Pearson Education, Inc. Energy Exchange: System vs. Surroundings System: the material or process within which we are studying the energy changes within. Example: Chemical Rxn. Surroundings: everything else with which the system can exchange energy with. Example: Solvent where rxn occurs. Left: System gives energy to the Surroundings. Right: System takes energy from the Surroundings. Surroundings Surroundings System Tro: Chemistry: A Molecular Approach, 2/e

System 9 Copyright 2011 Pearson Education, Inc. The First Law of Thermodynamics: Law of Conservation of Energy Thermodynamics is the study of energy and its interconversions The First Law of Thermodynamics: the Law of Conservation of Energy This means that the total amount of energy in the universe is constant You can therefore never design a system that will continue to produce energy without some source of energy Tro: Chemistry: A Molecular Approach, 2/e 10 Copyright 2011 Pearson Education, Inc. Energy Flow and

Conservation of Energy Conservation of energy requires that the sum of the energy changes (DE) in the system and the surroundings must be zero DEnergyuniverse = 0 = DEnergysystem + DEnergysurroundings D Is the symbol that is used to mean change final amount initial amount Tro: Chemistry: A Molecular Approach, 2/e 11 Copyright 2011 Pearson Education, Inc. Internal Energy The internal energy is the sum of the kinetic and potential energies of all of the particles that compose the system

The change in the internal energy of a system only depends on the amount of energy in the system at the beginning and end DE = Efinal Einitial Internal energy is a state function, a mathematical function whose result only depends on the initial and final conditions, not on the process used DEreaction = Eproducts Ereactants Tro: Chemistry: A Molecular Approach, 2/e 12 Copyright 2011 Pearson Education, Inc. State Function From the foothill to the peak, the change in potential energy is the same regardless of pathways. Potential energy is a state function. It depends only on the difference in states (elevation between the base and the peak)

Traveling distance is NOT a state function Tro: Chemistry: A Molecular Approach, 2/e 13 Copyright 2011 Pearson Education, Inc. the direction of energy flow during a process If the final condition has a larger amount of internal energy than the initial condition, the change in the internal energy will be + If the final condition has a smaller amount of internal energy than the initial

condition, the change in the internal energy will be Tro: Chemistry: A Molecular Approach, 2/e 14 Internal Energy Energy diagrams: showing Internal Energy Energy Diagrams final initial energy increase initial energy decrease final Copyright 2011 Pearson Education, Inc.

Energy Flow: System Surrounding When energy flows out of a system, it must all flow into the surroundings System loses energy to surroundings: DEsystem < 0 Surroundings gains energy, Surroundings DE + System DE DEsurroundings > 0 DEsystem= DEsurroundings Example: Hot water (system) lose heat to cool air in the room (surroundings) Tro: Chemistry: A Molecular Approach, 2/e

15 Copyright 2011 Pearson Education, Inc. Energy Flow: System Surroundings When energy flows into a system, it must all come from the surroundings System gains energy: DEsystem > 0 Surroundings loses energy: DEsurroundings < 0 Surroundings DE System DE + So DEsystem= DEsurroundings Example: Rechargeable battery (system) receives energy from

charger (surroundings) Tro: Chemistry: A Molecular Approach, 2/e 16 Copyright 2011 Pearson Education, Inc. Work and Heat in Energy Exchange Energy is exchanged between the system and surroundings through heat and work q = heat (thermal) energy w = work energy q and w are NOT state functions, their value depends on the process DE = q + w Tro: Chemistry: A Molecular Approach, 2/e 17 Copyright 2011 Pearson Education, Inc. Energy Exchange in real life

Example: Combustion reaction in engine: System (reaction) loses heat (q < 0) and does work (w < 0) to the surroundings. Charging battery: System (battery) loses heat (gets warm, q < 0) but gains work (w > 0) from the surroundings Tro: Chemistry: A Molecular Approach, 2/e 18 Copyright 2011 Pearson Education, Inc. Temperature Temperature is the measure of the amount of thermal energy within a sample of matter Heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature thermal equilibriu When a system absorbs heat q, its temperature

increases, and vice versa. Tro: Chemistry: A Molecular Approach, 2/e 19 Copyright 2011 Pearson Education, Inc. Heat Capacity The increase in temperature DT is directly proportional to heat absorbed q: More heat higher temperature increase, q DT The proportionality constant is called the heat capacity, C. C = q / DT units of C are J/C or J/K q = C DT Tro: Chemistry: A Molecular Approach, 2/e 20 Copyright 2011 Pearson Education, Inc.

Factors Affecting Heat Capacity The heat capacity of an object depends on the amount of matter usually measured by its mass 200 g of water requires twice as much heat to raise its temperature by 1C as does 100 g of water The heat capacity of an object also depends on the type of material 1000 J of heat energy will raise the temperature of 100 g sand by 12C, but only raise the temperature of 100 g water by 2.4 C Tro: Chemistry: A Molecular Approach, 2/e 21 Copyright 2011 Pearson Education, Inc. Specific Heat Capacity The specific heat (c) is the amount of heat energy required to raise the temperature of one gram of a substance by 1C Measure of a substances

intrinsic ability to absorb heat Unit = J/(gC) c = q/(mass x DT) Tro: Chemistry: A Molecular Approach, 2/e 22 Copyright 2011 Pearson Education, Inc. How to measure Heat Energy? The heat capacity of an object is proportional to its mass and the specific heat of the material Heat calculation from specific heat: Heat absorbed or released can be calculated as if we know the mass, the specific heat, and the temperature change of the object Heat = (mass) x (specific heat) x (temp. change) q = m x c x DT Tro: Chemistry: A Molecular Approach, 2/e 23

Copyright 2011 Pearson Education, Inc. Find Heat: How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from 8.0 C to 37.0 C? Specific heat of copper is 0.385 J/g C DT = 45.0C q = 53.7 J Copyright 2011 Pearson Education, Inc. Heat Transfer and Temperature Change Ti > Ti When two objects are placed in contact, heat flows from the material at the higher temperature to the material at the lower temperature

Heat flows until both materials reach the same final temperature COLD HOT Ti Ti heat transfer Heat lost by the hot material equals the heat gained by the cold material qhot = qcold Tf = Tf

heat transfer stops mhotCs,hotDThot = (mcoldCs,coldDTcold) Tro: Chemistry: A Molecular Approach, 2/e 25 Copyright 2011 Pearson Education, Inc. Measure Heat from Chem. Rxn: Bomb Calorimeter Calorimeter for Heat measurement Used to measure DE because it is a constant volume system The heat capacity of the calorimeter = heat absorbed by the calorimeter per degree rise in temperature. Also called the calorimeter constant Ccal, kJ/C Tro: Chemistry: A Molecular Approach, 2/e 26

Copyright 2011 Pearson Education, Inc. Enthalpy The enthalpy, H, of a system is the sum of the internal energy E of the system and the product of pressure and volume H is a state function H = E + PV The enthalpy change, DH, of a reaction is the heat evolved in a reaction at constant pressure DHreaction = qreaction at constant pressure Usually DH and DE are similar in value, the difference is largest for reactions that produce or use large quantities of gas Tro: Chemistry: A Molecular Approach, 2/e 27 Copyright 2011 Pearson Education, Inc.

Endothermic and Exothermic Reactions DH < 0 (or Hfinal < Hinitial): heat is released by the system Heat-releasing reactions: exothermic reactions Examples: Chemical heat packs Combustion reaction, Explosion reaction DH > 0 (or Hfinal > Hinitial): heat is absorbed by the system Heat-absorbing reactions: endothermic reactions Chemical cold packs contain NH4NO3 that dissolves in water in an endothermic process your hands get cold because the pack is absorbing your heat Tro: Chemistry: A Molecular Approach, 2/e 28 Copyright 2011 Pearson Education, Inc. Enthalpy Change in Reaction, HH The enthalpy change in a chemical reaction depends on the amount (#mole) the more reactants you use, the larger the enthalpy

change. More C3H8? More enthalpy change! HH is a conversion factor (unit: kJ/mol) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) Tro: Chemistry: A Molecular Approach, 2/e 29 DH = 2044 kJ Copyright 2011 Pearson Education, Inc. Example: How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)? C3H8(g) + 5O2(g)3CO2(g) + 4H2O(g) kg g DH = 2044 kJ mol

30 kJ Copyright 2011 Pearson Education, Inc. Measuring DH Measurement of heat at constant pressure: water to absorb/release heat Reaction in aqueous solution: Foam cup calorimeter (foam as insulator) Heat exchange between solution/water and reaction (1st law of thermodynamics): qrxn = qsoln since qsoln = masssoln x Csoln x Dtsoln qrxn = molL.R. x DHrxn DHreaction = __________________________ Tro: Chemistry: A Molecular Approach, 2/e

31 Copyright 2011 Pearson Education, Inc. More Example: What is DHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) if 0.158 g Mg reacts in 100.0 mL of solution and the solution changes the temperature from 25.6C to 32.8 C? Assume dsoln = 1.00 g/mL, Csoln = 4.18 J/gC rxn solution Mol Mg = 6.499E-3 mol Qsoln = 3.0E+3 J 32 DHrxn/mol Mg = -4.6E+5 J/mol Mg Copyright 2011 Pearson Education, Inc. Relationships Involving DHrxn When reaction is multiplied by a factor, DHrxn is

multiplied by that factor because DHrxn is extensive C(s) + O2(g) CO2(g) DH = 393.5 kJ 2 C(s) + 2 O2(g) 2 CO2(g) DH = ____________ If a reaction is reversed, then the sign of DH is changed CO2(g) C(s) + O2(g) DH = ____________ Tro: Chemistry: A Molecular Approach, 2/e 33 787.0 kJ, +393.5 kJ Copyright 2011 Pearson Education, Inc. Finding DHrxn for new reaction: Hesss Law If a reaction can be expressed as a series of steps, then the DHrxn for the overall reaction is the sum of the heats of reaction for each step A + 2B C

DH1 C+ EF DH2 FG DH3 A + 2B + C + E + F C + F + G Total reaction: A + 2B + E G Tro: Chemistry: A Molecular Approach, 2/e DH4 = DH1 + DH2 + DH3 34 Copyright 2011 Pearson Education, Inc. How to use Hesss Law from given DHs Compare the target reaction and given reactions Two basic manipulations: Reverse and/or Multiply

If one formula exists in more than one of the given rxns, do not attempt to manipulate eqn using this formula. Example: Find enthalpy change for A + 2B 3F Given: A + 2B 2C DH1 3G 2C DH2 FG 2C 3G DH3 -DH2 3G 3F -3 x DH3

A + 2B 3F DH = DH1 + (-DDH2) + (-D3DDH3) A + 2B 2C DH1 Tro: Chemistry: A Molecular Approach, 2/e 35 Copyright 2011 Pearson Education, Inc. Example: Calculating Hrxn from given H Values Find DHrxn for the rxn: 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Given: N2(g) + 3H2(g) 2NH3(g) DH1 = -91.8 kJ N2(g) + O2(g) 2NO(g) DH2 = 180.6 kJ 2H2(g) + O2(g) 2H2O(g)

DH3 = -483.6 kJ DHrxn= -906.0 kJ Copyright 2011 Pearson Education, Inc. Using Hess Law to Calculate an Unknown DH The catalytic converter in automobiles ultilizes the following reaction to convert pollutants CO and NO into harmless gases: CO (g) + NO (g) CO2 (g) + N2 (g) Given the following information, calculate the unknown DH: Equation A: CO (g) + O2 (g) CO2 (g) Equation B: N2 (g) + O2 (g) 2NO (g) DHB = 180.6 kJ DHA = 283.0 kJ DHrxn = 373.3 kJ Copyright 2011 Pearson Education, Inc. Standard Conditions

The standard state is the state of a material at a defined set of conditions pure gas at exactly 1 atm pressure pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest usually 25 C substance in a solution with concentration 1 M The standard enthalpy change, DH, is the enthalpy change when all reactants and products are in their standard states Tro: Chemistry: A Molecular Approach, 2/e 38 Copyright 2011 Pearson Education, Inc. Standard enthalpy of formation, DHf, Formation Reaction: Reactions of elements in their standard state to form 1 mole of a pure compound The standard states of all metals are solid except Hg; Nonmetals: Only Br2 is liquid, low atomic mass

nonmetals are gas whilst heavy nonmetals are solid The standard enthalpy of formation, DHf: the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements the elements must be in their standard states the DHf for a pure element in its standard state = 0 kJ/mol Tro: Chemistry: A Molecular Approach, 2/e 39 Copyright 2011 Pearson Education, Inc. Standard Enthalpies of Formation Tro: Chemistry: A Molecular Approach, 2/e 40 Copyright 2011 Pearson Education, Inc. Writing Formation Reactions

Write the Formation Reaction for CO(g) The formation reaction is the reaction between the elements in the compound C + O CO(g) The elements must be in their standard state C(s, graphite) + O2(g) CO(g) The equation must be balanced, but the coefficient of the product compound must be 1 C(s, graphite) + O2(g) CO(g) Tro: Chemistry: A Molecular Approach, 2/e 41 Copyright 2011 Pearson Education, Inc. Write the formation reactions for the following: CO2(g) C(s, graphite) + O2(g) CO2(g) Hg2SO4(s)

2 Hg(l) + S(s) + 2 O2(g) Hg2SO4(s) Tro: Chemistry: A Molecular Approach, 2/e 42 Copyright 2011 Pearson Education, Inc. More example: Writing Formation Equations Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DHf. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. Ag (s) + Cl2 (g) AgCl (s) DHf = 127.0 kJ 3 Ca (s) + C(graphite) + 2O2 (g) CaCO3 (s) DHf = 1206.9 kJ H2 (g) + C(graphite) + N2 (g) HCN (g) DHf = 135 kJ

Copyright 2011 Pearson Education, Inc. Calculating Standard Enthalpy Change for a Reaction Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products The DH for the reaction is then the sum of the DHf for the component reactions DHreaction = S n DHf(products) S n DHf(reactants) S means sum n is the coefficient of the reaction Tro: Chemistry: A Molecular Approach, 2/e 44 Copyright 2011 Pearson Education, Inc. Calculate the enthalpy change in the reaction 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) 1. Collect the DHf for each compound in the equation

2 C(s, gr) + H2(g) C2H2(g) DHf = +227.4 kJ/mol C(s, gr) + O2(g) CO2(g) DHf = 393.5 kJ/mol H2(g) + O2(g) H2O(l) DHf = 285.8 kJ/mol 2. DHreaction = S n DHf(products) S n DHf(reactants) DHrxn = 2600.4 kJ Tro: Chemistry: A Molecular Approach, 2/e 45 Copyright 2011 Pearson Education, Inc. Calculating Hrxn from Hf Values For the following reaction: 4NH3(g) + 5O2 (g) 4NO (g) + 6H2O (g) Calculate DHrxn from DHf values. DHf (kJ/mol): NH3(g) -45.9, NO(g) 90.3, H2O(g) -241.8

DHrxn = SmDHf (products) SnDHf (reactants) m and n are the coefficients for products and reactants, respectively. DHrxn = 906 kJ Copyright 2011 Pearson Education, Inc. Example: Determine the standard enthalpy of formation for nitromethane from DHrxn 2CH3NO2(l) + 3/2 O2(g) 2CO2(g) + N2(g) + 3H2O(g) DHrxn = -709.2 kJ/mol 1. Collect the DHf for each compound in the equation C(s, gr) + O2(g) CO2(g) DHf = 393.5 kJ/mol H2(g) + O2(g) H2O(g) DHf = 241.8 kJ/mol N2(g) and O2(g) are the element at the standard state, DHf = 0 2. Set up equation based on DHrxn = S n DHf(products) S n DHf(reactants) DHCH3NO2 = 401.6 kJ Tro: Chemistry: A Molecular Approach, 2/e 47 Copyright 2011 Pearson Education, Inc.

Practice: Determine the standard enthalpy of formation for TNT (C7H5N3O6). The heat of combustion for TNT is -3400kJ/mol. C7H5N3O6(s) + ?O2(g) 7CO2(g) + 3/2 N2(g) + 5/2 H2O(l) find DHf(C7H5N3O6) DHf[CO2(g)] = 393.5 kJ/mol DHrxn = -3400 kJ/mol DHf(H2O(l)) = 285.8 kJ/mol N2(g) and O2(g) as element at the standard state, DHf = 0 DHrxn = S n DHf(products) S n DHf(reactants) Tro: Chemistry: A Molecular Approach, 2/e 48 Copyright 2011 Pearson Education, Inc. Practice: How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? MMoctane = 114.2 g/mol

DHrxn = 5074.1 kJ Tro: Chemistry: A Molecular Approach, 2/e 49 Copyright 2011 Pearson Education, Inc. Table Selected Standard Enthalpies of Formation at 25C (298K) Formula DHf (kJ/mol) Formula Calcium Ca(s) CaO(s) CaCO3(s) Carbon C(graphite) C(diamond) CO(g) CO2(g) CH4(g) CH3OH(l)

HCN(g) CSs(l) Chlorine Cl(g) 0 635.1 1206.9 0 1.9 110.5 393.5 74.9 238.6 135 87.9 121.0 DHf (kJ/mol) Formula DHf (kJ/mol) 0

92.3 Hydrogen H(g) H2(g) Silver Ag(s) AgCl(s) 218.0 0 Sodium Nitrogen N2(g) NH3(g) NO(g) 0 45.9 90.3

Oxygen O2(g) O3(g) H2O(g) 0 143 241.8 Cl2(g) HCl(g) H2O(l) Na(s) Na(g) NaCl(s) 0 127.0 0 107.8 411.1

Sulfur S8(rhombic) 0 S8(monoclinic) 0.3 SO2(g) 296.8 SO3(g) 396.0 285.8 Copyright 2011 Pearson Education, Inc. The two-step process for determining DHrxn from DHf values. Copyright 2011 Pearson Education, Inc. Find Heat: How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from 8.0 C to 37.0 C? Sort Information

Strategize Given: T1 = 8.0 C, T2= 37.0 C, m = 3.10 g q, J Find: Conceptual Plan: Cs m, DT q Relationships: q = m Cs DT Follow the conceptual plan to solve the

problem Check Solution: Cs = 0.385 J/gC (Table 6.4) Check: the unit is correct, the sign is reasonable as the penny must absorb heat to make its temperature rise Tro: Chemistry: A Molecular Approach, 2/e 52 Copyright 2011 Pearson Education, Inc. Find Specific heat: A 100.-g of unknown metal alloy is heated to 99.8C and submerged into 200. g water at 25.9C. The final temperature of the water becomes 32.9C. What is the specific heat of the metal alloy? qm = qH2O Tmf TH 2O f DTmetal -D 67.1 C

qH 2O 5.858 103 J DTH 2O +7.0 C qmetal -D q H 2O -D 5.858 103 J cm 0.873 J / g C Tro: Chemistry: A Molecular Approach, 2/e 53 Copyright 2011 Pearson Education, Inc. Find Specific heat: A 100.-g of unknown metal alloy is heated to 99.8C and submerged into 200. g water at 25.9C. The final temperature of the water becomes 32.9C. What is the specific heat of the metal alloy? Given: mm=100.g, Tmi = 99.8C, mH20 =200.g, TH2Oi =25.9C, Tfinal= 32.9C Find: find Cm Plan: qm = qH2O

qm cm mm DTm qH 2O cH 2O mH 2O DTH 2O DTH 2O TH 2O f -D TH 2O i Tmf TH 2O f CH2O= 4.18 J/gC Solution: DTm Tm f -D Tm i DTm Tm f -D Tm i 32.9 -D 99.8 -D 67.1 C DTH 2O TH 2O f -D TH 2O i 32.9 -D 25.9 +7.0 C q H 2O cH 2O mH 2O DTH 2O 5.858 103 J qm -D q H 2O -D 5.858 103 J Tro: Chemistry: A Molecular Approach, 2/e 54 Copyright 2011 Pearson Education, Inc.

Find Specific heat (contd.): A 100.-g of unknown metal alloy is heated to 99.8C and submerged into 200. g water at 25.9C. The final temperature of the water becomes 32.9C. What is the specific heat of the metal alloy? Given: Find: mm = 100. g, Tmi = 99.8C, mH20 = 200. g, TH2Oi = 25.9C, Tfinal = 32.9C find Cm qm -D 5.858 103 J Solution: DTm -D 67.1 C qm cm mm DTm qm -D 5.858 103 J cm

+ 0 . 873 J / g C mm DTm 100.g (-D 67.1 C ) Tro: Chemistry: A Molecular Approach, 2/e 55 Copyright 2011 Pearson Education, Inc. Predict DT: A 32.5-g cube of aluminum initially at 45.8 C is submerged into 105.3 g of water at 15.4 C. What is the final temperature of both substances at thermal equilibrium? qAl = qH2O Tmf TH 2O f

Tmf 17.3 C Tro: Chemistry: A Molecular Approach, 2/e 56 Copyright 2011 Pearson Education, Inc. Predict DT: A 32.5-g cube of aluminum initially at 45.8 C is submerged into 105.3 g of water at 15.4 C. What is the final temperature of both substances at thermal equilibrium? Given: mAl = 32.5 g, Tal = 45.8 C, mH20 = 105.3 g, TH2O = 15.4 C Find: Tfinal, C Conceptual Plan: Relationships: Solution: Cs, Al mAl, Cs, H2O mqH2O DTAl = kDTH2O Al = qH2O Tfinal q = m Cs DT

Cs, Al = 0.903 J/gC, Cs, H2O = 4.18 J/gC(Table 6.4) the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc. 57 Check: (Continued): A 32.5-g cube of aluminum initially at 45.8 C is submerged into 105.3 g of water at 15.4 C. What is the final temperature of both substances at thermal equilibrium? Given: mAl = 32.5 g, Tal = 45.8 C, mH20 = 105.3 g, TH2O = 15.4 C Find: Tfinal, C Conceptual Plan: Relationships: Solution: Cs, Al mAl, Cs, H2O mqH2O DTAl = kDTH2O Al = qH2O

Tfinal q = m Cs DT Cs, Al = 0.903 J/gC, Cs, H2O = 4.18 J/gC(Table 6.4) the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc. 58 Check: Molar Energy Change: When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92 C to 28.33 C. If Ccal = 4.90 kJ/C, find DE for burning 1 mole of sugar. Given: 1.010 g C12H22O11, T1 = 24.92 C, T2 = 28.33 C, Ccal = 4.90 kJ/C Find: DErxn, kJ/mol

Conceptual Plan: Ccal, DT Relationships: qcal qrxn qrxn, mol DE qcal = Ccal x DT = qrxn MM C12H22O11 = 342.3 g/mol Solution: Check: the units are correct, the sign is as expected for combustion

Tro: Chemistry: A Molecular Approach, 2/e 59 Copyright 2011 Pearson Education, Inc. When 1.010 g of sugar (C12H22O11) is burned in a bomb calorimeter, the temperature rises from 24.92 C to 28.33 C. If heat capacity of calorimeter Ccal = 4.90 kJ/C, find DE for burning 1 mole of sugar. qcal = 16.71 kJ qrxn = -16.71 kJ Mol sugar = 0.0029507 mol DE = -5.66 x 103 kJ Tro: Chemistry: A Molecular Approach, 2/e 60 Copyright 2011 Pearson Education, Inc.

Example: How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)? Given: 13.2kg C3H8, C3H8(g) + 5O2(g)3CO2(g) + 4H2O(g) DH = 2044 kJ Find: q, kJ/mol Conceptual Plan: combustion of 1 mole C3H8 gives 2044 kJ heat. kg g mol kJ Relationships: 1 kg = 1000 g, Molar Mass = 44.09 g/mol Solution: Tro: Chemistry: A Molecular Approach, 2/e 61

Copyright 2011 Pearson Education, Inc. Example: Determining the Enthalpy Change of an Aqueous Reaction PROBLEM: 50.0 mL of 0.500 M NaOH is placed in a coffee-cup calorimeter at 25.00C and 25.0 mL of 0.500 M HCl is carefully added, also at 25.00C. After stirring, the final temperature is 27.21C. Calculate qsoln (in J) and the change in enthalpy, DH, (in kJ/mol of H2O formed). Assume that the total volume is the sum of the individual volumes, that d = 1.00 g/mL, and that c = 4.184 J/gK PLAN: Heat flows from the reaction (the system) to its surroundings (the solution). Since qrxn = qsoln, we can find the heat of the reaction by calculating the heat absorbed by the solution. Copyright 2011 Pearson Education, Inc. SOLUTION: (a) To find qsoln: Total mass (g) of the solution = (25.0 mL + 50.0 mL) x 1.00 g/mL = 75.0 g DTsoln = 27.21C 25.00C = 2.21C = 2.21 K

qsoln = csoln x masssoln x DTsoln = (4.184 J/gK)(75.0 g)(2.21 K) = 693 J (b) To find DHrxn we first need a balanced equation: HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) Copyright 2011 Pearson Education, Inc. Example (contd): For HCl: 25.0 mL HCl x 1 L x 0.500 mol 103 mL 1L x 1 mol H2O 1 mol HCl = 0.0125 mol H2O For NaOH: 50.0 mL NaOH x 1 L x 103 mL 0.500 mol x 1 mol H2O = 0.0250 mol H2O

1L 1 mol NaOH HCl is limiting, and the amount of H2O formed is 0.0125 mol. DHrxn = qrxn mol H2O = 693 J x 1 kJ 103J 0.0125 mol = 55.4 kJ/mol H2O Copyright 2011 Pearson Education, Inc. Example 6.8: What is DHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) if 0.158 g Mg reacts in 100.0 mL of solution and changes the temperature from 25.6 C to 32.8 C?

Given: Find: Conceptual Plan: 0.158 g Mg, 100.0 mL, T1 = 25.6 C, T2 = 32.8 C, assume d = 1.00 g/mL, Cs = 4.18 J/gC H, kJ/mol Cs, DT, m qsoln qrxn qrxn, mol DH Relationships: qsoln = m x Cs, soln x DT = qrxn, MM Mg= 24.31 g/mol Solution: Check:

the heat released from reaction (-) is transferred to the solution, so solution gains heat (+) Tro: Chemistry: A Molecular Approach, 2/e 65 Copyright 2011 Pearson Education, Inc. Example 6.8: What is DHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) if 0.158 g Mg reacts in 100.0 mL of solution and changes the temperature from 25.6 C to 32.8 C? Given: Find: Conceptual Plan: 0.158 g Mg, 100.0 mL, T1 = 25.6 C, T2 = 32.8 C, assume d = 1.00 g/mL, Cs = 4.18 J/gC H, kJ/mol Cs, DT, m qsoln

qrxn qrxn, mol DH Relationships: qsoln = m x Cs, soln x DT = qrxn, MM Mg= 24.31 g/mol Solution: Check: the sign is correct and the value is reasonable because the reaction is exothermic Tro: Chemistry: A Molecular Approach, 2/e 66 Copyright 2011 Pearson Education, Inc. Determining the Enthalpy Change of an Aqueous Reaction 50.0 mL of 0.500 M NaOH is placed in a coffee-cup calorimeter at rxn 25.00C and 25.0 mL of 0.500 M HCl is carefully added, also at

solution 25.00C. After stirring, the final temperature is 27.21C. Calculate qsoln (in J) and the change in enthalpy, DH, (in kJ/mol of H2O formed). Assume that the total volume is the sum of the individual volumes, that d = 1.00 g/mL, and that c = 4.184 J/gK Heat flows from the reaction (the system) to its surroundings (the PLAN: solution): qrxn = qsoln. Start with heat absorbed by the solution. Find Limiting reactant: HCl; the amount of product H2O = 0.0125 mol. qsoln = 693 J qrxn = -693 J DHrxn = qrxn/mol L.R. = -55.4 kJ/mol H2O Copyright 2011 Pearson Education, Inc. More example: Calculating Hrxn from given H Values Find DHrxn for the rxn: 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Given: N2(g) + 3H2(g) 2NH3(g) DH1 = -91.8 kJ

N2(g) + O2(g) 2NO(g) DH2 = 180.6 kJ 2H2(g) + O2(g) 2H2O(g) DH3 = -483.6 kJ NH3: Rxn# 1 reverse and x 2 4NH3(g) 2N2(g) + 6H2(g) DH = 183.6 kJ O2: in more than one rxn, skip NO: Rxn# 2 times 2 H2O: Rxn# 3 times 3 2N2(g) + 2O2(g) 4NO(g) DH = 361.2 kJ 6H2(g) + 3O2(g) 6H2O(g) DH = -1450.8 kJ Sum of all reactions: DHrxn= 183.6 kJ+361.2 kJ + (-1450.8 kJ) = 906.0 kJ 4NH3(g) + 2N2(g) + 2O2(g) + 6H2(g) + 3O2(g) 2N2(g) + 6H2(g) + 4NO(g) + 6H2O(g)

Copyright 2011 Pearson Education, Inc. Example 6.12: How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? Given: Find: Conceptual Plan: 1.0 x 1011 kJ DHrxn = 5074.1 kJ mass octane, kg Write the balanced equation per mole of octane DHfs DHrxn kJ from above mol C8H18

g C8H18 kg C8H18 Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g Solution: C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(g) Look up the DHf for each material in Appendix IIB Check: the units and sign are correct, the large value is expected Tro: Chemistry: A Molecular Approach, 2/e 69 Copyright 2011 Pearson Education, Inc.

Example 6.12: How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? Given: Find: Conceptual Plan: 1.0 x 1011 kJ DHrxn = 5074.1 kJ mass octane, kg Write the balanced equation per mole of octane DHfs DHrxn kJ from above mol C8H18

g C8H18 kg C8H18 Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g Solution: Check: C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(g) the units and sign are correct, the large value is expected Tro: Chemistry: A Molecular Approach, 2/e 70 Copyright 2011 Pearson Education, Inc. Example 6.12: How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? Given: Find:

Conceptual Plan: 1.0 x 1011 kJ DHrxn = 5074.1 kJ mass octane, kg Write the balanced equation per mole of octane DHfs DHrxn kJ from above mol C8H18 g C8H18 kg C8H18 Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g

Solution: Check: C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(g) the units and sign are correct, the large value is expected Tro: Chemistry: A Molecular Approach, 2/e 71 Copyright 2011 Pearson Education, Inc. More examples: Using Hesss Law to Calculate an Unknown DH PROBLEM: Two gaseous pollutants that form in auto exhausts are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following reaction: CO (g) + NO (g) CO2 (g) + N2 (g) DH = ?

Given the following information, calculate the unknown DH: Equation A: CO (g) + O2 (g) CO2 (g) DHA = 283.0 kJ Equation B: N2 (g) + O2 (g) 2NO (g) DHB = 180.6 kJ PLAN: Manipulate Equations A and/or B and their DH values to get to the target equation and its DH. All substances except those in the target equation must cancel. Copyright 2011 Pearson Education, Inc. More example: Writing Formation Equations PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DHf. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. PLAN: Write the elements as reactants and 1 mol of the compound as the product formed. Make sure all substances are in their standard states. Balance the equations and find the value of

DHf in Table 6.3 or Appendix B. Copyright 2011 Pearson Education, Inc. SOLUTION: (a) Silver chloride, AgCl, a solid at standard conditions. Ag (s) + Cl2 (g) AgCl (s) DHf = 127.0 kJ (b) Calcium carbonate, CaCO3, a solid at standard conditions. 3 Ca (s) + C(graphite) + 2O2 (g) CaCO3 (s) DHf = 1206.9 kJ (c) Hydrogen cyanide, HCN, a gas at standard conditions. H2 (g) + C(graphite) + N2 (g) HCN (g) DHf = 135 kJ Copyright 2011 Pearson Education, Inc.

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