CHAPTER 6 All Bold Numbered Problems

Chapter 6 Outline Energy -vs- Heat Specific Heat First Law of Thermodynamics q (Heat) Hess’s Law

THERMOCHEMISTRY or Thermodynamics 3

Energy & Chemistry Burning sugar (sugar reacts with KClO3, a strong oxidizing agent) Burning peanuts supply sufficient energy to boil a cup of water.

Energy and Chemistry 2 H2(g) + O2(g) --> 2 H2O(g) + heat and light This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H2 ---> 4 H+ + 4 e- Reduction: 4 e- + O2 + 2 H2O ---> 4 OH-

Energy and Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 samples because of their difference in temperature. Other types of energy light electrical kinetic and potential

Kinetic and Potential Energy Potential energy: energy stored in chemical due to its structure. The energy of a motionless body due to its position.

Kinetic and Potential Energy Kinetic energy: energy of motion. Translational Rotational Vibrational (IR)

Thermodynamics Thermodynamics is the science of heat (energy) transfer. Heat energy is associated with molecular motions.

Energy and Chemistry All of thermodynamics depends on the law of THE CONSERVATION OF ENERGY. The total energy of a system is constant.

UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) The S.I sytem uses the unit called the JOULE 1 cal = 4.18 Joules James Joule 1818-1889

Specific Heat Capacity Thermochemistry is the science of heat (energy) flow. A difference in temperature leads to energy transfer. The heat “lost” or “gained” is related to a) sample mass b) change in T c) specific heat capacity Specific heat capacity = heat lost or gained by substance (J) (mass, g)(T change, K)

Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.18 Al 0.902 glass 0.84 Aluminum

What drives chemical reactions? CHEMICAL REACTIVITY What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. Have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer in a battery

CHEMICAL REACTIVITY But ENERGY TRANSFER also allows us to predict reactivity. In general, reactions that transfer energy to their surroundings are product-favored. So, let us consider heat transfer in chemical processes.

Heat Energy Transfer in Chemical Processes CO2 (s, -78 oC) ---> CO2 (g, -78 oC)

FIRST LAW OF THERMODYNAMICS heat energy transferred (to the system) DE = q + w work done (by the surroundings) energy change Energy is conserved!

DH = qp (for chemist in a lab which means only T changes) The First Law of Thermodynamics is the law of conservation of energy. DE = q + w DE is the change in Energy, q is heat, w is work Enthalpy (H), defined H = E + PV. DH = DE + PDV at constant pressure (only V and T change). If w = - PDV (because V does not change in a typical lab experiment, only T changes) and DH = q + w + PDV DH = qp (for chemist in a lab which means only T changes)

ENTHALPY Most chemical reactions occur at constant P. qp = DH DH = change in heat content of the system DH = Hfinal - Hinitial How do we measure q in the lab?

Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.18 Al 0.902 glass 0.84 If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? Specific heat capacity = heat lost or gained by substance (J) or (mass, g)(T change, K) C = Specific Heat, units of J/(gK)

Specific Heat Capacity If 25.0 g of Al cool from 310. oC to 37 oC, how many joules of heat energy are lost by the Al? heat gain/lost = q = (C)(mass)(DT) where DT = Tfinal - Tinitial q = (0.902 J/g•K)(25.0 g)(37 - 310.)K q = - 6160 J

Heat Energy Transfer in Chemical Processes Heat flows FROM the SYSTEM into the SURROUNDINGS is an EXOTHERMIC process. q is “-” The reaction feels hot. Energy given Off!

Heat Energy Transfer in Chemical Processes Heat flows INTO the SYSTEM from the SURROUNDINGS is an ENDOTHERMIC process. q is “+” The reaction feels cold. Energy Added!

Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? q = - 6160 J q = - 6.16 kJ Notice that the negative sign on q signals heat “lost by” or transferred out of Al. Is this Exo or Endothermic?

Specific Heat Capacity A 500.0 g piece of metal is heated to 75.0 0C. It is placed into 300. mL of water at 22.0 0C and the final temperature is 31.0 0C. Calculate the specific heat of the metal. qwater + qmetal = 0 qwater = -qmetal (300. g)(4.18 J/gK)(9.0oC) = -(500.0 g)(c)(-44.0oC) c = 0.51 J/gK

Specific Heat Capacity To 150 g of water at 25oC is added 45 g of Al at 115oC, what will be the final temperature? Al 0.902

Heat Transfer and Changes of State Changes of state involve energy Ice -----> Water 333 J/g (heat of fusion) + energy Exo or Endo? Sign of q?

Heat Transfer and Changes of State Liquid ---> Vapor Requires energy (heat). This is the reason a) you cool down after swimming b) you use water to put out a fire. + energy

Heat and Changes of State 29 Heat and Changes of State What quantity of heat is required to change 10.0 g of ice at -50.0 oC to steam at 200.0 oC? Endo or Exo? Sign of q? Heat of fusion of ice = 333 J/g Heat of vaporization = 2260 J/g +333 J/g +2260 J/g

Heating/Cooling Curve for Water 5 4 3 2 1

Heat and Changes of State 1. To heat ice q = (10.0 g)(2.09 J/g•K)(50.0K) = 1050J 2. To melt ice q = (10.0 g)(333 J/g) = 3330J 3. To heat water q = (10.0 g)(4.18 J/g•K)(100.K) = 4180J 4. To evaporate water q = (10.0 g)(2260 J/g) = 22600J 5. To heat steam q = (10.0 g)(2.03 J/g•K)(100.0K) = 2030J Total heat energy = 33190.J = 33.190kJ

Heat and Changes of State Calculate the amount of heat energy necessary to change 25.0 g of copper solid at 925o C to liquid at 1083o C. Melting point = 1083o C, csolid = 0.382 J/g.K, Heat of fusion = 205 J/g. q1 = (25.0 g)(.382 J/gK)(1083oC-925oC)= 1510 J q2 = (25.0 g)(205 J/g) = 5120 J qT = q1 + q2 = 6630 J

Endo- and Exothermic qsystem > 0 qsystem < 0 EXOTHERMIC T(system) goes up T(system) goes down ENDOTHERMIC Heat goes in the system EXOTHERMIC Heat leaves the system

USING ENTHALPY Consider the decomposition of water H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g) Endo or Exo? Endothermic reaction Heat is a “reactant” DH = + 242 kJ

USING ENTHALPY Making H2 from H2O involves two steps. Each step requires energy. Liquid H2O H2 + O2 gas H2O vapor

USING ENTHALPY Example of HESS’S LAW Making H2 from H2O involves two steps. H2O(liq) + 44 kJ ---> H2O(g) H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g) ----------------------------------------------------------------------- H2O(liq) + 286 kJ --> H2(g) + 1/2 O2(g) Example of HESS’S LAW If a rxn. is the sum of 2 or more others, the net DH is the sum of the DH’s of the other rxns.

USING ENTHALPY DHnet = DH1 + DH2 = -395.7 kJ Calc. DH for: S(s) + 3/2 O2(g) --> SO3(g) Given: S(s) + O2(g) --> SO2(g) DH1 = -320.5 kJ SO2(g) + 1/2 O2(g) --> SO3(g) DH2 = -75.2 kJ The two equations add up to give the desired equation, so - DHnet = DH1 + DH2 = -395.7 kJ

 DH along one path =  DH along another path ENERGY S solid +O 2 direct path DH = 1 -320.5 kJ + 3/2 O 2 DH = SO gas 2 -395.7 kJ + 1/2 O 2 SO gas 3 DH = -75.2 kJ 2  DH along one path =  DH along another path

DH along one path = DH along another path This equation is valid because DH is a STATE FUNCTION State functions depend only on the state of the system and not how it got there. Examples: V, T, P, energy — and your bank account! Unlike V, T, and P, one cannot measure absolute H. We can only measure DH.

Standard Enthalpy Values Most DH values are labeled ΔHo Measured under standard conditions P = 1 atmosphere or 1 bar, (105Pascals) Concentration = 1 mol/L T = 25 oC with all species in standard states e.g., C = graphite and O2 = gas, etc.

Standard Enthalpy Values NIST (National Institute for Standards and Technology) gives values of DHof = standard molar enthalpy of formation This is the enthalpy change when 1 mole of compound is formed from elements under standard conditions. See Table 6.2 and Appendix L

DHof, standard molar enthalpy of formation H2(g) + 1/2 O2(g) --> H2O(g) DHof = -241.8 kJ/mol By definition, DHof = 0 for elements in their standard states.

Using Standard Enthalpy Values Use DH°f’s to calculate enthalpy change for: H2O(g) + C(graphite) --> H2(g) + CO(g) (product is called “water gas”)

Using Standard Enthalpy Values H2O(g) + C(graphite) --> H2(g) + CO(g) From reference books we find: H2(g) + 1/2 O2(g) --> H2O(g) DH°f of H2O vapor = - 241.8 kJ/mol C(s) + 1/2 O2(g) --> CO(g) DH° f of CO = - 110.5 kJ/mol

Using Standard Enthalpy Values H2O(g) --> H2(g) + 1/2 O2(g) DHo = +241.8kJ C(s) + 1/2 O2(g) --> CO(g) DHo = -110.5 kJ -------------------------------------------------------------------------------- H2O(g) + C(graphite) --> H2(g) + CO(g) DHonet = +131.3 kJ To convert 1 mol of water to 1 mol each of H2 and CO requires 131.3 kJ of energy. The “water gas” reaction is ENDOthermic.

Using Standard Enthalpy Values Calculate D H of reaction? In general, when ALL enthalpies of formation are known, DHorxn =  DHof (products) -  DHof (reactants)

Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., DHorxn for: CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) DHorxn =  DHof (prod) -  DHof (react) Use Appendix L

Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) DHorxn =  DHof (prod) -  DHof (react) DHorxn = {DHof (CO2) + 2 DHof (H2O)} - { 3/2 DHof (O2) + DHof (CH3OH) } = {(-393.5 kJ) + 2 (-241.8 kJ)} - { 0 + (-200.7 kJ) } DHorxn = - 676.4 kJ per mol of methanol

Using Standard Enthalpy Values OF2(g) + H2O(g) --> O2(g) + 2 HF(g) DHorxn =- 318.0kJ DHof (H2O) = - 241.8 kJ and DHof (HF) = - 271.1 kJ Calculate DHof (OF2) DHorxn = {DHof (O2) + 2 DHof (HF) } - { DHof (OF2) + DHof (H2O) } -318.0 kJ = {( 0 ) + 2 (-271.1 kJ) } - {DHof (OF2) + (-241.8 kJ) } DHof (OF2) = 17.6 kJ /mole

Measuring Heats of Reaction with a BOMB CALORIMETRY Coffee Cup Calorimeter

Measuring Heats of Reaction CALORIMETRY Calculate heat of combustion of octane. C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O • Burn 1.00 g of octane Temp rises from 25.00 to 33.20 oC Calorimeter contains 1300. g water Heat capacity of bomb = 427 J/K

CALORIMETRY Step 1 Calc. heat transferred from reaction to water. q = (4.18 J/g•K)(1300. g)(8.20 K) = 44,600 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(DT) = (427 J/K)(8.20 K) = 3500 J Step 3 Total heat evolved = 44,600 J + 3500 J = 48,100 J Heat of combustion of 1.00 g of octane = - 48.1 kJ

Measuring Heats of Reaction CALORIMETRY Calculate heat of combustion of ethanol. C2H5OH + 3 O2 --> 2 CO2 + 3 H2O • Burn 0.888 g of ethanol Temp rises from 25.00 oC to 35.00 oC Calorimeter contains 210. g water Heat capacity of bomb = 0.950 kJ/K

CALORIMETRY Step 1 Calc. heat transferred from reaction to water. q = (4.18 J/g•K)(210. g)(10.00 K) = 8.80 kJ Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(DT) = (0.950 kJ/K)(10.00 K) = 9.50 kJ Step 3 Total heat evolved = 8.80 kJ + 9.50 kJ = 18.30 kJ (-18.30 kJ/0.888g)(46.0 g/mole) = - 948 kJ/mole

Sample Problem A 0.105 g sample of C2H4 is burned in a calorimeter with a heat capacity of 0.47 kJ/oC and containing 2,000. g of water. The temperature increased from 25.00oC to 27.14oC. Calculate the heat of combustion of C2H4 in kJ/mole.

CALORIMETRY Step 1 Calc. heat transferred from reaction to water. q = (4.18 J/g•K)(2000. g)(2.14 K) = 17,900 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(DT) = (0.47 kJ/K)(2.14 K) = 1.0 kJ Step 3 Total heat evolved = 17.9 kJ + 1.0 kJ = 18.9 kJ (-18.9 kJ/0.105g)(28.0 g/mole) = - 5040 kJ/mole

Practice Problems 1. How many kcal is 75.0 kJ? 2. Convert 88.6 kcal/mole to kJ/mole. 3. Calculate the kinetic energy of a 2000. lb car traveling at 50. miles per hour. 4. Is the kinetic energy of a 500. g object moving at 50. km/hr greater, less than, or the same as that of a 1.00 kg object moving at 25 km/hr? 5. How much heat is required to raise the temperature of 204 grams of lead from 22.8oC to 64.9oC? (C= 0.038 cal/gK)

Practice Problems 6. If the specific heat of silver is 0.0573 cal/gK, how much heat would it take to raise the temperature of 1 kg of silver from -50.oC to 150.oC? 7. To what temperature will one pound of nickel be raised, beginning at 25.0oC, if 750 calories are absorbed by it? (c = 0.106 cal/gK) 8. A 50. g chunk of unknown metal X is heated to 98.5oC and then dropped into 450 g of water initially at 25.00oC. The water temperature is observed to rise to 26.47oC. Calculate the specific heat of X.

Practice Problems 9. 2 H2O (l) --> 2 H2 (g) + O2 (g) DH = 273.2 kcal How many mLs of water will be decomposed to hydrogen and oxygen by 85 kcal? 10. CaCO3 (s) --> CaO (s) + CO2 (g) DH = 42.5 kcal How much heat is required to react 186 g CaCO3? 11. C6H12O6 (s) + 6 O2 (g) --> 6 H2O (g) + 6 CO2 (g) DH = -673 kcal. How much heat is evolved in the oxidation of 240 g of C6H12O6? 12. MgCO3(s)+2 HCl(g)->MgCl2(s)+CO2(g) + H2O(g) DH = -5.6 kcal. How many grams of MgCl2 will be produced if 80.0 kilocalories are released?

Practice Problems 13. 3 Cu2S(s) + 16 HNO3(l) --> 6 Cu(NO3)2(s) + 3 S(s) + 4 NO(g) + 8 H2O(l) DH = -543 kcal/mole How much energy will be released by the reaction of 325 grams of Cu2S with HNO3. 14. 2 B5H9 (l) + 12 O2 (g) --> 9 H2O (l) + 5 B2O3 (s) DH = -4188 kcal/mole. How much B5H9 must be burned to produce 4500 kcal of energy? 15. Upon dissolving 8.63 g of ammonium dihydrogen phosphate in 100. mL of water, it was found that DT= - 2.7oC. Calculate the value of the molar heat of reaction.

Practice Problems 16. A 0.25 mole sample of ammonium nitrate was dissolved in 500. mL of water at 21.0oC. As the solute dissolved the temperature of the solution dropped to a minimum value of 18.0oC. Calculate the heat of reaction in kcal/mole. 17. Given the following data: DH(kcal) MnO2 + CO --> MnO + CO2 -36.0 Mn3O4 + CO --> 3 MnO + CO2 -13.0 3 Mn2O3 + CO --> 2 Mn3O4 + CO2 -34.0 Find DH in kcal for the following reaction: 2 MnO2 + CO --> Mn2O3 + CO2

Practice Problems 18. Given the following two reactions: DH(kcal) 2 Na(s) + 2HCl(g) --> 2 NaCl(s) + H2(g) -152.34 H2(g) + Cl2(g) --> 2 HCl(g) -44.12 What is DHfo,in kcal/mole for NaCl(s)? 19. Calculate the DHfo of glycerol, (CH2OH)2CHOH, knowing that its heat of combustion is -397.0 kJ/mole. 20. Find the DH for the reaction: 2 SO2(g) + 2 H2O(g) + O2(g) --> 2 H2SO4 (l)

Practice Problems 21. Determine the enthalpy of formation in kcal/mole for chloroethane, C2H5Cl, from the following data: DH(kcal) H2 + 1/2 O2 --> H2O -68.3 C + O2 --> CO2 -94.1 C2H5Cl --> C2H4 + HCl 17.2 2 CO2 + 2 H2O --> C2H4 + 3 O2 337.2 H2 + Cl2 --> 2 HCl -44.2

Practice Problems 22. The combustion of 0.450 g of ethene (C2H4) causes a temperature rise of 2.00oC in a bomb calorimeter that has a specific heat capacity of 600. cal/oC and containing 500. g of water. What is the heat of combustion in kJ/mole for ethene?

Practice Problems 23. A bomb calorimeter has a heat capacity of 385 J/oC and containing 200. g of water. A temperature rise of 3.577oC is observed when the calorimeter is used in the combustion of a 0.7600 g sample of powdered tantalum in excess oxygen to product Ta2O5. Find DHfo of Ta2O5. 24. How many calories would be required to change the temperature of 750 g of water from 15.0 oC to 90.0 oC? 25. How many joules of heat would be liberated if the temperature of 300. g of iron were changed from 75 oC to 17 oC?

Practice Problems 26. The heat of formation of water is -68.4 kcal/mole. How many joules would be liberated by burning 12 g of hydrogen? 27. 4 Fe + 3 O2 --> 2 Fe2O3 + 398 kcal 4 Al + 3 O2 --> 2 Al2O3 + 798 kcal Calculate DH for the reaction Fe2O3 + 2 Al --> Al2O3 + 2 Fe

Practice Problems 28. A 0.757 g sample of C was placed in a calorimeter with a specific heat capacity of 1550 cal/oC and containing 1,000. G of water and burned. The temperature rose from 22.54 oC to 24.87 oC. Calculate the heat of combustion in kcal/mole?

Practice Problems Answers 1. 17.9 kcal 2. 370. kJ/mole 3. 2 x 105 J 4. Greater 5. 330 cal 6. 10,000 cal 7. 41oC 8. .8 J/gK 9. 11 mL 10. 79.0 kcal 11. 9.0 x 102 kcal 12. 1400 g 13. 370. Kcal 14. 140 g 15. 15,000 J/mole 16. 6.0 kcal/mole 17. -52.0 kJ 18. -98.23kcal/mole

Practice Problems Answers 19. -1750.8 kJ 20. -550.682 kJ 21. -26.9 kJ 22. -572 kJ/mole 23. -2080 kJ/mole 24. 56000 cal 25. 7800 J 26. 1.7 x 106 J 27. -200 kcal 28. -94.2 kcal/mole

Specific Heat Capacity How much heat is required to raise the temperature of 57 g of water from 25.6oC to 66.5oC? q = (57 g)(4.18 J/g K)(40.9oC) q = 9700 J

Energy and Reactions Endothermic Exothermic C+D W+X H DH H DH A+B Y+Z A + B + energy -->C + D W + X -->Y + Z + energy

Energy and Reactions 1. The reaction of 14 g of hydrogen will produce how much energy? 3 H2 + N2 --> 2 NH3 + 92 kJ 14 g ? kJ 14 g H2 mole H2 92 kJ 2.0 g H2 3 mole H2 = 210 kJ

Energy and Reactions 2. If 126 kJ are produced how many grams of ammonia are produced? 3 H2 + N2 --> 2 NH3 + 92 kJ ? g 126 kJ 126 kJ 2 mole NH3 17.0 g NH3 92 kJ mole NH3 = 47g NH3

Hess’s Law N2 (g) + O2 (g) --> 2 NO (g) DH = 180.8 kJ Find DH for the following equation: 2 NO (g) + O2 (g) --> 2 NO2 (g) 2 NO (g) --> N2 (g) + O2 (g) DH = -180.8 kJ N2 (g) + 2 O2 (g) --> 2 NO2 (g) DH = 67.7 kJ 1 2 NO (g) + O2 (g) --> 2 NO2 (g) DH = -113.1 kJ

Hess’s Law 2 C2H4 + 2 H2 + 7 O2 --> 4 CO2 + 6 H2O DH = -3394 kJ 2 C2H6 + 7 O2 --> 4 CO2 + 6 H2O DH = -3120. kJ Find DH for the following equation: C2H4 + H2 --> C2H6 2 C2H4 + 2 H2 + 7 O2 --> 4 CO2 + 6 H2O DH = -3394 kJ 4 CO2 + 6 H2O --> 2 C2H6 + 7 O2 DH = +3120. kJ 1 1 7/2 2 3 -1697 kJ 2 3 1 7/2 +1560. kJ C2H4 + H2 --> C2H6 DH = -137 kJ

Hess’s Law 1/2 H2 + N2 + 5/2 O2 --> HNO3 + NO2 DH = -101 kJ 2 NH + H2 --> N2H4 DH = -567 kJ 2 N2O5 --> 2 N2 + 5 O2 DH = 23 kJ Find DH for the following equation: N2H4 + 2 N2O5 --> 2 HNO3 + 2 NO2 + 2 NH N2H4 --> 2 NH + H2 DH = 567 kJ 2 N2O5 --> 2 N2 + 5 O2 DH = 23 kJ 1 2 5 2 2 -202 kJ N2H4 + 2 N2O5 --> 2 HNO3 + 2 NO2 + 2 NH DH = 388 kJ

Using Standard Enthalpy Values Calculate the DH for the following reaction: Al2O3 (s) + 3 CO (g) --> 3 CO2 (g) + 2 Al (s) DHorxn = { 3 DHof (CO2) + 2 DHof (Al) } - { DHof (Al2O3) + 3 DHof (CO) } = { 3 (-393.5 kJ) + 2 ( 0 )} - { -1675.7 kJ + 3 (-110.5 kJ) } DHorxn = 826.7 kJ

Using Standard Enthalpy Values C2H4 + 3 O2 --> 2 CO2 + 2 H2O (g) DH = -1323.0 kJ 2 C + 2 H2 --> C2H4 DH = 52.3 kJ H2 + 1/2 O2 --> H2O (g) DH = -241.8 kJ Calculate DHf for CO2. 2 C + 2 H2 --> C2H4 DH = 52 kJ H2O (g) --> H2 + 1/2 O2 DH = 241.8 kJ 1/2 3/2 1 1 -661.5 kJ 1 1 1 1/2 26.0 kJ f f C + O2 --> CO2 DHf = -393.7 kJ

Using Standard Enthalpy Values C2H4 + 3 O2 --> 2 CO2 + 2 H2O (g) DH = -1323.0 kJ 2 C + 2 H2 --> C2H4 DH = 52.3 kJ H2 + 1/2 O2 --> H2O (g) DH = -241.8 kJ Calculate DHf for CO2. f f DHorxn = { 2 DHof (CO2) + 2 DHof (H2O) } - {DHof (C2H4) + 3 DHof (O2) } -1323.0 kJ = { 2 DHof (CO2) + 2 ( -241.8 kJ )} - { 52.3 kJ + 3 ( 0 ) } DHof (CO2) = -393.6 kJ