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Energy and Thermochemistry

Energy and Thermochemistry. Energy. The ability to do work 2 types Potential: stored energy Kinetic: energy in motion. Thermochemistry. Changes of heat content and heat transfer Follow Law of Conservation of Energy Or, 1 st Law of Thermodynamics

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Energy and Thermochemistry

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  1. Energy and Thermochemistry

  2. Energy • The ability to do work • 2 types • Potential: stored energy • Kinetic: energy in motion

  3. Thermochemistry • Changes of heat content and heat transfer • Follow Law of Conservation of Energy • Or, 1st Law of Thermodynamics • Energy can neither be created nor destroyed • Thus, total energy of universe constant

  4. Temperature & Heat • Heat not same as temperature • Heat = energy transferred to one system by another due to temperature difference • Temperature = measure of heat energy content & ability to transfer heat • Thermometer • Higher thermal energy, greater motion of constituents • Sum of individual energies of constituents = total thermal energy

  5. Systems and Surroundings • System = the object in question • Surrounding(s) = everything outside the system • When both system and surrounding at same temperature  thermal equilibrium • When not • Heat transfer to surrounding = exothermic • (you feel the heat)  hot metal! • Heat transfer to system = endothermic • (you feel cold)  cold metal!

  6. Math! • Joules (J) used for energy quantities • But usually kJ (1000 J) used • Ye Royal Olde School used calorie (cal) • cal = amt of heat required to raise the temperature of 1.00 g of water by 1C • 1 cal = 4.184 J (SI-unit) • But…Calorie (Cal) = 1000 cal • Used in nutrition science and on food labels

  7. Specific heat capacity Quantity of heat required to raise the temp of 1 gram of any substance by 1 K Molar heat capacity Quantity of heat required to raise the temp of 1 mole of any substance by 1 K Heat Capacity

  8. Calculating heat transfer • FYI • Specific heat capacity of metals is very low •  < 1.000 J/(gK) • What does this tell us about heat transfer in metals?

  9. Let’s do an example • In your backyard, you have a swimming pool that contains 5.19 x 103 kg of water. How many kJ are required to raise the temperature of this water from 7.2 °C to 25.0 °C?

  10. Example solved • Trick:  T in K = T in °C

  11. Practice • How many kJ are required to raise the temperature of 25.8 g of quicksilver from 22.5 °C to 28.0 °C? CHg = 0.1395 J/(gK)

  12. But what if there’s a change of state? • Temperature constant throughout change of state • Added energy overcomes inter-molecular forces

  13. Change of state • What do the flat areas represent?

  14. qtot = qs + qsl + ql + qlg + qg • qsl = heat of fusion • Heat required to convert solid at melting pt. to liq • Ice = 333 J/g • qlg = heat of vaporization • Heat required to convert liq. at boiling pt. to gas • Water = 2256 J/g

  15. Practice • How much heat is required to vaporize 250.0 g of ice at -25.0 °C to 110.0 °C? • Given: • Specific heat capacity of ice = 2.06 J/gK • Specific heat capacity of water = 4.184 J/gK • Specific heat capacity of steam = 1.92 J/gK • Let’s do this

  16. Calorimetry • The process of measuring heat transfer in chemical/physical process • qrxn + qsoln = 0 • qrxn = -qsoln • Rxn = system • Soln = surrounding • What you’ll do in lab • Heat given off by rxn • Measured by thermometer • Figure out qrxn indirectly

  17. Enthalpy • = heat content at constant pressure • If H = “+”, process endothermic • If H = “-”, process exothermic • Enthalpy change dependent on states of matter and molar quantities • For example: • Is vaporizing ice an exothermic or endothermic process? • Thus, will H be “+” or “-”?

  18. Hess’s Law • If a rxn is the sum of 2 or more other reactions, H = sum of H’s for those rxns • So, Htot = H1 + H2 + H3 + … + Hn

  19. Let’s solve a problem • C(s) + 2S(s)  CS2(l); H = ? • Given: • C(s) + O2(g)  CO2(g); H = -393.5 kJ/mol • S(s) + O2(g) SO2(g); H = -296.8 kJ/mol • CS2(l) + 3O2(g)  CO2(g) + 2SO2(g); H = -1103.9 kJ/mol • How do we manipulate the 3 rxns to achieve the necessary net rxn? • Does H change if the rxns are reversed and/or their mole ratios are changed? • Let’s talk about this on the next slide

  20. Let’s work it out 1. Switch this rxn: CS2(l) + 3O2(g)  CO2(g) + 2SO2(g); H = -1103.9 kJ Thus, CO2(g) + 2SO2(g) CS2(l) + 3O2(g) ; H = + 1103.9 kJ Thus, -Hfwd = +Hrev 2. Double this rxn: S(s) + O2(g) SO2(g); H = -296.8 kJ Thus, 2S(s) + 2O2(g) 2SO2(g); H = (-296.8 kJ) x 2 = -593.6 kJ Since H is per mole, changing the stoichiometric ratios entails an equivalent change in H 3. Keep this rxn: C(s) + O2(g)  CO2(g); H = -393.5 kJ 4. Add those on same side of rxns/eliminate those on opposite sides of rxn: CO2, 2SO2, 3O2 5. Net rxn: C(s) + 2S(s)  CS2(l) H = 1103.9 kJ - 593.6 kJ – 393.5 kJ = 116.8 kJ Is it an exo- or endothermic rxn?

  21. Practice • Given: • CH4(g) C(s) + 2H2(g); H = 74.6 kJ/mol • C(s) + O2(g)  CO2(g); H = -393.5 kJ/mol • H2(g) + O2(g)  H2O(g); = -241.8 kJ/mol • CH4(g) + 2O2(g) CO2(g) + 2H2O(g); Hrxn = ?

  22. Standard Energies of Formation • Standard molar enthalpies of formation = Hf = enthalpy change for formation of 1 mol of cmpd directly from component elements in standard states • Standard state = most stable form of substance in physical state that exists @ 1 bar pressure & a specific temp., usually 25C (298K) • 1 bar = 100kPa • 101.325 kPa = 1 atm • So 1 bar  1 atm (an SI unit) • Example • C(s) + O2(g)  CO2(g); H = Hf = -393.5 kJ • Hf = 0 for elements in standard state

  23. Enthalpy Change for a Rxn • Must know all std molar enthalpies • Hrxn = Hfprods - Hfreactants • Given to you in a table in the back of the book • Again, keep in mind the mole ratios for each species involved!

  24. Example • Hrxn for 10.0 g of nitroglycerin? • 2C3H5(NO3)3(l)  3N2(g) + ½O2(g) + 6CO2(g) + 5H2O(g) • C3H5(NO3)3(l) = -364 kJ/mol • CO2(g) = -393.5 kJ/mol • H2O(g) = -241.8 kJ/mol

  25. Solution

  26. Practice • Determine H°rxn for: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) • Given: • NH3(g) = -45.9 kJ/mol • NO(g) = 91.3 • H2O(g) = -241.8

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