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Thermochemistry

Thermochemistry. Heat. energy transferred between two objects as a result of the temperature difference between them. Temperature. A measure of kinetic energy. 1st Law of Thermodynamics. The energy of the universe is constant. i.e. the energy of the universe is conserved.

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Thermochemistry

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  1. Thermochemistry

  2. Heat • energy transferred between two objects as a result of the temperature difference between them.

  3. Temperature • A measure of kinetic energy

  4. 1st Law of Thermodynamics • The energy of the universe is constant. • i.e. the energy of the universe is conserved

  5. E = Efinal Einitial • E if energy leaves system • + E if energy enters system • Note the E of a system doesn’t depend on how system got there -- i.e. it is a state function

  6. State Function • A function or property whose value depends only on the present state (condition) of the system, not on the path used to arrive at that condition.

  7. E = q + w Heat gain or loss Work done = -PDV Matches our earlier convention that Ein is + and Eout is –

  8. Enthalpy • H = qP = E + PV • H = Hfinal Hinitial • = Hproducts Hreactants

  9. Quantity of heat supplied Some specific heats are Al 0.902 J/g oK Cu 0.385 J/g oK H2O 4.184 J/g oK Tells how much heat is required to change the temp of a substance. Temperature change (always Tf-Ti)

  10. A 55.0 g piece of metal was heated in boiling water to a temperature of 99.8oC and dropped into an insulated beaker with 225 mL of water (d = 1.00 g/ml) at 21.0 oC. The final temperature of the metal and water is 23.1oC. Calculate the specific heat of the metal assuming that no heat was lost to the surroundings.

  11. Octane, C8H18, a primary constituent of gasoline, burns in air. • C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9 H2O(l) • Suppose that a 1.00 g sample of octane is burned in a calorimeter that contains 1.20 kg of water. The temperature of the water and the bomb rises from 25.00oC to 33.20oC. If the specific heat of the bomb, Cbomb, is known to be 837 J/oC, calculate the molar heat of reaction of C8H18.

  12. A quantity of ice at 0oC is added to 90.0 g of water at 80oC. After the ice melted, the temperature of the water was 25oC. How much ice was added? • specific heat of ice 2.06 J/goC 0.91 kJ/moloC • specific heat of water 4.184 J/goC 7.54 kJ/moloC • specific heat of steam 2.0 J/goC 0.92 kJ/moloC • heat of fusion 333 J/g 6.01 kJ/mol • heat of vaporization 2226 J/g 40.67 kJ/mol

  13. 50.0 g of ice at -20.0 oC are added to 342.0 g of water at 86.0 oC. What will be the final temperature of the sample? • specific heat of ice 2.06 J/goC 0.91 kJ/moloC • specific heat of water 4.184 J/goC 7.54 kJ/moloC • specific heat of steam 2.0 J/goC 0.92 kJ/moloC • heat of fusion 333 J/g 6.01 kJ/mol • heat of vaporization 2226 J/g 40.67 kJ/mol

  14. A 33.14 g sample of copper and aluminum was heated to 119.25oC and dropped into a calorimeter containing 250.0 g of water at 21.00oC. The temperature rose to 23.05oC. Assuming no heat was lost to the surroundings, what is the percent copper in the sample?

  15. Enthalpy • Enthalpy transferred out of reactants  exothermic H =  • Enthalpy transferred into products  endothermic H = +

  16. Enthalpy • Hforward = Hreverse (For reversible reactions) • H2O(g)  H2(g) + 1/2 O2(g) H = +241.8 kJ • H2(g) + 1/2 O2(g)  H2O(g) H = 241.8 kJ

  17. Enthalpy • The H is proportional to the amount of substance undergoing change. • H2O(g)  H2(g) + 1/2 O2(g) H = +241.8 kJ • 2 H2O(g)  2 H2(g) + 1 O2(g) H = +483.6 kJ

  18. Enthalpy • The physical state of reactants and products is important. • H2O(g)  H2(g) + 1/2 O2(g) H = +241.8 kJ • H2O(l)  H2(g) + 1/2 O2(g) H = +285.8 kJ

  19. Enthalpy • Enthalpy is a state function -- it doesn’t matter how you go from one place to another -- enthalpy and enthalpy changes are the same!! • The H value is the same no matter how you get from AB

  20. Hess’s Law • The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. • Valid because enthalpy is a state function.

  21. Determine the H for the sublimation of ice to water vapor at 0oC. • H2O(s)  H2O(l) H = 6.02 kJ/reaction • H2O(l)  H2O(g) H = 40.7 kJ/reaction • ----------------------------------------------------- • H2O(s)  H2O(g) H = 46.7 kJ/reaction

  22. Calculate the enthalpy change for the formation of methane, CH4, from solid carbon (as graphite) and hydrogen gas. • C(s) + 2 H2(g)  CH4(g) • The enthalpies for the combustion of graphite, hydrogen gas and methane are given. • C(s) + O2(g)  CO2(g) 393.5 kJ • H2(g) + ½ O2(g)  H2O(l) 285.8 kJ • CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l) 890.3 kJ

  23. Calculate the enthalpy change for the reaction • S(s) + O2(g)  SO2(g) • given • 2 SO2(g) + O2(g)  2 SO3(g) H = 196 kJ • 2 S(s) + 3 O2(g)  2 SO3(g) H = 790 kJ

  24. Standard Heat of Formation • The enthalpy change, Hfo, for the formation of 1 mol of a substance in the standard state from the most stable forms of its constituent elements in their standard states.

  25. Benzene, C6H6, is an important hydrocarbon. Calculate its enthalpy of combustion; that is, find the value of Ho for the following reaction. • C6H6(l)+15/2 O2(g)  6 CO2(g)+3 H2O(l) • Given • Hfo [C6H6(l)] = +49.0 kJ/mol • Hfo [CO2(g)] = 393.5 kJ/mol • Hfo [H2O(l)] = 285.8 kJ/mol

  26. Nitroglycerin is a powerful explosive, giving four different gases when detonated. • 2 C3H5(NO3)3(l)  3 N2(g) + ½ O2(g) + 6 CO2(g) + 5 H2O(g) • Given the enthalpy of formation of nitroglycerin, Hfo, is 364 kJ/mol, calculate the energy liberated when 10.0 g of nitroglycerin is detonated.

  27. Enthalpies from Bond Energies • Calculate the enthalpy of formation of water vapor from bond energies. • 2 H2(g) + O2(g)  2 H2O(g) • (The experimental value is 241.8kJ/mol)

  28. Oxygen difluoride, OF2, is a colorless, very poisonous gas that reacts rapidly and exothermically with water vapor to produce O2 and HF. Calculate the DHof for OF2. • OF2(g) + H2O(g)  2 HF(g) + O2(g) DHorxn = -318 kJ • The heats of formation for H2O(g) and HF(g) are -241.8 kJ/mol and -271.1 kJ/mol respectively.

  29. Stoichiometry using Enthalpy • Consider the following reaction: • 2 Na(s) + Cl2(g) 2 NaCl(s) H = 821.8 kJ • Is the reaction exothermic or endothermic? • Calculate the amount of heat transferred when 8.0 g of Na(s) reacts according to this reaction.

  30. We generally expect that reactions evolving heat should proceed spontaneously and those that absorb heat should require energy to occur. • Mix barium hydroxide and ammonium chloride Ba(OH)28H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(g) + 10 H2O(l)

  31. Determination of H using Hess’s Law • Hrxn is well known for many reactions, but it is inconvenient to measure Hrxn for every reaction. • However, we can estimate Hrxn for a reaction of interest by using Hrxn values that are published for other more common reactions.  • The Standard Enthalpy of Reaction(Hrxn) of a series of reaction steps are added to lead to reaction of interest (indirect method). Standardconditions (25°C and 1.00 atm pressure). (STP for gases T= 0°C) 

  32. Hess’s Law “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” - 1840, Germain Henri Hess (1802–50), Swiss

  33. Calculation of H by Hess’s Law 3 C(graphite) + 4 H2 (g)  C3H8 (g)H= -104 3 C(graphite) + 3 O2 (g) 3 CO2 (g)H=-1181 4 H2 (g) + 2 O2 (g) 4 H2O (l)H=-1143 C3H8 (g)  3 C(graphite) + 4 H2 (g) H= +104 C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) • Appropriate set of Equations with their H values are obtained (or given), which containing chemicals in common with equation whose H is desired. • These Equations are all added to give you the desired equation. • These Equations may be reversed to give you the desired results (changing the sign of H). • You may have to multiply the equations by a factor that makes them balanced in relation to each other. • Elimination of common terms that appear on both sides of the equation .

  34. Calculation of H by Hess’s Law 3 C(graphite) + 3 O2 (g)  3 CO2 (g)H=-1181 4 H2 (g) + 2 O2 (g)  4 H2O (l)H=-1143 C3H8 (g)  3 C(graphite) + 4 H2 (g) H= +104 • Hrxn = • + 104 kJ • 1181 kJ • 1143 kJ • 2220 kJ C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

  35. Calculation of H by Hess’s Law Calculate heat of reaction W + C (graphite)  WC (s) ΔH = ? Given data: 2 W(s)+ 3 O2 (g)  2 WO3 (s) ΔH = -1680.6 kJ C (graphite) + O2(g)  CO2(g)ΔH = -393.5 kJ 2 WC (s)+ 5 O2(g)  2 WO3 (s) + CO2(g)ΔH = -2391.6 kJ ½(2 W(s) + 3 O2 (g)  2 WO3 (s) ) ½(ΔH = -1680.6 kJ) W(s) + 3/2 O2 (g)  WO3 (s) ) ΔH = -840.3 kJ C (graphite) + O2(g)  CO2(g)ΔH = -393.5 kJ ½(2 WO3 (s) + CO2(g)  2 WC (s) + 5 O2(g)) ½ (ΔH = +2391.6 kJ) WO3 (s) + CO2(g)  WC (s) + 5/2 O2(g)ΔH = + 1195.8 kJ) W + C (graphite)  WC (s) ΔH = - 38.0

  36. Hess’s Law Problem: Chloroform, CHCl3, is formed by the following reaction: Desired ΔHrxn equation: CH4 (g) + 3 Cl2 (g) → 3 HCl (g) + CHCl3 (g) Determine the enthalpy change for this reaction (ΔH°rxn), using the following: 2 C (graphite) + H2 (g) + 3Cl2 (g) → 2CHCl3 (g) ΔH°f = – 103.1 kJ/mol CH4 (g) + 2 O2 (g) → 2 H2O (l) + CO2 (g) ΔH°rxn = – 890.4 kJ/mol 2 HCl (g) → H2 (g) + Cl2 (g) ΔH°rxn = + 184.6 kJ/mol C (graphite) + O2 (g) → CO2(g) ΔH°rxn = – 393.5 kJ/mol H2 (g) + ½ O2 (g) → H2O (l)ΔH°rxn = – 285.8 kJ/mol answers: a) –103.1 kJ b) + 145.4 kJ c) – 145.4 kJ d) + 305.2 kJ e) – 305.2 kJ f) +103.1 kJ This is a hard question. To make is easer give: C (graphite) + ½ H2(g) + 3/2 Cl2(g) → CHCl3(g) ΔH°f = – 103.1 kJ/mol

  37. Methods of determining H Calorimetry (experimental) Hess’s Law: using Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps (indirect method). Standard Enthalpy of Formation (Hf ) used with Hess’s Law (direct method) Bond Energies used with Hess’s Law  Experimental data combined with theoretical concepts 

  38. (3) Determination of H using Standard Enthalpies of Formation (Hf )  Enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. StandardEnthalpy of formationHf are measured under standard conditions (25°C and 1.00 atm pressure).  C + O2 CO2∆Hf = -393.5 kJ/  

  39. Calculation of H CH4(g) + O2(g) CO2(g) + H2O(g) C + 2H2(g) CH4(g)ΔHf = -74.8 kJ/ŋ We can use Hess’s law in this way: H = nHf(products) - mHf(reactants) where n and m are the stoichiometric coefficients. C(g) + O2(g) CO2(g) ΔHf = -393.5 kJ/ŋ 2H2(g) + O2(g) 2H2O(g) ΔHf = -241.8 kJ/ŋ   - n CH4(g) + n O2(g) n CO2(g) + n H2O(g) H = [1(-393.5 kJ) + 1(-241.8 kJ)] - [1(-74.8 kJ) + 1(-0 kJ)] = - 560.5 kJ

  40. Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) H = nHf(products) - mHf(reactants) H = [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ Table of Standard Enthalpy of formation, Hf

  41. (4) Determination of H using Bond Energies • Most simply, the strength of a bond is measured by determining how much energy is required to break the bond. • This is the bond enthalpy. • The bond enthalpy for a Cl—Cl bond, D(Cl—Cl), is measured to be 242 kJ/mol.

  42. Average Bond Enthalpies (H) • Average bond enthalpies are positive, because bond breaking is an endothermic process. NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than the C—H bond in chloroform, CHCl3.

  43. Enthalpies of Reaction (H ) • Yet another way to estimate H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed. • In other words, • Hrxn = (bond enthalpies of bonds broken)  • (bond enthalpies of bonds formed)

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