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Lecture 19

Lecture 19. Enthalpy of Chemical Change 6.13-6.18 8-October Assigned HW 6.46, 6.50, 6.52, 6.60, 6.62, 6.66, 6.68, 6.70, 6.74, 6.76 Due: Monday 18-Oct. Review. At constant volume, q = Δ U At constant pressure, q = Δ H Enthalpy ( Δ H) is the heat term we care about the most.

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Lecture 19

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  1. Lecture 19 Enthalpy of Chemical Change 6.13-6.18 8-October Assigned HW 6.46, 6.50, 6.52, 6.60, 6.62, 6.66, 6.68, 6.70, 6.74, 6.76 Due: Monday 18-Oct

  2. Review • At constant volume, q = ΔU • At constant pressure, q = ΔH • Enthalpy (ΔH) is the heat term we care about the most. • Molar heat capacities – predict how much energy is needed to heat a compound in the same state • Changes of state take a lot of energy • Fusion, freezing, vaporization, condensation, sublimation and deposition

  3. Reaction Enthalpies A complete thermochemical equation contains all chemical information as well as thermodynamic information CH4(g) + 2O2(g) 2H2O(l) + CO2(g) ΔH = -890 kJ Is the reaction Endothermic or Exothermic? How much heat is produced from every mole of O2? How much heat would be produced if 2 moles of methane was combusted? 2CH4(g) + 4O2(g) 4H2O(l) + 2CO2(g) ΔH = 2 [CH4(g) + 2O2(g) 2H2O(l) + CO2(g)] 2[ΔH] = 2 (-890kJ)

  4. Reaction Enthalpies When 0.231g of phosphorus reacts with chlorine to form phosphorus trichloride in a constant pressure calorimeter (C = 216 J K-1), the temperature rises from 20 °C to 31.06 °C. Write the thermochemical equation. Need balanced chemical reaction and enthalpy thermometer Heat flow (q) Constant pressure reaction chamber

  5. Reaction Energies – ΔH vs. ΔU It is at times necessary to convert between these State Functions 1. If a reaction takes place ONLY in the liquid or solid phases, very little change in volume will be observed 2. If a reaction significantly alters the number of molecules in the gas phase, we must consider PΔV.

  6. Reaction Energies – ΔH vs. ΔU 1.00 mole of Adenine, C5H5N5, is combusted in a constant volume calorimeter at 298K. 2559 kJ of heat is produced. Calculate ΔH for this reaction.

  7. Standard Reaction Enthalpies Reaction enthalpies depend on the conditions at which the reaction is carried out CH4(g) + 2O2(g) 2H2O(l) + CO2(g) ΔH° = -890 kJ CH4(g) + 2O2(g) 2H2O(g) + CO2(g) ΔH° = -802 kJ To alleviate confusion in the scientific community, we define standards states and standard reaction enthalpies: Standard State: the pure form of a compound at exactly 1 bar Standard Reaction Enthalpy (ΔH°): the enthalpy when reactants in their standard states turn into products in their standard states. There is no standard temperature – although commonly reported at 298.15K

  8. Standard Reaction Enthalpies CH4(g) + 2O2(g) 2H2O(l) + CO2(g) ΔH° = -890 kJ CH4(g) + 2O2(g) 2H2O(g) + CO2(g) ΔH° = -802 kJ

  9. Hess’s Law Since Enthalpy is a State Function (path independent), we can determine reaction enthalpies from individual reactions that sum to the desired reaction. A + B  C A + 2B  D C + B  D

  10. Hess’s Law Since Enthalpy is a State Function (path independent), we can determine reaction enthalpies from individual reactions that sum to the desired reaction. CH4(g) + 2O2(g) 2H2O(l) + CO2(g) ΔH° = -890 kJ H2O(l) H2O(g) ΔH°vap = 44 kJ H2O(l) H2O(g) ΔH°vap = 44 kJ CH4(g) + 2O2(g) 2H2O(g) + CO2(g) ΔH° =

  11. Hess’s Law Since Enthalpy is a State Function (path independent), we can determine reaction enthalpies from individual reactions that sum to the desired reaction. General Rules: Multiply ΔH by whatever coefficient you multiply the reaction by If you invert the reaction (products become reactants), switch the sign of ΔH. H2O(l) H2O(g) ΔH°vap = 44 kJ 2H2O(l) 2H2O(g) ΔH°vap = 88 kJ H2O(l) H2O(g) ΔH°vap = 44 kJ H2O(g)  H2O(l) ΔH°vap = -44 kJ

  12. Hess’s Law – Examples Determine the ΔH° for the oxidation of C  CO2 C(s) + O2(g) CO(g) ΔH° = -110 kJ CO(g)+ O2(g) CO2(g) ΔH° = -283 kJ C(s) + O2(g) CO2(g) ΔH° =

  13. Hess’s Law – Examples The synthesis of propane from C(s) and H2(g) is difficult to measure. However, these three thermochemical reactions are known. Determine the ΔH° for the reaction. 3C(s) + 4 H2(g) C3H8(g) ΔH° = C3H8(g) + 5O2(g) 4 H2O(l) + 3CO2(g) ΔH° = -2220 kJ ΔH° = -394 kJ C(s) + O2(g) CO2(g) ΔH° = -286 kJ H2(g) + O2(g) H2O(l)

  14. Standard Combustion Enthalpies Note that ΔHc° scales with the number and type of bonds that are broken

  15. Standard Formation Enthalpies (ΔHf°) The enthalpy that it takes to create a compound from it’s elements in their MOST STABLE FORM 2C(gr) + 3H2(g) + O2(g) C2H5OH(l) ΔHf° = -277.69 kJ ΔHf° = 0

  16. Standard Formation Enthalpies (ΔHf°) The enthalpy of a reaction can be determined from ΔHf° of the products relative to the reactants 2C(gr) + 3H2(g) + O2(g) C2H5OH(l) ΔH° = -277.69 kJ

  17. Standard Formation Enthalpies (ΔHf°) 2 NH2CH2COOH(s) + 3 O2(g) 3 CO2(g) + 3 H2O(l) + H2NCONH2(s)

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