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Chapter 6: Thermochemistry

Chapter 6: Thermochemistry. CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University. Overview. Thermodynamics Summary Example Problems Hess’s Law. Thermodynamics Summary. So far we have learned the following equations. D E = q + w w = - P D V q = D E + P D V

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Chapter 6: Thermochemistry

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  1. Chapter 6: Thermochemistry CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University

  2. Overview Thermodynamics Summary Example Problems Hess’s Law

  3. Thermodynamics Summary So far we have learned the following equations. DE = q + w w = - PDV q = DE + PDV At constant volume (DV = 0):qv = DE At constant pressure:qp = DH Enthalpy change:DH = Hproducts – Hreactants

  4. Example Problem The reaction between hydrogen and oxygen to yield water vapor has DH° = –484 kJ. How much PV work is done, and what is the value of DE (in kilojoules) for the reaction of 0.50 mol of H2 with 0.25 mol of O2 at atmospheric pressure if the volume change is –5.6 L?

  5. Example Problem The explosion of 2.00 mol of solid TNT with a volume of approximately 0.274 L produces gases with a volume of 489 L at room temperature. How much PV (in kilojoules) work is done during the explosion? Assume P = 1 atm, T = 25°C. 2 C7H5N3O6(s)  12 CO(g) + 5 H2(g) + 3 N2(g) + 2 C(s)

  6. Enthalpies of Reaction: Key Concepts Reversing a reaction changes sign of DHfor reaction. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ 3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) DH = +2219 kJ Multiplying a reaction increases DHby the same factor. 3 C3H8(g) + 15 O2(g)  9 CO2(g) + 12 H2O(l) DH = –6657 kJ

  7. Example Problems How much heat (in kilojoules) is evolved or absorbed in each of the following reactions? Burning of 15.5 g of propane: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ Reaction of 4.88 g of barium hydroxide octahydrate with ammonium chloride: Ba(OH)2·8 H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(aq) + 10 H2O(l)DH = +80.3 kJ

  8. Hess’s Law 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. 3 H2(g) + N2(g)  2 NH3(g) DH° = –92.2 kJ

  9. Hess’s Law Reactants and products in individual steps can be added and subtracted to determine the overall equation. (a) 2 H2(g) + N2(g) → N2H4(g) DH°1 = ? (b) N2H4(g) + H2(g) → 2 NH3(g) DH°2 = –187.6 kJ (c) 3 H2(g) + N2(g) → 2 NH3(g) DH°3 = –92.2 kJ DH°1 = DH°3 – DH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ

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