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Enthalpy C 6 H 12 O 6 (s) + 6O 2 (g) --> 6CO 2 (g) + 6H 2 O(l) + 2803 kJ

Enthalpy C 6 H 12 O 6 (s) + 6O 2 (g) --> 6CO 2 (g) + 6H 2 O(l) + 2803 kJ 2C 57 H 110 O 6 + 163O 2 (g) --> 114 CO 2 (g) + 110 H 2 O(l) + 75,520 kJ The heat lost or gained by a system undergoing a process at constant pressure is related to the change in ENTHALPY ( D H ) of the system

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Enthalpy C 6 H 12 O 6 (s) + 6O 2 (g) --> 6CO 2 (g) + 6H 2 O(l) + 2803 kJ

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  1. Enthalpy C6H12O6(s) + 6O2 (g) --> 6CO2 (g) + 6H2O(l) + 2803 kJ 2C57H110O6 + 163O2 (g) --> 114 CO2(g) + 110 H2O(l) + 75,520 kJ The heat lost or gained by a system undergoing a process at constant pressure is related to the change in ENTHALPY (DH) of the system DH = Hfinal - Hinitial = qp Note: DH is a state function Most physical and chemical changes take place under the constant pressure of the Earth’s atmosphere.

  2. For an exothermic process at constant pressure DH < 0 For an endothermic process at constant pressure DH > 0 DH < 0 DH > 0

  3. Enthalpy, H is defined as H = E + PV For a system undergoing pressure-volume work, the change in enthalpy in the system is DH = DE + DPV If the pressure is a constant external pressure, Pext, DH = DE + PextDV From the 1st law: DE = q + w For constant P: DH = qp + w + PextDV Since w = - PextDV DH = qp - PextDV + PextDV = qp

  4. If volume is held constant, no pressure-volume work can be done on that system or by that system. work = - PextDV = 0 at constant volume DE = q + w, For constant volume processes, DE = qv A constant pressure calorimeter measures DH A constant volume calorimeter (like a bomb calorimeter) measures DE.

  5. Relationship between DH and DE C(s) + 1/2 O2(g) --> CO(g) DH = -110.5 kJ Determine the change in internal energy accompanying this reaction. DH = DE + D (PV) DE = DH - D (PV) D(PV) = D (nRT) = RT(Dng) DE = DH - RT(Dng) or DH = DE + RTDng For this reaction Dng = 0.5 mol; hence DE = -111.7 kJ For reactions that do not involve gases; DH ≈ DE

  6. Enthalpy of Physical Change Phase transitions involve change in energy Vaporization is endothermic; condensation exothermic. Since phase transitions typically take place at constant pressure, the heat transfer is the change in enthalpy. DHvap = Hvapor - Hliquid liquid --> gas DHfusion = Hliquid - Hsolid solid --> liquid DHfreezing = - DHfusion

  7. DHforward = - DHreverse

  8. DHsublimation = DHfusion + DHvaporization

  9. Heating curve of water H2O(s) --> H2O(l) DH = 6.01 kJ H2O(l) --> H2O(g) DH = 40.7 kJ

  10. Enthalpy of Chemical Change The enthalpy change for a chemical reaction is given by DH = SH(products) - SH(reactants) For example: 2H2(g) + O2 (g) --> 2H2O(g) DH = - 483.6 kJ Thermochemical reaction

  11. In a chemical reaction, the enthalpy change during the reaction indicates whether the reaction releases energy or consumes energy. If DH < 0, the reaction releases heat and is EXOTHERMIC If DH > 0, the reaction absorbs heat and is ENDOTHERMIC

  12. The magnitude of DH for a reaction is directly proportional to the amount of reactants consumed by the reaction. CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l) DH = -890 kJ Calculate the amount of heat that would be released when 4.50 g of CH4(g) is burned in an oxygen atmosphere at constant pressure. 4.50g CH4 => 0.28 mole CH4 => (0.28 mol CH4)(-890 kJ/mol CH4) = -250 kJ

  13. The enthalpy change for a reaction is equal in magnitude but opposite in sign to DH for the reverse reaction CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l) DH = -890 kJ CO2(g) + 2H2O(l) --> CH4(g) + 2 O2(g) DH = +890 kJ The enthalpy change for a reaction depends on the phase of the reactants and products. 1) CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l) DH = -890 kJ 2) CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(g) DH = -802 kJ 3) 2H2O(g) --> 2H2O(l) DH = -88kJ

  14. Standard Reaction Enthalpies CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l) DH = -890 kJ CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(g) DH = -802 kJ

  15. In order to compare DH’s accompanying reactions need to define a standard state Standard State: reactant and products in their pure forms at a pressure of 1 atm (or 1 bar). A solute in a liquid solution is in its standard state when its concentration is 1 mol L-1 While temperature is not part of the definition of the standard state, standard reaction enthalpies are usually reported for a temperature of 298.15 K.

  16. Standard Reaction Enthalpies (DHo): enthalpy change accompanying a reaction when reactants in their standard states change to products in their standard states. CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l) DHo = -890 kJ

  17. Hess’s Law If a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of the enthalpy change for the individual steps.

  18. (2) CO(g) +O2 (g) --> CO2 (g) DH = -283.0 kJ (3) C(s) +O2 (g) --> CO (g) DH = ? 1 1 2 2 Calculate the enthalpy change of combustion of C to CO The enthalpy of combustion of C to CO2 is -393.5 kJ/mol, and the enthalpy of combustion of CO to CO2 is -283.0 kJ/mol CO. (1) C(s) + O2 (g) --> CO2 (g) DH = -393.5 kJ

  19. CO2 (g) --> CO(g) +O2 (g) DH = 283.0 kJ C(s) +O2 (g) --> CO (g) DH = -110.5 kJ 1 1 2 2 C(s) + O2 (g) --> CO2 (g) DH = -393.5 kJ

  20. Standard Enthalpy of Combustion Change in enthalpy per mole of a substance that is burned in a combustion reaction under standard conditions

  21. Calculate the mass of propane that you would need to burn to obtain 350 kJ of heat which is just enough energy to heat I L of water from room temperature (20oC) to boiling at sea level. Assume that all the heat generated is absorbed by the water. C3H8(g) + 5 O2(g) --> 3 CO2(g) + 4H2O(l) DHo = -2220 kJ Combustion of 1 mole of propane generates 2220 kJ of heat Mass or propane required = (350 kJ) (1 mole C3H8 /2220kJ) (44.09 g/mol) = 6.95 g

  22. If you were to use butane instead of propane in the previous example, how much butane (in grams) would you need? 2 C4H10(g) + 13 O2(g) --> 8 CO2(g) + 10 H2O(l) DHo = -5756 kJ Answer: 7.07 g If you were to use ethanol instead of propane in the previous example, how much ethanol (in grams) would you need? C2H5OH(l) + 3 O2(g) --> 2 CO2(g) + 3 H2O(l) DHo = -1368 kJ Answer: 11.8 g

  23. Enthalpies of Formation The enthalpy of formation, DHf, or heat of formation, is defined as the change in enthalpy when one mole of a compound is formed from its stable elements. The standard enthalpy of formation (DHfo) of a compound is defined as the enthalpy change for the reaction that forms 1 mole of compound from its elements, with all substances in their standard states. 2C(s) + 1/2 O2(g) + 3 H2 (g) --> C2H5OH(l) DHfo = -277.69 kJ

  24. Standard enthalpies of formation of substance can be used to determine standard reaction enthalpies.

  25. The standard enthalpy of formation of the most stable form of an element under standard conditions is ZERO. DHfo for C(graphite), H2(g), O2(g) are zero The stoichiometry for formation reactions indicate the formation of 1 mole of the desired compound, hence enthalpies of formation are always listed as kJ/mol.

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