1. Calorimetry and Heats of Reaction Very often, the most accurate measurements of heats of reaction are obtained using bomb calorimeter experiments. These experiments give us a q V or ∆U value (the change in internal energy of a system). For many applications its more useful to have a q P or ∆H value (the change in enthalpy of a system). (Many rxns take place at constant P.) We must be able to “convert” a ∆U value to a corresponding ∆H value.

2. Slide 2 of 57 7-6 Heats of Reaction:  U and  H Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 Reactants → Products U i U f  U = U f - U i  U = q rxn + w In a system at constant volume (bomb calorimeter):  U = q rxn + 0 = q rxn = q v But we live in a constant pressure world! How does q p relate to q v ?

3. Slide 3 of 57 Two different paths leading to the same internal energy change in a system FIGURE 7-13 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

4. Slide 4 of 57 Heats of Reaction Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 q V = q P + w We know that w = - P  V and  U = q v, therefore:  U = q P - P  V q P =  U + P  V These are all state functions, so define a new function. Let enthalpy beH = U + PV Then  H = H f – H i =  U +  (PV) If we work at constant pressure and temperature:  H =  U + P  V = q P

5. Comparing ∆H and ∆U When a physical or chemical change involves gases, the ideal gas equation, PV = nRT, is often used to calculate the difference between ∆H and ∆U. For a chemical change our initial and final states will be comprised of different substances. For a physical change (such as a phase change) this will not be the case.

6. ∆H and ∆U Difference ∆(PV) for constant P and T: ∆(PV) = P(V final - V initial ) = P∆V We could also use ∆n Gas RT. This assumes, reasonably in most cases, that the volume change for a system is likely to result mostly from a change in the amount (# moles) of gas. P∆V = P(V final - V initial ) = (n Gas,Final – n Gas,Initial )RT = ∆n Gas RT Mention T changes?

7. ∆H and ∆U Difference – Chemical Change The next slide shows how q V and q P would vary for a simple chemical change. 2 CO(g) + O 2 (g) → 2 CO 2 (g) at 298 K Here ∆n Gas T = 2mol – (2mol+1mol) = -1mol Clearly, there is no PV work at constant volume but, there is PV work at constant P. Everything else is much less clear – and requires some careful consideration!

8. Slide 8 of 57 Comparing Heats of Reaction Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 FIGURE 7-14 Comparing heats of reaction at constant volume and constant pressure for the reaction 2 CO(g) + O 2 (g) 2 CO 2 (g) q V =  U =  H - P  V = -563.5 kJ/mol w = P  V = P(V f – V i ) = RT(n f – n i ) = -2.5 kJ q P =  H = -566 kJ/mol

9. Tabulated Heats of Reaction Heats of reaction tabulations are similar to prices encountered at the supermarket fish, meat or produce counters. These are usually given as $ per kg or $. kg -1. Similarly, heats of reaction are usually tabulated as kJ/mol or kJ. mol -1 although units such as kJ/kg or kJ/L can be more useful. When writing thermochemical eqtns for the combustion of hydrocarbons a “1” usually appears before C X H Y. Why?

10. Slide 10 of 57 7-9 Fuels as Sources of Energy Fossil fuels. – Combustion is exothermic. – Non-renewable resource. – Environmental impact. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

11. Combustion Reactions – Thermochemical Equations Class example: The heats (enthalpies) of combustion of propane, C 3 H 8 (g), and butane, C 4 H 10 (l), are -2220 kJ/mol and -2877 kJ/mol respectively. Write complete and balanced thermochemical equations to represent the two combustion reactions.

12. Bomb Calorimetry, ∆U and ∆H

13. Copyright  2011 Pearson Canada Inc. 7 - 13

15. Enthalpy and Physical Changes Heat flows are associated with chemical reactions and temperature changes of a solid, a liquid and a gas. Heat flows are also associated with phase changes. In Newfoundland and Labrador we know, for example, that heat must be supplied to melt ice in the spring. Rowers (and others) must be patient! Heat + H 2 O(s) → H 2 O(l) Endothermic process

16. Slide 16 of 57 Enthalpy Change (  ) Accompanying a Change in State of Matter Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 H 2 O (l) → H 2 O(g)  H = 44.0 kJ at 298 K Molar enthalpy of vaporization: Molar enthalpy of fusion: H 2 O (s) → H 2 O(l)  H = 6.01 kJ at 273.15 K

17. George Street Phase Change?

18. Phase Changes and PV Work (Horton’s and Hockey?) Class example: Calculate the amount of PV work done when: (a) one mole of liquid water at 373 K is transformed into steam at a pressure of 1.000 atm. Assume that the steam behaves as an ideal gas. (b) one mole of ice at 273 K is melted at 1.000 atm pressure. What two pieces of information are missing?

20. PVPV

21. Slide 21 of 57 Standard States and Standard Enthalpy Changes Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 Define a particular state as a standard state. Standard enthalpy of reaction,  H ° – The enthalpy change of a reaction in which all reactants and products are in their standard states. Standard State – The pure element or compound at a pressure of 1 bar and at the temperature of interest.

22. Slide 22 of 57 Enthalpy Diagrams FIGURE 7-15 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

23. Enthalpy Diagrams The energy changes associated with chemical reactions can be represented using either balanced thermochemical equations or enthalpy diagrams. In either case the simplest examples involve the formation of a single pure compound from its constituent elements. H 2 (g) + ½ O 2 (g) → H 2 O(l) ∆H o f,298K = -286 kJ

24. Water Electrolysis The reaction on the previous slide is highly exothermic and does not pollute - an ideal way to heat our homes if there was a readily available source of H 2 (g). The opposite reaction must be endothermic (Conservation of Energy). If we supply energy: H 2 O(l) → H 2 (g) + ½ O 2 (g) ∆H o 298K = +286 kJ The reaction can be used to produce both elemental hydrogen and oxygen via electrolysis.

25. H 2 O(l) H 2 (g) + ½ O 2 (g) H (Enthalpy) (kJ∙mol -1 ) ΔH 0 = - 286 kJ Reactants Product Combustion of Hydrogen/Formation of Water Exothermic Reaction ! (Apollo 13/Fuel Cells)

26. H 2 O(l) H 2 (g) + ½ O 2 (g) H (Enthalpy) (kJ.mol -1 ) ΔH 0 = + 286 kJ Products Reactant Electrolysis of Water/Production of Hydrogen Endothermic Reaction !