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# Energetics1

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### Energetics1

1. 1. Definitions Standard Enthalpy Change  the change of enthalpy that accompanies the formation of 1 mole of a substance in its standard state from its constituent elements in their standard states. Exothermic Rxn  Typical examples; combustion, neutralisation, respiration Endothermic Rxn  Typical examples; melting, evaporating, photosynthesis.
2. 2. Signs  Exothermic Endothermic
3. 3. Energy Level Diagrams
4. 4. Exo and Endo Energy Level Diagramshttp://www.docbrown.info/page03/3_51energy.htm#3.
5. 5. Energetics Calculation of Enthalpyhttp://www.youtube.com/watch?v=lxTmei2yrBg
6. 6. Experiments Enthalpy of Neutralisation of HCl with NaOH. Enthalpy of dissolution. Enthalpy of precipitation.
7. 7. Define Enthalpy of combustion Enthalpy of formation Draw energy level diagram for the process of evaporation.
8. 8. Hess’s Law Students should be able to use simple enthalpy cycles and enthalpy level diagrams and to manipulate equations. Students will not be required to state Hess’s law…..but I’ll tell you anyway  Energy can not be created or destroyed. It can only be converted from one form to another – first law of thermodynamics
9. 9. Conventions -ve for exothermic +ve for endothermic If forward reaction is exo the reverse reaction is endo and of identical magnitude. Hess’s Law states that the enthalpy change for a reaction is independent of the route the reaction takes. The overall enthalpy change depends only on the initial and final stages. Direct measurement of enthalpy is impossible.
10. 10. Enthalpy types Combustion  Energy released when one mole of a compound is burned in excess oxygen. Formation  Energy change when one mole of a compound is formed under standard conditions from its constituent elements. Bond Enthalpies – we will discuss later 
11. 11. Hess’ Law Defined Hess’ Law: H for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate H for a reaction.
12. 12. Shortcuts  Learn Them  Enthalpies of formation :  ΔH = Σproducts – Σ reactants Enthalpies of combustion :  ΔH = Σreactants– Σ products Average Bond Enthalpies  Δ H = [bonds broken] – [bonds made]
13. 13. Hess’ Law: An Example
14. 14. Using Hess’ Law  When calculating HN2 (g) + 2O2 (g) 2NO2 (g) for a chemical reaction as a single step, we can use 2NO2 (g) combinations of reactions as q “pathways” to N2 (g) + 2O2 (g) determine H for our “single step” reaction.
15. 15. Example (cont.) Our reaction of interest is: N2(g) + 2O2(g) 2NO2(g) H = 68 kJ • This reaction can also be carried out in two steps: N2 (g) + O2 (g) 2NO(g) H = 180 kJ 2NO (g) + O2 (g) 2NO2(g) H = -112 kJ
16. 16. Example (cont.) If we take the previous two reactions and add them, we get the original reaction of interest: N2 (g) + O2 (g) 2NO(g) H = 180 kJ2NO (g) + O2 (g) 2NO2(g) H = -112 kJ N2 (g) + 2O2 (g) 2NO2(g) H= 68 kJ
17. 17. Changes in Enthalpy Consider the following expression for a chemical process: H = Hproducts - Hreactants If H >0, then qp>0. The reaction is endothermic If H <0, then qp<0. The reaction is exothermic
18. 18. Example (cont.) Note the important things about this example, the sum of H for the two reaction steps is equal to the H for the reaction of interest. We can combine reactions of known H to determine the H for the “combined” reaction.
19. 19. Hess’ Law: Details Once can always reverse the direction of a reaction when making a combined reaction. When you do this, the sign of H changes. N2(g) + 2O2(g) 2NO2(g) H = 68 kJ 2NO2(g) N2(g) + 2O2(g) H = -68 kJ
20. 20. Details (cont.) The magnitude of H is directly proportional to the quantities involved (it is an “extensive” quantity). As such, if the coefficients of a reaction are multiplied by a constant, the value of H is also multiplied by the same integer. N2(g) + 2O2(g) 2NO2(g) H = 68 kJ N2(g) + 4O2(g) 4NO2(g) H = 136 kJ
21. 21. Using Hess’ Law When trying to combine reactions to form a reaction of interest, one usually works backwards from the reaction of interest. Example: What is H for the following reaction? 3C (gr) + 4H2 (g) C3H8 (g)
22. 22. Example (cont.) 3C (gr) + 4H2 (g) C3H8 (g) H=? • You’re given the following reactions: C (gr) + O2 (g) CO2 (g) H = -394 kJC3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220 H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ
23. 23. Example (cont.) Step 1. Only reaction 1 has C (gr). Therefore, we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation. Initial: C (gr) + O2 (g) CO2 (g) H = -394 kJ Final: 3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ
24. 24. Example (cont.)  Step 2. To get C3H8 on the product side of the reaction, we need to reverse reaction 2. Initial:C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220 Final:3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220
25. 25. Example (cont.)  Step 3: Add two “new” reactions together to see what is left: 3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220 2 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 k
26. 26. Example (cont.) Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ 3C (gr) + 4H2 (g) C3H8 (g) Need to multiply second reaction by 4
27. 27. Example (cont.) Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ 3C (gr) + 4H2 (g) C3H8 (g)
28. 28. Example (cont.)• Step 4 (cont.): 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ 3C (gr) + 4H2 (g) C3H8 (g) H = -106 kJ
29. 29. Changes in Enthalpy Consider the following expression for a chemical process: H = Hproducts - Hreactants If H >0, then qp>0. The reaction is endothermic If H <0, then qp<0. The reaction is exothermic
30. 30. Another Example Calculate H for the following reaction:H2(g) + Cl2(g) 2HCl(g) Given the following:NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ N2 (g) + 3H2 (g) 2NH3 (g) H = -92 kJ N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ
31. 31. Another Example (cont.) Step 1: Only the first reaction contains the product of interest (HCl). Therefore, reverse the reaction and multiply by 2 to get stoichiometry correct. NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ 2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = 352 kJ
32. 32. Another Example (cont.) Step 2. Need Cl2 as a reactant, therefore, add reaction 3 to result from step 1 and see what is left. 2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = 352 kJ N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ
33. 33. Another Example (cont.)  Step 3. Use remaining known reaction in combination with the result from Step 2 to get final reaction.N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ ( N2 (g) + 3H2(g) 2NH3(g) H = -92 kJ) H2(g) + Cl2(g) 2HCl(g) H=? Need to take middle reaction and reverse it
34. 34. Another Example (cont.)  Step 3. Use remaining known reaction in combination with the result from Step 2 to get final reaction.N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ 1 2NH3(g) 3H2 (g) + N2 (g) H = +92 kJ H2(g) + Cl2(g) 2HCl(g) H = -185 kJ
35. 35. Changes in Enthalpy Consider the following expression for a chemical process: H = Hproducts - Hreactants If H >0, then qp>0. The reaction is endothermic If H <0, then qp<0. The reaction is exothermic
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