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IBDP SL Kinetics

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- 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. Signs Exothermic Endothermic
- 3. Energy Level Diagrams
- 4. Exo and Endo Energy Level Diagramshttp://www.docbrown.info/page03/3_51energy.htm#3.
- 5. Energetics Calculation of Enthalpyhttp://www.youtube.com/watch?v=lxTmei2yrBg
- 6. Experiments Enthalpy of Neutralisation of HCl with NaOH. Enthalpy of dissolution. Enthalpy of precipitation.
- 7. Define Enthalpy of combustion Enthalpy of formation Draw energy level diagram for the process of evaporation.
- 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. 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. 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. 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. 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. Hess’ Law: An Example
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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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