Chapter6LEcturepart2

# When you look up hcn there are values for g and for l

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When you look up HCN there are values for (g) and for (l) and for (aq). They are all different. What will you get on an exam? You might get values you need, but more likely you will get a photocopy of part of the table in Appendix B.

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6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies of Formation Another example: CH 4 (g) + 4Cl 2 (g) CCl 4 (l) + 4HCl(g) H rxn = ? -75 4(0) -139 4(-92) H rxn = -139 +4(-92) – [-75 + 4(0)] = -432 kJ Note: 1. You don’t need to waste time looking up Cl 2 (g) if you are sure that (g) is the standard state. 2. The CCl 4 (l) is what carbon tetrachloride is at standard conditions. However, you can find values for other states, notably (g). They tell you what is the value if you add the liquid value plus the enthalpy of vaporization at that T. (It will evaporate even at room T; you don’t have to go to the boiling T.) So be careful to check the state in the equation. And C can be graphite or diamond. Assume graphite, which has 0 kJ. 3. Cl 2 is an oxidizing agent, like O 2 . Check that 1C goes from -4 to +4, and 8Cl go from 0 to -1. Many oxidation reactions are very exothermic.
6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies of Formation Another example of using H f Where does H f come from? You burn C 3 H 6 O (propanone) and find: C 3 H 6 O(l) + 4O 2 (g) 3CO 2 (g) + 3H 2 O(l) H rxn = -1791 kJ But ? 4(0) 3(-393.5) 3(-285.8) Add products, subtract reactants, 3(-393.5) + 3(-285.8) – [ H f + 4(0)] = -1791 to find that the ? = -247 kJ = H f of C 3 H 6 O So if you have some H f you can build up huge tables, much bigger than our Appendix B. But where do those H rxn come from?

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6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies of Reaction How much are the surroundings heated when a mixture of 10.0 g of NH 3 and 20.0 g of O 2 react to form NO and H 2 O(g), assuming we start and end at standard conditions. The water vapor is not the standard state of the element, but that is OK. It is just humidity in air. Balance the equation first: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) H f : 4(-46) 5(0) 4(90) 6(-242) Products – reactants = -908 kJ for the reaction as written, with all those moles. How many moles did we actually react? What was the limiting reagent? 3 2 2 3 3 2 2 2 1mole 10.0 g NH 0.588mole 17.0g 1mole 20.0gO 0.625mole 32.0g But we need 5 moles O for each 4 mole NH . For0.588mole NH weneed1.25 0.588 0.753of O SotheO islimiting.Therefore,theheatingis 908k 0.625moleO 2 J 114kJenergyreleasedasheat 5moleO  
6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies of Reaction We mixed 10.0 g of NH 3 and 20.0 g of O 2 to form NO and H 2 O(g). We had some NH 3 left over. We calculated that we had -114 kJ of heating of the surroundings. But actually that was what we measured. How did we do that? We measured the T increase of water (and the container it was in).

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