Chapter6LEcturepart2

# Should we use to make a useful table of reactions to

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should we use to make a useful table of reactions to combine to make other reactions? Use reactions that form molecules from elements. Like the CO 2 from C and O 2 we saw above and repeat here. (Or the equation to make CH 4 or the one for NH 3 .) The combustion reaction below is also a formation reaction. C(s) + O 2 (g) CO 2 (g) H = - 393.5 kJ If we use this, we do not need to bother writing the equation. We just write the value, and people can figure out what the reaction must be. H f = - 393.5 kJ This is so important that it gets a special symbol, the H f above. The subscript f says it is a formation enthalpy. It is always understood that you form one mole.

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6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies of Formation Note that if you always make 1 mole on the product side, you may need a half mole of something on the left as a reactant, like C(s) + ½O 2 (g) CO(g) H f = -110.5 kJ We did not allow this in the first four chapters, but now we do. You might now ask what is the thing telling us. It really should have a horizontal line through it, like this o. But this is a nuisance to type, so most of us just use the degree sign, unless we are making a formal book. What it means is that the enthalpy of the formation reaction is accurate at 25 C and 1 bar (or 1 atmosphere) of surrounding external pressure. You can use the value anywhere near room temperature and pressure, but only if the same physical phase applies. That is why we have to write (g), (s), (l), (aq) for every molecule. --------------------------------------------------------------
6 Copyright: 2010 Prof. Magde Chapter 6: Using Enthalpies of Formation You don’t really have to think about the underlying formation reactions. The atoms always cancel out. So there is shorthand way to do it. You look up values in Appendix B. You just use the formation enthalpies like this: 2SO 2 (g) + O 2 (g) 2SO 3 (g) H f : 2(-296.8) 0 2(-396) kJ H rxn : Products – Reactants 2(-396) – 2(-296.8) = -198.4 kJ for the reaction as written, for two moles SO 3 !!! The negative sign means the reaction heats the surroundings. It is exothermic. It gives off heat, although I don’t like using the word heat as a noun.

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6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies of Formation Note: You look up H f in Appendix B of our text. But what you calculate for your reaction is H rxn . They are totally different. Another example: CH 4 (g) + NH 3 (g) HCN(g) + 3H 2 (g) H rxn = ? -74.87 -45.9 135 3(0) kJ Products – reactants: H rxn = 135 + 3(0) – [(-74.9) + (-45.9)] = 256 kJ Note: Since H 2 (g) is the element in its standard state, it has a formation enthalpy of 0. It is already formed, so to speak. You have to be careful of the physical states.
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