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It makes no sense to me that these thermodynamic changes are given in kJ/mol. Reactions can have several compounds on both sides and its not clear what exactly the molar variable refers to…

I've been given a homework where this is the convention followed. My temptation is to just ignore the /mol as senseless, and proceed just considering the energy change. But I'm sure I'm wrong.

Can anyone help me?

Thanks.

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  • $\begingroup$ $\frac{KJ}{mol}$ means amount of heat [an action, e.g.: gained] per mole. I don't know why that should confuse you. Could you bring up more examples? $\endgroup$ – M.A.R. Mar 4 '15 at 18:20
  • $\begingroup$ Yeah I'm not speaking of formation reactions. Say for example: H2S+1.5 O2 -> H2O + SO2. I'm given the enthalpy as kJ/mol. $\endgroup$ – DLV Mar 4 '15 at 18:49
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It makes perfect sense that the units KJ/mol are used.

If you burn 2 cords of wood you get twice the heat as from 1 cord of wood.

Enthaply (and Gibbs free energy) of the reaction is proportional to the number of moles that react.

If the reaction is written as:

3A + 2B => 6C

Then the change per mole would mean: per 3 moles of A = per 2 moles of B = per 6 moles of C.

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  • $\begingroup$ In the case of molar entropy I have JK^-1mol^-1. This is useful because I can multiply the quantity by the number of moles I am considering. Here I can't see how I could do that. "Per mole". Per mole of what? Thanks. $\endgroup$ – DLV Mar 4 '15 at 19:35
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    $\begingroup$ So in the example in the answer, "per mole of what" is per 3 moles of A or per 2 moles of B or per 6 moles of C that react, all of which are equivalent $\endgroup$ – DavePhD Mar 4 '15 at 20:01
  • $\begingroup$ So if I were to work with 1 mole of B, I will multiply the heat of reaction by 0.5 ? $\endgroup$ – DLV Mar 5 '15 at 2:58
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    $\begingroup$ If 1 mole of B reacts, then the enthalpy change would be 0.5 of the molar enthalpy of the reaction, as you are saying, yes $\endgroup$ – DavePhD Mar 5 '15 at 3:02
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    $\begingroup$ Yes, it is somewhat confusing because each reactant and product can have a different number of moles in a given reaction $\endgroup$ – DavePhD Mar 5 '15 at 3:09

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