0
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • $\begingroup$ $\frac{KJ}{mol}$ 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, 2015 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, 2015 at 18:49

1 Answer 1

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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, 2015 at 19:35
  • 1
    $\begingroup$ 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, 2015 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, 2015 at 2:58
  • 1
    $\begingroup$ 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, 2015 at 3:02
  • 1
    $\begingroup$ 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, 2015 at 3:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.