## Units for calculating standard reaction enthalpies...

Regina Chi 2K
Posts: 51
Joined: Fri Sep 26, 2014 2:02 pm

### Units for calculating standard reaction enthalpies...

I was a bit confused on what units to use when (such as kJ versus kJ/mol). I know that for calculating regular reactions that deal with entropy and internal energy, we would just have J/K and J respectively. However, I need clarification on what units to use when calculating the following:

1. Hess's Law
2. Standard Reaction Enthalpies
3. Bond Enthalpies
4. Standard Reaction Entropies
5. Standard Reaction Gibb's Free Energy
6. Reaction Enthalpies

I know that determining the units may depend on what the question is asking. If that is the case, it would be great if specific situations are provided! :)

Anuk Burli 2C
Posts: 29
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Units for calculating standard reaction enthalpies...

I know Hess's Law is generally given in kJ, bond enthalpies tend to be given in kj/mol, I think Gibb's free energy is given in KJ in a hess's law type of calculation, standard reaction enthalpies in kj/mol, standard reaction entropies in j/mol.

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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### Re: Units for calculating standard reaction enthalpies...

There is no fixed "method" to determine the units other than to keep track of them. Values in the appendix are always per-mole, so when you add and subtract them, you get kJ/mol (J/mol*K for entropy)

In a Hess's Law type problem, on the other hand, if the energy of reaction is written for example:
2A + 3B ---> C, deltaG = 60 kJ
This means:
30 kJ/mol A consumed
20 kJ/mol B consumed or
60 kJ/mol C produced

So the "kJ/mol" value is only meaningful with respect to a substance, which will usually be made obvious in the problem. Eg. what is deltaG upon consuming 5 moles of A? We pick the energy with respect to A and do (5 mol)*(30 kJ/mol) = 150 kJ

Always write your units in your calculation. If you started with kJ/mol but somewhere multiplied by the number of moles given, the answer is kJ. If you didn't multiply by any amount, it stays kJ/mol.

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