Page **1** of **1**

### Units for Gibbs Free Energy

Posted: **Mon Feb 12, 2018 6:24 pm**

by **Ashley Davis 1I**

So when using difference of sums to calculate something such as enthalpy, we multiplied enthalpies of formation by however many moles of each molecule were present in the equation. Like if there were 2 moles of CO

, we would enter 2 mol(-394 kJ/mol), and the mol would cancel out, leaving us with a final answer in kJ. Is this not true for Gibbs Free Energy? I noticed that when calculating K (using the equation where ΔGº = -RTlnK), my units weren't cancelling out. So if we're calculating ΔGº using difference of sums, and there are more than 1 mol of something in the reaction, do we not multiply it by the number of moles?

If this question doesn't make sense, please let me know!

### Re: Units for Gibbs Free Energy

Posted: **Mon Feb 12, 2018 7:04 pm**

by **Nicole Anisgard Parra 2H**

When calculating K using ΔGº = -nRTlnK, your units should cancel out, as K is unitless. ΔGº utilizes units of joules (if your value is in kilojoules from a prior calculation, you need to concert it to joules), n is units of moles, R is units of joules per kelvin per mole (J/K*n), and T utilizes units of kelvin. Thus, when you are calculating K, all of your units should cancel out :) hope that helps!

### Re: Units for Gibbs Free Energy

Posted: **Mon Feb 12, 2018 8:38 pm**

by **Ashley Davis 1I**

Oh, sorry, I actually meant for ΔGº = RTlnK. because then where would the moles cancel out for the R constant?

### Re: Units for Gibbs Free Energy

Posted: **Tue Feb 13, 2018 7:12 pm**

by **Nicole Anisgard Parra 2H**

Oh! Your units for delta G should be in kilojoules or joules per mole, as it is calculated from a delta H and a delta S value that is written as kj or j per mol. This would cancel out with the mole unit in R :) sorry about that, hope that helps