[maakii]

Hi, I was doing a question on calculating deltaG, using the formula ΔG= -RT ln K, where K is the equilibrium constant.

However, I was wondering why must K have units of (mol dm^-3) or bar^-1 in order for the calculation to work? Why can't we use normal SI units such as Pa or mol m^-3? Thanks in advance!

[dailan]

K is equilibrium constant, so it is dimensionless.
In this case, R=8.314 J.K^-1.mol^-1
T is measured in Kelvin

[Yggdrasil]

When calculating ΔG using the formula you gave, you are actually calculating the standard change in free energy (ΔG^o).  This is the free energy change when all reactants and products are at 1 mol dm^-3 concentration or at 1 bar partial pressure.

Recall that ΔG depends on the concentration of the reactants and products.  A more complete equation is as follows:

For the hypothetical reaction A --> C

ΔG = ΔG^o + RT ln ( ([C]/[C]^o) / ([A]/[A]^o) )

where [A]^o and [C]^o are the reference concentrations with which ΔG^o is calculated.  You need to divide the concentrations of A and B by reference concentrations because the argument of a transcendental function (such as a log) has to be dimensionless.

If you define ΔG^o in terms of different reference concentrations, then you could plug in different units for the concentrations you plug into K.

I hope that explanation makes sense.

[edit: fixed tags]

[maakii]

ah silly me, I forgot that ΔG^o was under standard conditions..It makes sense now thank you!