We know that Gibbs free energy (G) equals H - T*S (G=H-T*S), but how is G related to equilibrium constant k?
Since at equlibrium condition, k=Q (Q is the reaction quotient), and deltaG=0 , we can put all of these numbers into the formula deltaG = deltaG° + RT lnQ and get the following:
0 = deltaG° + RT lnK
Move the deltaG to the left side, we get:
-deltaG° = RT lnK
Times -1 on both sides we get:
deltaG° = -RT lnK
Both sides devided by -RT and that will be:
(deltaG°)/-RT = lnK
OK, to lnK, we can use logarithm formula and get: ( This is just like y=lnX, X=ey )
K = e-deltaG°/RT
We have establish this equation to find equilibrium constant k when deltaG and temperature at equilibrium is given, so let's do a practice:
In the reaction , what is the what is the value of K for this reaction if
=209.2
,
=0
, and
=
32.89
at 298K?
Solution:
In this question, we are given of all the reactants and product, so we need to find the total
first through the formula
= -32.89
- (2*0
+ 209.2
) =-242.1
(Be careful about significant figures)
However, since R= 8.314 K*J/mol, we have to convert the we get from kJ to J in order to match the units with R:
=-242.1
=−2.421×105
Finally just put everything we get into the formula K = e-deltaG°/RT :
K=e-(−2.421×105 )/((8.314 K*J/mol) *(298K)) =K=e2.421×105
)/((8.314 K*J/mol) *(298K)) = 2.7×1042
Remeber K has no units