## How to get equilibrium constant from Gibbs free energy

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

• Subject : Science
• Topic : Chemistry
• Posted By : Jason
• Created on : 03-30-2011

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