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 \rm C_2H_2({g})+2H_2({g})\rightleftharpoons C_2H_6({g}) , what is the what is the value of K for this reaction if  \Delta G^\circ_{\rm f}\rm C_2H_2({g})=209.2\rm kJ/mol,\Delta G^\circ_{\rm f}\rm H_2({g})=0 \rm kJ/mol, and \Delta G^\circ_{\rm f}\rm C_2H_6({g})=-32.89\rm kJ/mol at 298K?


In this question, we are given \Delta G^\circ_{\rm f} of all the reactants and product, so we need to find the total \Delta G^\circ_{\rm f} first through the formula \Delta G^\circ = \Delta G^\circ_{\rm f}{\rm (products)}-\Delta G^\circ_{\rm f}{\rm (reactants)} 

\Delta G^\circ  =  -32.89\rm kJ/mol - (2*0\rm kJ/mol + 209.2\rm kJ/mol) =-242.1  \rm kJ   (Be careful about significant figures)

However, since R= 8.314 K*J/mol,  we have to convert the \Delta G^\circ we get from kJ to J in order to match the units with R:

\Delta G^\circ  =-242.1  \rm kJ=−2.421×105  \rm J

Finally just put everything we get into the formula K = e-deltaG°/RT  :

K=e-(−2.421×105  \rm J)/((8.314 K*J/mol) *(298K)) =K=e2.421×105  \rm J)/((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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