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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=e^y )

K = e^-delta^G^°/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?

Solution:

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 $\Delta G^\circ = \Delta G^\circ_{\rm f}{\rm (products)}-\Delta G^\circ_{\rm f}{\rm (reactants)}$