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 delta*G* = delta*G*° + *RT* lnQ* and get the following:*

0 = delta*G*° + *RT* ln*K*

*Move the deltaG to the left side, we get:*

-delta*G*° = *RT* ln*K*

*Times -1 on both sides we get:*

delta*G*° = -*RT* ln*K*

*Both sides devided by -RT and that will be:*

*(delta G°)/-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 * , 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 =−2.421×10^{5}

Finally just put everything we get into the formula *K* = e^{-delta}^{G}^{°/RT :}

*K=e ^{-(−2.421×105 )/((8.314 K*J/mol) *(298K)) =}K=e^{2.421×105 )/((8.314 K*J/mol) *(298K)) }^{ } = 2.7×10^{42 }*

*Remeber K has no units*

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

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