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This question is asking to calculate equilibrium constant for oxidation of glucose from thermodynamics quantities. So the first expression came to my mind was $-\Delta G^\circ=RT\ln K$. I found the value of $\Delta G^\circ=-2\,278.78$ but as I proceed I've got $\ln K=1\,161.94$ which I couldn't reach to the $K$ value. Is there any other ways to solve this question?

$$\ce{C6H12O6(s) + 6O2(g) <=> 6CO2(g) + 6H2O(l)}$$

$\Delta G^\circ_\mathrm r=-2878.78\ \mathrm{kJ}$

At equilibrium, $\Delta G=0$; $Q=K$,

$\Delta G=\Delta G^\circ+RT \ln K$

$2878.78\times10^3\ \mathrm J=\left(8.314\ \mathrm{J/(mol\ K)}\right) (298\ \mathrm K) \ln K$

$\ln K=1161.93596$

Answer from the book is $K=5\times10^{504}$

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  • $\begingroup$ Plugging the values into $K = e^{\tfrac{-\Delta G^0}{RT}}$ gives me an equilibrium constant of 2.51. I am unsure what your question is. Do you have problems solving the mathematical equation? $\endgroup$
    – Yoda
    Oct 22, 2015 at 12:51
  • $\begingroup$ @Anders MB I've posted the question together with my workouts. $\endgroup$
    – Theresa
    Oct 22, 2015 at 13:29
  • $\begingroup$ Your answer is pretty close to the book answer. When you found the change in free energy, did you use the free energy of formation for liquid water or for water vapor? $\endgroup$
    – Chet Miller
    Oct 22, 2015 at 14:19
  • $\begingroup$ @chester miller, I plugged in formation of liquid water to find the free energy change. I'm not sure if my answer is close to the book, but I've no idea on how they can reach to such a large K value because we wouldn't get an answer from natural log of 5e+504. Same goes to my own answer, there's no way to natural antilog 1161.94. So I'm wondering if there's any other way to relate free energy with equilibrium constant. $\endgroup$
    – Theresa
    Oct 22, 2015 at 17:08
  • $\begingroup$ e^1162 is about 10^504 so what's the problem really? It's terribly simple calculation $\endgroup$
    – Mithoron
    Oct 22, 2015 at 17:24

1 Answer 1

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Your problem seems to be one of calculation using large numbers, so look at it this way: $$ \ln K = 1161.93596 $$ So $$ \log_{10}K = \ln K \log_{10}(e) = 504.65\\ \Rightarrow K = 10^{504.65} = 10^{0.65}\times 10^{504} = 4.5 \times 10^{504} $$ without your calculator exploding.

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