Finding equilibrium constant from standard free energy 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? 

$$\ce{C6H12O6(s) + 6O2(g) &lt;=&gt; 6CO2(g) + 6H2O(l)}$$

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

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 $K=5\times10^{504}$
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.