Correlation between the equilibrium constant and delta G

![enter image description here](https://i.stack.imgur.com/Nr1qi.png)


I understand this up until the last two steps. How does $\ln(K\_\mathrm{eq})$ become $2{,}303\log(K\_\mathrm{eq})$ and how does $-RT\cdot2{,}303$ become $-1{,}42$?





> 
> How does ln(Keq) become 2,303log(Keq) 
> 
> 
> 


Because $e^{2.303} = 10$ and 


$\ln(x) = \log (10^{\ln(x)})$ therefore


$\ln(x)=\log((e^{2.303})^{\ln(x)})$


$\ln(x)=\log (x^{2.303})$


$\ln(x)=2.303\log(x)$



> 
> how does -RT\*2,303 become -1,42?
> 
> 
> 


That would only be true at a particular temperature in a particular set of units


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The first step is simple: 


$$\frac{{\ln x}}{{\log x}} = \ln 10 \approx 2.303$$


thus


$$ - RT\ln {K\_{{\text{eq}}}} \approx - RT \times 2.303 \times \log {K\_{{\text{eq}}}}$$


The second step is terrible because $\text{kcal}$ is used and the units are missing:


$R = 1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$


$T = 310{\text{ K}}$ (my guess)


$$\begin{aligned}
 \Delta G = & \; {-} RT \times 2.303 \times \log {K\_{{\text{eq}}}} \\ 
 = & \; {-}1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mol}^{ - 1}} \times 310{\text{ K}} \times {\text{2}}{\text{.303}} \times \log {K\_{{\text{eq}}}} \\ 
 = & \; {-}1.42{\text{ kcal mol}^{ - 1}} \times \log {K\_{{\text{eq}}}} \\ 
\end{aligned} $$


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