# Correlation between the equilibrium constant and delta G 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

• I forgot to say that this calculation is done at 37 degrees celcius. But thanks, I understand it now. – Paze Mar 6 '15 at 14:48

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}