The Gibbs energy change for a reaction at standard conditions, $\Delta_r G^\theta$, can be calculated using the equation:

$ \Delta \mathrm{G}^\theta = -\mathrm{RT} \ln \mathrm{K}_{\mathrm{eq}} $

Where $\mathrm{K}_{\mathrm{eq}}$ is the equilibrium constant given by $\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}$. For the reaction:

$ \mathrm{A + B} \rightleftharpoons \mathrm{C + D} $

The forward rate constant, $k_f$, is $10^3$, and the reverse rate constant, $k_r$, is $10^2$. Therefore,

$ \mathrm{K}_{\mathrm{eq}} = \frac{10^3}{10^2} = 10 $

Substitute the values into the Gibbs energy change formula:

$ \Delta \mathrm{G}^\theta = -\mathrm{RT} \ln 10 $

where $R = 8.3 \, \mathrm{J \, K^{-1} \, mol^{-1}}$ and $T = 300 \, \mathrm{K}$ (since the temperature is $27^\circ \mathrm{C}$ which converts to 300 K). The natural logarithm of 10 is approximately 2.3.

$ \Delta \mathrm{G}^\theta = -(8.3 \times 300 \times 2.3) $

Calculating gives:

$ \Delta \mathrm{G}^\theta = -5711 \, \mathrm{J \, mol^{-1}} $

Converting to kJ:

$ \Delta \mathrm{G}^\theta = -5.7 \, \mathrm{kJ \, mol^{-1}} $

Rounding to the nearest integer, the standard Gibb's energy change is approximately $-6 \, \mathrm{kJ \, mol^{-1}}$.