= Solution
Use consistent units for the <Gibbs free energy>:
$$
\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ
=-15.0-300(0.020)=-21.0\,\mathrm{kJ\,mol^{-1}}.
$$
Then $\Delta G^\circ=-RT\ln K$ gives
$$
\boxed{K=\exp\!\left[\frac{21000}{8.314(300)}\right]=4.53\times10^3.}
$$
This is a dimensionless thermodynamic <equilibrium constant> defined through activities. For an ideal dilute $1:1$ association, the numerical concentration association constant is about $4.53\times10^3\,\mathrm{M^{-1}}$ when concentrations use the $1\,\mathrm M$ standard state. Both favourable enthalpy and positive entropy favour association here; association need not always have a negative entropy change because solvent release can outweigh the loss of translational freedom.
Back to article page
