- Using your first approach, van't Hoff :
$$ \begin{eqnarray} && \ln \frac{k_2}{k_1} = -\frac{∆H^\circ}{R} (\frac{1}{T_2} - \frac{1}{T_1}) \end{eqnarray} $$
For an exothermic reaction $\Delta H <0$
Now , Let $T_2>T_1$ therefore $K_2<K_1$ i.e. equilibrium constant decreases.
- Using your second approach,
$$ \begin{eqnarray} \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \end{eqnarray} $$ $$ \begin{eqnarray} \Delta G = \Delta H - T \Delta S \end{eqnarray} $$ $$ \begin{eqnarray} \Delta G = \Delta G^\circ + RT \ln(Q) \end{eqnarray} $$ At equilibrium $ \Delta G = 0 $ and $Q= K_{eq}$ and hence $$ \begin{eqnarray} && K=e^{\frac{-\Delta G^\circ}{RT}} =e^{\frac{-(\Delta H^\circ - T \Delta S^\circ)}{RT}} \end{eqnarray} $$
For Temperature $T_1$ and Equilibrium Constant $K_1$ $$ \begin{eqnarray} K_1=e^{\frac{-(\Delta H^\circ - T_1 \Delta S^\circ)}{RT_1}} \end{eqnarray} $$ Similarly, For Temperature $T_1$ and Equilibrium Constant $K_1$ $$ \begin{eqnarray} K_2=e^{\frac{-(\Delta H^\circ - T_2 \Delta S^\circ)}{RT_2}} \end{eqnarray} $$
Dividing the above two equations, you get $$ \begin{eqnarray} && \frac{k_2}{k_1} = e^{-\frac{∆H^\circ}{R} (\frac{1}{T_2} - \frac{1}{T_1})} \end{eqnarray} $$
The Van't Hoff Equation !!! And it would give the same result as before.
I didn't conclude Dependence of $K_{eq}$ and T from $\Delta G^\circ$ because if a reaction is exothermic, it only tells that $\Delta H^\circ <0$ and doesn't tell anything about $\Delta G^\circ$ ..