how to increase Keq(equilibrium) with temperatrue
You can easily see how the equilibrium constant varies with temperature using the Van’t Hoff equation:
$$\frac{\mathrm{d}\ln{K_\text{eq}}}{\mathrm{d}T}=\frac{\Delta H^\circ}{R{T^2}}$$

Rearranging the equation a little bit, we get that:
$$\mathrm{d}\ln K_\text{eq} = \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$

Now we can integrate each side of this equation: 
 $$\int_{K_1}^{K_2} \mathrm{d}\ln K = \int_{T_1}^{T_2} \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$
Because $\Delta H^\circ$ and $R$ are constants, this will be equivalent to :
$$\int_{K_1}^{K_2} \mathrm{d}\ln K = \frac{\Delta H^\circ}{R} \int_{T_1}^{T_2} \frac{1}{T^2}\mathrm{d}T$$
Solving the integral we will get this :
$$\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

For $\Delta H^\circ &gt; 0$ we have two cases: when $T_2&gt;T_1$ and $T_2&lt;T_1$.
If $T_2&gt;T_1$ then $\frac{1}{T_2}&lt;\frac{1}{T_1}$ and hence $\frac{1}{T_2} - \frac{1}{T_1} &lt; 0$ . Knowing that $\Delta H^\circ&gt;0$, this means that $\frac{-\Delta H^\circ}{R}$ will also be negative, and the product of these 2 will be positive:
 $$\frac{-\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)&gt;0$$
And from the Van’t Hoff equation we get that : $$\ln\left(\frac{K_2}{K_1}\right)&gt;0 \Leftrightarrow \frac{K_2}{K_1} &gt; 1 \Leftrightarrow K_2 &gt; K_1$$
i.e. The equilibrium constant increases as the temperature is increased.
Using the same principle for $T_2 &lt; T_1$, we get that $K_2 &lt; K_1$