Van 't Hoff equation: derive it and find K of the Haber reaction at 700 K

Derive the van 't Hoff equation $\dfrac{d\ln K}{dT} = \dfrac{\Delta_rH^\circ}{RT^2}$. For $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$, $\Delta_fG^\circ(\ce{NH3}, 298\ \text{K}) = -16.45$ kJ mol⁻¹ and $\Delta_rH^\circ = -92.2$ kJ mol⁻¹. Calculate $K$ at 298 K and at 700 K (take $\Delta_rH^\circ$ as independent of temperature). Explain why the Haber process is nevertheless run at about 700 K and 200 atm. ($R = 8.314$ J K⁻¹ mol⁻¹)