Some reactions, such as the formation of KClO3 cannot be measured directly. Using the following equations, find the ΔHf of KClO3:

KClO_3_{(s)}+3Mg_{(s)}\rightarrow KCl_{(s)}+3MgO_{(s)}\: \: \: \Delta H:-1852 kJ\: mol^-^1

K_{(s)}+\frac{1}{2}Cl_{2(s)}\rightarrow KCl_{(s)}\: \: \: \Delta H:-437 kJ\: mol^-^1

Mg_{(s)}+\frac{1}{2}O_{2(g)}\rightarrow MgO_{(s)}\: \: \: \Delta H:-602 kJ\: mol^-^1

clockwise = anticlockwise

{\color{Magenta} x}{\color{Red} -1852} = {\color{Blue} -437 +3(-602)}

x=-391kJ