Enthalpy of Zinc and Sulfur Reaction I was working on my homework which was to write a description of what would occur if zinc and sulfur reacted and what steps would a scientist have to make in order to make them react. When I researched I bit, I saw that it would be a exothermic reaction, and I wanted to show this by using enthalpy. I haven't been taught this however, and as a result, I'm confused. I know that enthalpy equals the heat transfer and there is some sort relationship with bond energy, but I don't know how to apply this to the reaction. 

Zinc Powder + Sulfur Powder = Zinc Sulfide + Energy 

$\ce{Zn_{(s)} + S_{(s)} -&gt; ZnS_{(s)}}$ + $205.98 \text{ kJ}$

Because $205&gt;0$ this is an exothermic reaction. But where did this $205.98\text{ kJ}$ come from?
The free energies of formation of the elements in their standard states is zero, by definition.  You can look up the (variously reported) standard enthalpy of formation of the product, −204.6 kJ/mol (exothermic!).  It is roughly the binding energy of the crystal lattice less the ionization energies of the inputs.  So, by the numbers, do you obtain spalerite or wurtzite?  ZnS has crystal polymorphs and polytopes with differently populated unit cells

$\ce{Zn_2S_2}$ Wurtzite-2H 
$\ce{Zn_4S_4}$ Wurtzite-4H and Sphalerite 
$\ce{Zn_6S_6}$ Wurtzite-6H 
$\ce{Zn_15S_15}$ Wurtzite-15R