Fe(r)+S(тв)—>FeS(тв)-98кДж

The reaction you've provided is the formation of iron(II) sulfide (FeS) from iron (Fe) and sulfur (S) with a standard enthalpy change of -98 kJ. This indicates that the reaction is exothermic, releasing 98 kJ of energy.

The balanced chemical equation for the reaction is:

\[ \text{Fe (s)} + \text{S (s)} \rightarrow \text{FeS (s)} \]

In this case:

• Reactants: Iron (Fe) in solid form and sulfur (S) in solid form.
• Product: Iron(II) sulfide (FeS) in solid form.
• Enthalpy Change: The negative value (-98 kJ) signifies that energy is released when the reaction occurs.

If you need to delve into specific details or concepts related to this reaction, like thermodynamics or reaction mechanisms, feel free to ask!