# Question 71b3e

May 14, 2017

$\Delta {H}_{f}^{\circ} = - {\text{36 kJ mol}}^{- 1}$

#### Explanation:

The standard enthalpy of formation, $\Delta {H}_{f}^{\circ}$, tells you the change in enthalpy that occurs when $1$ mole of a given compound is produced from its constituent elements in their more stable form.

In this case, you know that the reaction gives off $\text{0.855 kJ}$ of heat when $\text{2.1 g}$ of iron(II) sulfide are produced. This is equivalent to saying that this reaction has

$\Delta {H}_{\text{rxn"^@ = - "0.855 kJ}}$

The minus sign is used to denote heat given off.

Use the molar mass of the compound to convert the number of grams to moles

2.1 color(red)(cancel(color(black)("g"))) * "1 mole FeS"/(87.91color(red)(cancel(color(black)("g")))) = "0.02389 moles FeS"

So, you can say that when $1$ mole of iron(II) sulfide is produced, the reaction gives off

1 color(red)(cancel(color(black)("moles FeS"))) * "0.855 kJ"/(0.02389color(red)(cancel(color(black)("moles FeS")))) = "35.79 kJ"#

Therefore, the standard enthalpy of formation of iron(II) sulfide will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {H}_{f}^{\circ} = - {\text{36 kJ mol}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of iron(II) sulfide produced by the initial reaction.

SIDE NOTE The actual value for the standard enthalpy of formation of iron(II) sulfide is $- {\text{102 kJ mol}}^{- 1}$, so you might want to double-check the values given to you.

https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation