# The standard heat of formation of Fe2O3(s) is 824.2kJ/mol. Calculate heat change for the reaction 4Fe(s)+3O2(g)=2Fe2O3(s)?

Sep 10, 2017

$\Delta {H}_{R x n}^{o} = - 1648.4 K j$
Note, the $\Delta {H}_{\text{f}}^{o} \left(F {e}_{2} {O}_{3} \left(s\right)\right)$ = $- 824.2 \text{Kj"/"mole}$ (exothermic)

#### Explanation:

The Standard Heat of Formation ($\Delta {H}_{f}^{o}$) is given in terms of Kj per 1 mole. By definition $\Delta {H}_{f}^{o}$ is the heat liberated or gained on formation of one mole of substance from basic elements in their standard states. The Standard Enthalpy of Formation for elements in their basic standard states is equal to 'zero' Kj/mole. The form/state can be identified in Thermodynamics Tables in the appendix of most college level general chemistry text books.

For $F {e}_{2} {O}_{3}$ the Standard State Equation would be represented as follows ...

$2 F {e}^{o} \left(s\right)$ + $\frac{3}{2} {O}_{2} \left(g\right)$ => $F {e}_{2} {O}_{3} \left(s\right)$; $\Delta {H}_{f}^{o}$= - 824.2 "Kj"/"mole"

For the reaction in the post it is showing 2 moles of $F {e}_{2} {O}_{3}$. Therefore, the $\Delta {H}_{R x n}^{o}$ = $2 \times \Delta {H}_{f}^{o}$ = 2cancel("moles")(-824.2 "Kj"/cancel("mole")) = $- 1648.4 K j$