Answer:
The answer is "−847 J/K".
Explanation:
The given expression is:
2Al(s)+ Fe2O3(s) → Al2O3(s)+ 2Fe(s)
Δ[tex]H^{\circ}_{rxn}=[/tex] ∑(Δ[tex]H^{\circ}_{products}-H^{\circ}_{reactants}[/tex])
by the above definition Δ[tex]H^{\circ}_{element}= 0\cdot KJ \cdot Mol^{-1}[/tex] For Such a Component under standard conditions from its standard state, that also applies here. But, we start taking the overview and follow the conventions of signing:
[tex]\to (-1669)-(-822) \frac{KJ}{mol}\\\\\to (-1669+822) \frac{KJ}{mol}\\\\\to -847\frac{KJ}{mol}\\\\[/tex]
Δ[tex]H^{\circ}_{rxn}=[/tex] -847 [tex]\frac{KJ}{mol} \ mol^{-1} \texttt{ we mean \mole of Reaction as written....}\\[/tex]
