Question

Using the thermodynamic information , calculate the standard reaction entropy of the following chemical reaction: Round your answer to zero decimal places.
2Al(s)+ Fe2O3(s) → Al2O3(s)+ 2Fe(s)

Answer 1

Answer:

The answer is "−847 J/K".

Explanation:

The given expression is:

2Al(s)+ Fe2O3(s) → Al2O3(s)+ 2Fe(s)

Δ$H^{\circ}_{rxn}=$ ∑(Δ$H^{\circ}_{products}-H^{\circ}_{reactants}$)

by the above definition Δ$H^{\circ}_{element}= 0\cdot KJ \cdot Mol^{-1}$ 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:

$\to (-1669)-(-822) \frac{KJ}{mol}\\\\\to (-1669+822) \frac{KJ}{mol}\\\\\to -847\frac{KJ}{mol}$

Δ$H^{\circ}_{rxn}=$ -847 $\frac{KJ}{mol} \ mol^{-1} \texttt{ we mean \mole of Reaction as written....}$