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

The answer is "−847 J/K".

Step-by-step 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) (KJ)/(mol)\\\\\to (-1669+822) (KJ)/(mol)\\\\\to -847(KJ)/(mol)\\\\`

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