Question

Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)Entropy data:Fe2O3(s): 90 J/K molC(s): 5.7 J/K molFe(s): 27.2 J/K molCO(g): 198 J/K molA. +129.5 J/K molB. -129.5 J/K molC. +541.3 J/K molD. +320.9 J/K mol

Answer 1

Explanation:

We are given: entropy of Fe2O3 = 90J/K.mol

: entropy of C = 5.7J/K.mol

: entropy of Fe = 27.2J/K.mol

: entropy of CO = 198J/K.mol

$\begin{gathered} \Delta S\text{ = S}_{products}-S_{reactants} \\ \\ \text{ = \lparen3}\times198+2\times27.2)-(3\times5.7+90) \\ \\ \text{ = 541.3J/K.mol} \end{gathered}$

Answer:

The correct answer is C.