Given that, $3C(s) + 2Fe_{2}O_{3}(s) \rightarrow 4Fe(s) + 3CO_{2}(g)\mathrm{\Delta}H^{o} = - 93657\$cal $C(s) + O_{2}(g) \rightarrow CO_{2}(g);\ \mathrm{\Delta}H^{o} = - 94050$ cal If both the values of $\mathrm{\Delta}H$ are at $25℃$ then calculate$\mathrm{\Delta}H_{Fe_{2}O_{3}}^{o}$

Question

Dr. Abubacker · Accepted Answer

(c)
Calculate it for the equation
$2Fe(s) + \frac{3}{2}O_{2}(g) ightarrow Fe_{2}O_{3}(s)$
Eqs. $3(ii) - (i)$, we have
$4Fe(s) + 3O_{2}(g) ightarrow 2Fe_{2}O_{3}(s)$
$\mathrm{\Delta}H{^\circ} = 3( - 94050) - ( - 93657)$
$= - 18493\ cal = 188.493\ kcal$
$\because 2\ moles\ Fe_{2}O_{3}\ has\ \mathrm{\Delta}H{^\circ} = - 188.493$ kcal
$	herefore 1\ mole\ Fe_{2}O_{3}\ has\mathrm{\Delta}H{^\circ} = - \frac{188.493}{2}$
$= - 94.25$ kcal/mol

Q. Given that, $3C(s) + 2Fe_{2}O_{3}(s) \rightarrow 4Fe(s) + 3CO_{2}(g)\mathrm{\Delta}H^{o} = - 93657\$cal $C(s) + O_{2}(g) \rightarrow CO_{2}(g);\ \mathrm{\Delta}H^{o} = - 94050$ cal If both the values of $\mathrm{\Delta}H$ are at $25℃$ then calculate$\mathrm{\Delta}H_{Fe_{2}O_{3}}^{o}$

Explanation

Reviewed by Dr. Abubacker

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