For the reaction $2 A l + F e_{2} O_{3} ⟶ 2 F e + A l_{2} O_{3}$ , the standard heat enthalpy of $F e_{2} O_{3}$ and $A l_{2} O_{3}$ are $- 196.5$ and $- 399.1$ kcal respectively. $Δ H^{\circ}$ for the reaction is:

- 252.4 k c a l

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- 135.5 k c a l

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- 202.6 k c a l

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none of the above

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Solution

The correct option is B $- 202.6 k c a l$
$Δ H_{R e a c t i o n}^{\circ} - Δ H_{P r o d u c t s}^{\circ} - Δ H_{R e a c t a n t s}^{\circ}$
$= 2 \times Δ H_{F e}^{\circ} + Δ H_{A l_{2} O_{3}}^{\circ} - [2 \times Δ H_{A l}^{\circ} + Δ H_{F e_{2} O_{3}}^{\circ}]$
$∵ Δ H^{\circ}$ of free elements is zero, $i . e ., Δ H_{F e}^{\circ} = 0; Δ H_{A l}^{\circ} = 0$
$= 2 \times 0 + (- 399.1) - [2 \times 0 + (- 196.5)]$
$∴ Δ H^{\circ} = - 202.6 k c a l$

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