Given the information below, what is the standard enthaply of formation for $A l_{2} O_{3} (s)$ ?
$2 A l_{2} O_{3} (s) \to 4 A l (s) + 3 O_{2} (g); Δ H_{R X N}^{o} = 3352 k J$

- 6704

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- 3352

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- 1676

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1676

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Solution

The correct option is D $- 1676$
Solution:- (C) $- 1676$
Given:-
$2 {A l}_{2} O_{3}_{(s)} ⟶ 4 {A l}_{(s)} + 3 {O_{2}}_{(g)} Δ H = 3352 k J$
$4 {A l}_{(s)} + 3 {O_{2}}_{(g)} ⟶ 2 {A l}_{2} O_{3}_{(s)} Δ H = - 3352 k J$
Enthalpy of formation of $2$ mole of ${A l}_{2} O_{3} = - 3352 k J$
Enthalpy of formation of $1$ mole of ${A l}_{2} O_{3} = \frac{- 3352}{2} = - 1676 k J$
As we know that standard enthalpy of formation is same as enthalpy of formation of $1$ mole of product.
Therefore, standard enthalpy of formation of ${A l}_{2} O_{3}$ is $- 1676 k J$ .

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