$4 S (s) + 6 O_{2} \to 4 S O_{3} (g)$
$Δ H$ for this reaction is $- 1583 K J$ . The enthalpy of formation of sulphur trioxide?

- 395.8 K J

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395.8 K J

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495.5 K J

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595.5 K J

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Solution

The correct option is D $- 395.8 K J$
Solution:- (A) $- 395.8 k J$
Given:-
$4 S_{(s)} + 6 {O_{2}}_{(g)} ⟶ 4 {S O_{3}}_{(g)}; Δ H = - 1583 k J$
As we know that the enthalpy of formation is equal to the enthalpy of reaction when $1$ mole of product is formed.
Therefore,
$\frac{1}{4} \times [4 S_{(s)} + 6 {O_{2}}_{(g)} ⟶ 4 {S O_{3}}_{(g)}; Δ H = - 1583 k J]$
$S_{(s)} + \frac{3}{2} {O_{2}}_{(g)} ⟶ {S O_{3}}_{(g)}; Δ H = - 395.8 k J$
Hence the enthalpy of formation sulphur trioxide is $- 395.8 k J$ .

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