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Heat of Formation

4Ss+6O2→ 4SO3...

Question

$4 S (s) + 6 O_{2} \to 4 S O_{3} (g); Δ H$ for this reaction is -1583.2 KJ. The enthalpy of formation of sulphur trioxide is:

-395.8 kJ

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395.8 kJ

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495.5 kJ

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595.5 kJ

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Solution

The correct option is A -395.8 kJ
Formation enthalpy is valid for $1$ mol of substrate , so the formation enthalphy of $S$ is,
$Δ H_{f} = \frac{- 1583.2}{4} = - 395.8 K J$

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4 S (s) + 6 O_{2} \to 4 S O_{3} (g)

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