$S (s) + O_{2} (g) \to {S O}_{2} (g); Δ H^{\circ} = - 297 k J$
$2 {S O}_{2} (g) + O_{2} (g) \to 2 {S O}_{3} (g); Δ H^{\circ} = - 198 k J$
From the above data which has given, calculate the heat of reaction for the formation of $S O_{3}$ ?
$S (s) + \frac{3}{2} O_{2} (g) \to {S O}_{3} (g)$

-495 kJ

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-396 kJ

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-198 kJ

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+198 kJ

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Solution

The correct option is C -396 kJ
$S + O_{2} \to S O_{2}$ $Δ H^{o} = - 297 k J$
$S O_{2} + O - 2 \to 2 S O - 3$ $Δ H^{o} = - 198 k J$
First on multiplying equation first by 2:
$2 S + 2 O_{2} \to 2 S O_{2}$ $2 \times Δ H^{o} = - 297 \times 2 = 594 k J$
Now, adding both equation: ( New + 2nd )
$2 S + 3 O_{2} \to 2 S O_{3}$ $Δ_{r e a c t i o n} H^{o} = - 594 - 198 = - 792 k J$
Now, determining the enthalpy for 1 mol of $S O_{3}$ to determine Enthalpy of formation of $S O_{3}$ , divide the equation by 2.
$2 S (s) + \frac{3}{2} O_{2} (g) \to S O_{3} (g)$ $Δ_{f o r m a t i o n} H^{o} = \frac{- 792}{2} = - 396 k J$

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