Question

given: so₂(g) + ½o₂(g) → so₃(g) δh° = -99 kj
what is the enthalpy change for the reaction below:
2 so₃(g) → o₂(g) + 2 so₂(g)
-198 kj
+99 kj
-99 kj
+49.5 kj
+198 kj

Explanation:

Step1: Reverse the given reaction

When a reaction is reversed, the sign of $\Delta H$ changes.
Original reaction: $SO_{2}(g)+\frac{1}{2}O_{2}(g)\to SO_{3}(g)\quad\Delta H^{0}=- 99\ kJ$
Reversed reaction: $SO_{3}(g)\to SO_{2}(g)+\frac{1}{2}O_{2}(g)\quad\Delta H^{0}=+99\ kJ$

Step2: Multiply the reversed reaction by 2

When a reaction is multiplied by a factor $n$, its $\Delta H$ is also multiplied by $n$.
Multiply the reversed reaction by 2: $2SO_{3}(g)\to 2SO_{2}(g)+O_{2}(g)$
$\Delta H^{0}=2\times(+99\ kJ)= + 198\ kJ$

Answer:

+198 kJ