Question

given: $so_{2}(g)+\frac{1}{2}o_{2}(g)\to so_{3}(g) delta h^{circ}=-99 kj$ what is the enthalpy change for the reaction below: $2so_{3}(g)\to o_{2}(g)+2so_{2}(g)$ -198 kj +198 kj +99 kj +49.5 kj -99 kj

Explanation:

Step1: Reverse the given reaction

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

Step2: Multiply the reversed reaction by 2

When a reaction is multiplied by a factor \(n\), its \(\Delta H\) is also multiplied by \(n\).
Multiplying \(SO_{3}(g)\to SO_{2}(g)+\frac{1}{2}O_{2}(g)\) (\(\Delta H = + 99\space kJ\)) by 2 gives \(2SO_{3}(g)\to 2SO_{2}(g)+O_{2}(g)\)
\(\Delta H=(+99\space kJ)\times2=+198\space kJ\)

Answer:

+198 kJ