Question

Lei de Hess
Ajudem Pfv

Answer 1

Precisa inverter a primeira equação, para que o $\text{SO}_2$ fique nos reagentes:

$(\text{I})~~\text{S}_{(\text{romb})}+\text{O}_{2(\text{g})} \Longrightarrow \text{SO}_{2(\text{g})} \ \ \ \ \ \ \ \Delta\text{H}_\text{f}^{0}=-71,0~\text{kcal}$

Invertendo:

$\text{SO}_{2(\text{g})} \Longrightarrow \text{S}_{(\text{romb})}+\text{O}_{2(\text{g})} \ \ \ \ \ \ \ \Delta\text{H}_\text{f}^{0}=71,0~\text{kcal}$

Note que o sinal do $\Delta\text{H}_\text{f}^{0}$ mudou.

Agora some as equações:

$(\text{I})~~\text{SO}_{2(\text{g})} \Longrightarrow \text{S}_{(\text{romb})}+\text{O}_{2(\text{g})} \ \ \ \ \ \ \ \ \ \  \Delta\text{H}_\text{f}^{0}=71,0~\text{kcal}$

$(\text{II})~~\text{S}_{(\text{romb})}+\dfrac{3}{2}\text{O}_{2(\text{g})} \Longrightarrow \text{SO}_{3(\text{g})} \ \ \ \ \ \ \ \Delta\text{H}_\text{f}^{0}=-94,0~\text{kcal}$

$\text{SO}_{2(\text{g})}+\text{S}_{(\text{romb})}+\dfrac{3}{2}\text{O}_{2(\text{g})} \Longrightarrow \text{S}_{(\text{romb})}+\text{O}_{2(\text{g})}+\text{SO}_{3(\text{g})} \ \ \ \ \ \Delta\text{H}=71-93~\text{kcal}$

Simplificando $\text{S}_{(\text{romb})$ em ambos os lados e $\dfrac{3}{2}\text{O}_{2(\text{g})}-\text{O}_{2(\text{g})}=\dfrac{1}{2}\text{O}_{2(\text{g})}$, obtemos:

$\text{SO}_{2(\text{g})}+\dfrac{1}{2}\text{O}_{2(\text{g})} \Longrightarrow \text{SO}_{3(\text{g})} \ \ \ \ \boxed{\Delta\text{H}=-23~\text{kcal}}$