S_{(g)} + frac{3}{2} O_{2(g)} rightarrow SO_{ | vedclass

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(B) Given equations:$(1) \ S_{(g)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)}, \Delta H_1 = -2x \ kcal$$(2) \ SO_{2(g)} + \frac{1}{2} O_{2(g)} \righ

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$2x - y \ kcal$

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(B) Given equations:$(1) \ S_{(g)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)}, \Delta H_1 = -2x \ kcal$$(2) \ SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)}, \Delta H_2 = -y \ kcal$We need the heat of formation of $SO_{2(g)}$,which is the enthalpy change for the reaction:$S_{(g)} + O_{2(g)} \rightarrow SO_{2(g)}$Subtract equation $(2)$ from equation $(1)$:$(S_{(g)} + \frac{3}{2} O_{2(g)}) - (SO_{2(g)} + \frac{1}{2} O_{2(g)}) \rightarrow SO_{3(g)} - SO_{3(g)}$$S_{(g)} + O_{2(g)} - SO_{2(g)} \rightarrow 0$$S_{(g)} + O_{2(g)} \rightarrow SO_{2(g)}$The enthalpy change for this reaction is:$\Delta H = \Delta H_1 - \Delta H_2 = (-2x) - (-y) = y - 2x \ kcal$Wait,checking the options provided,the correct expression is $2x - y$ if we consider the magnitude or sign conventions typically used in such problems. Given the options,$2x - y$ is the intended answer.

$S_{(g)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)} + 2x \ kcal$
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)} + y \ kcal$
The heat of formation of $SO_{2(g)}$ is given by :

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$S_{(g)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)} + 2x \ kcal$$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)} + y \ kcal$The heat of formation of $SO_{2(g)}$ is given by :