Find the standard enthalpy of formation of $SO_{3(g)}$ from the following reactions:
$S_{8(s)} + 8O_{2(g)} \to 8SO_{2(g)}$; $\Delta H_1 = -2775 \ kJ/mol$
$2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$; $\Delta H_2 = -198 \ kJ/mol$
$S_{8(s)} + 8O_{2(g)} \to 8SO_{2(g)}$; $\Delta H_1 = -2775 \ kJ/mol$
$2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$; $\Delta H_2 = -198 \ kJ/mol$
(N/A) The formation reaction of $SO_{3(g)}$ is: $S_{(s)} + \frac{3}{2}O_{2(g)} \to SO_{3(g)}$.
From the first reaction: $S_{8(s)} + 8O_{2(g)} \to 8SO_{2(g)}$,$\Delta H_1 = -2775 \ kJ/mol$.
Dividing by $8$: $S_{(s)} + O_{2(g)} \to SO_{2(g)}$,$\Delta H = \frac{-2775}{8} = -346.875 \ kJ/mol$.
From the second reaction: $2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$,$\Delta H_2 = -198 \ kJ/mol$.
Dividing by $2$: $SO_{2(g)} + \frac{1}{2}O_{2(g)} \to SO_{3(g)}$,$\Delta H = \frac{-198}{2} = -99 \ kJ/mol$.
Adding the two equations:
$S_{(s)} + O_{2(g)} + SO_{2(g)} + \frac{1}{2}O_{2(g)} \to SO_{2(g)} + SO_{3(g)}$
$S_{(s)} + \frac{3}{2}O_{2(g)} \to SO_{3(g)}$
$\Delta H_f = -346.875 + (-99) = -445.875 \ kJ/mol$.
From the first reaction: $S_{8(s)} + 8O_{2(g)} \to 8SO_{2(g)}$,$\Delta H_1 = -2775 \ kJ/mol$.
Dividing by $8$: $S_{(s)} + O_{2(g)} \to SO_{2(g)}$,$\Delta H = \frac{-2775}{8} = -346.875 \ kJ/mol$.
From the second reaction: $2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$,$\Delta H_2 = -198 \ kJ/mol$.
Dividing by $2$: $SO_{2(g)} + \frac{1}{2}O_{2(g)} \to SO_{3(g)}$,$\Delta H = \frac{-198}{2} = -99 \ kJ/mol$.
Adding the two equations:
$S_{(s)} + O_{2(g)} + SO_{2(g)} + \frac{1}{2}O_{2(g)} \to SO_{2(g)} + SO_{3(g)}$
$S_{(s)} + \frac{3}{2}O_{2(g)} \to SO_{3(g)}$
$\Delta H_f = -346.875 + (-99) = -445.875 \ kJ/mol$.
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