[tex]\tt{\huge{\blue{Explanation:}}}[/tex]
Our goal here is to calculate the enthalpy change for the decomposition of Fe₂(SO₄)₃ to form Fe₂O₃ and SO₃. The reaction does not occur directly, however, so we must use an indirect route using the information given by Equations (a), (b), and (c). The equations are:
(a) 2Fe + (3/2)O₂ → Fe₂O₃ ∆H = -1,650 kJ/mol
(b) 3S + (9/2)O₂ → 3SO₃ ∆H = -792 kJ/mol
(c) 2Fe + 3S + 6O₂ → Fe₂(SO₄)₃ ∆H = -2,583 kJ/mol
The overall reaction is
Fe₂(SO₄)₃ → Fe₂O₃ + 3SO₃
[tex]\tt{\huge{\red{Solution:}}}[/tex]
First, we need 1 mole of Fe₂O₃ as a product and this is provided by Equation (a). Next, we need 3 moles of SO₃ as a product and this is provided by Equation (b). Looking at the overall reaction, we need 1 moles of Fe₂(SO₄)₃ as reactant. So we reverse Equation (c) to get
(d) [tex]\begin{array}{cccccccc} \text{Fe}_2(\text{SO}_4)_3 & \rightarrow & 2\text{Fe} & + & 3\text{S} & + & 6\text{O}_2 & \Delta H = \text{2,583 kJ/mol} \end{array}[/tex]
Adding Equations (a), (b), and (d) together, we get
[tex]\begin{array}{llllllllll} 2\text{Fe} & + & \frac{3}{2}\text{O}_2 & \rightarrow & \text{Fe}_2\text{O}_3 & & & & & \Delta H = -\text{1,650 kJ/mol} \\ 3\text{S} & + & \frac{9}{2}\text{O}_2 & \rightarrow & 3\text{SO}_3 & & & & & \Delta H = -\text{792 kJ/mol} \\ & & \text{Fe}_2(\text{SO}_4)_3 & \rightarrow & 2\text{Fe} & + & 3\text{S} & + & 6\text{O}_2 & \Delta H = \text{2,583 kJ/mol} \\ \hline & & \text{Fe}_2(\text{SO}_4)_3 & \rightarrow & \text{Fe}_2\text{O}_3 & + & 3\text{SO}_3 & & & \Delta H = \text{141 kJ/mol} \end{array}[/tex]
Hence, the heat of reaction is
[tex]\boxed{\Delta H = \text{141 kJ/mol}}[/tex]
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Note: Kindly swipe the screen to the left to see the continuation of the answers on the right side.
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