question 7 (1 point)\nuse the following information to answer the next question.\niron (iii) oxide (fe₂o₃)…

question 7 (1 point)\nuse the following information to answer the next question.\niron (iii) oxide (fe₂o₃) called rust, can be formed from\nthe oxidation of iron as follows:\n6fe(s) + 9/2o₂ (g) → 3fe₂o₃(s)\nwhat would be the enthalpy reaction, if\n(i) 3fe₂o₃(s) → 2fe₃o₄ + 1/2o₂ (g) δh° = 232.2 kj\n(ii) 2fe₃o₄(s) → 6 fe(s) + 4o₂ (g) δh° = 2234 kj\n-2002 kj\n+ 2002 kj\n2466 kj\n-2466 kj\n-2699 kj

question 7 (1 point)\nuse the following information to answer the next question.\niron (iii) oxide (fe₂o₃) called rust, can be formed from\nthe oxidation of iron as follows:\n6fe(s) + 9/2o₂ (g) → 3fe₂o₃(s)\nwhat would be the enthalpy reaction, if\n(i) 3fe₂o₃(s) → 2fe₃o₄ + 1/2o₂ (g) δh° = 232.2 kj\n(ii) 2fe₃o₄(s) → 6 fe(s) + 4o₂ (g) δh° = 2234 kj\n-2002 kj\n+ 2002 kj\n2466 kj\n-2466 kj\n-2699 kj

Answer

Explanation:

Step1: Reverse the second reaction

When we reverse the reaction (2Fe_3O_4(s)\to6Fe(s) + 4O_2(g)) ((\Delta H^{\circ}=2234\ kJ)), we get (6Fe(s)+4O_2(g)\to2Fe_3O_4(s)) and (\Delta H^{\circ}=- 2234\ kJ)

Step2: Add the two reactions

Add the reaction (3Fe_2O_3(s)\to2Fe_3O_4+\frac{1}{2}O_2(g)) ((\Delta H^{\circ}=232.2\ kJ)) and (6Fe(s)+4O_2(g)\to2Fe_3O_4(s)) ((\Delta H^{\circ}=-2234\ kJ))

[ \begin{align*} 3Fe_2O_3(s)+6Fe(s)+4O_2(g)&\to2Fe_3O_4+\frac{1}{2}O_2(g)+2Fe_3O_4(s)\ 6Fe(s)+\frac{9}{2}O_2(g)&\to3Fe_2O_3(s) \end{align*} ]

The enthalpy change (\Delta H=\Delta H_1+\Delta H_2)

(\Delta H = 232.2+( - 2234))

[ \begin{align*} \Delta H&=232.2-2234\ &=-2001.8\approx - 2002\ kJ \end{align*} ]

Answer:

-2002 kJ