Question

question 7 (1 point) ✓ saved
use the following information to answer the next question.
iron (iii) oxide (fe₂o₃) called rust, can be formed from
the oxidation of iron as follows:
6fe(s) + 9/2o₂ (g) → 3fe₂o₃(s)
what would be the enthalpy reaction, if
(i) 3fe₂o₃(s) → 2fe₃o₄ + 1/2o₂ (g) δh⁰ = 232.2 kj
(ii) 2fe₃o₄(s) → 6 fe(s) + 4o₂ (g) δh⁰ = 2234 kj
-2002 kj
+ 2002 kj
2466 kj
-2466 kj
-2699 kj

Explanation:

Step1: Reverse the second equation

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

Step2: Add the two equations

Add \(3Fe_2O_3(s)\to2Fe_3O_4+\frac{1}{2}O_2(g)\) (\(\Delta H^{\circ}=232.2\space kJ\)) and \(6Fe(s)+4O_2(g)\to2Fe_3O_4(s)\) (\(\Delta H^{\circ}=-2234\space kJ\))
The sum of the equations:
\(6Fe(s)+\frac{9}{2}O_2(g)\to3Fe_2O_3(s)\)
The sum of the enthalpies: \(\Delta H=\Delta H_1+\Delta H_2\)
\(\Delta H = 232.2+( - 2234)\)
\(\Delta H=-2001.8\approx - 2002\space kJ\)

Answer:

-2002 kJ