Question

question 7 (1 point)
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) δh° = 232.2 kj
(ii) 2fe₃o₄(s) → 6 fe(s) + 4o₂ (g) δh° = 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 \(\Delta H^{\circ}=2234\ kJ\), we get \(6Fe(s)+4O_2(g)\to2Fe_3O_4(s)\) and \(\Delta H^{\circ}=- 2234\ kJ\)

Step2: Add the two equations

We have equation (i): \(3Fe_2O_3(s)\to2Fe_3O_4+\frac{1}{2}O_2(g)\) with \(\Delta H^{\circ}=232.2\ kJ\) and the reversed equation \(6Fe(s)+4O_2(g)\to2Fe_3O_4(s)\) with \(\Delta H^{\circ}=-2234\ kJ\)

If we add them:

$$
LATEXBLOCK0
$$

Cancel out \(2Fe_3O_4\) on both sides. And we want \(6Fe(s)+\frac{9}{2}O_2(g)\to3Fe_2O_3(s)\)

The total \(\Delta H\) is \(\Delta H=\Delta H_{(i)}+\Delta H_{(reversed\ (ii))}\)

$$
LATEXBLOCK1
$$

Answer:

-2002 kJ