Calculate Enthalpy Change for Iron Reduction

ChemistryThermochemistryMediumSTEM

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Question

7.88 In the recovery of iron from iron ore, the reduction of the ore is actually accomplished by reactions involving carbon monoxide. Use the following thermochemical equations,

$Fe_2O_3(s) + 3CO(g) \\longrightarrow 2Fe(s) + 3CO_2(g)$ $\\Delta H^{\\circ} = -28 \\text{ kJ}$

$3Fe_2O_3(s) + CO(g) \\longrightarrow 2Fe_3O_4(s) + CO_2(g)$ $\\Delta H^{\\circ} = -59 \\text{ kJ}$

$Fe_3O_4(s) + CO(g) \\longrightarrow 3FeO(s) + CO_2(g)$ $\\Delta H^{\\circ} = +38 \\text{ kJ}$

to calculate $\\Delta H^{\\circ}$ for the reaction

$FeO(s) + CO(g) \\longrightarrow Fe(s) + CO_2(g)$

Animated Video Solution

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Step by Step Written Solution

1
Step 1

Hi Vearn, let's use Hess's Law to calculate the enthalpy change for the reduction of iron oxide by carbon monoxide.

Applying Hess's Law

2
Step 2

We are given three equations. Let's label them one, two, and three so we can refer back to them easily. Our goal is to reach the target equation listed at the bottom.

$$ (1) \ Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g) \quad \Delta H_1 = -28 \text{ kJ}$$
$$ (2) \ 3Fe_2O_3(s) + CO(g) \rightarrow 2Fe_3O_4(s) + CO_2(g) \quad \Delta H_2 = -59 \text{ kJ}$$
$$ (3) \ Fe_3O_4(s) + CO(g) \rightarrow 3FeO(s) + CO_2(g) \quad \Delta H_3 = +38 \text{ kJ}$$
$$ \text{Target: } FeO(s) + CO(g) \rightarrow Fe(s) + CO_2(g)$$
3
Step 3

Let's analyze our target equation and see how we can combine the given ones.

Strategy

Our target needs one mole of iron oxide FeO on the reactant side.

Looking at equation three, FeO is on the product side with a coefficient of three.

4
Step 4

So, we need to reverse equation three and divide it by three. When we reverse it, the sign of delta H flips from positive to negative.Dividing by three gives us one third of the original heat value.

$$ \text{Modified (3): } FeO(s) + \frac{1}{3}CO_2(g) \rightarrow \frac{1}{3}Fe_3O_4(s) + \frac{1}{3}CO(g)$$
$$ \Delta H_{3}' = \frac{-38}{3} \text{ kJ} \approx -12.67 \text{ kJ}$$
5
Step 5

Next, we need iron on the product side. Equation one has two moles of iron as a product.

Strategy continued...

$$ (1) \ Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g) \quad \Delta H_1 = -28 \text{ kJ}$$
6
Step 6

Since our target only needs one mole of iron, we will divide equation one by two. This also divides the enthalpy change by two.

$$ \text{Modified (1): } \frac{1}{2}Fe_2O_3(s) + \frac{3}{2}CO(g) \rightarrow Fe(s) + \frac{3}{2}CO_2(g)$$
$$ \Delta H_{1}' = \frac{-28}{2} \text{ kJ} = -14 \text{ kJ}$$
7
Step 7

Now we have iron three oxide and magnetite left in our modified equations that aren't in the target. We need to cancel them out using equation two.

Final Adjustment

$$ (2) \ 3Fe_2O_3(s) + CO(g) \rightarrow 2Fe_3O_4(s) + CO_2(g) \quad \Delta H_2 = -59 \text{ kJ}$$

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About This Question

Subject
Chemistry
Topic
Thermochemistry
Difficulty
Medium
Exam
STEM
Question Type
Open Ended

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