Determine [tex]\Delta H^{\circ}[/tex] for the following thermochemical equation:

[tex]\[
\begin{array}{l}
Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g); \Delta H^{\circ} = -26.8 \, \text{kJ} \\
FeO(s) + CO(g) \rightarrow Fe(s) + CO_2(g); \Delta H^{\circ} = -16.5 \, \text{kJ}
\end{array}
\][/tex]

[tex]\[
Fe_2O_3(s) + CO(g) \rightarrow 2FeO(s) + CO_2(g)
\][/tex]



Answer :

Sure, let's determine the ΔH° for the reaction [tex]\( \text{Fe}_2\text{O}_3(\text{s}) + \text{CO}(\text{g}) \rightarrow 2 \text{FeO}(\text{s}) + \text{CO}_2(\text{g}) \)[/tex] using Hess's Law. We will combine the given reactions in a way that will lead to the desired reaction.

The given reactions and their enthalpy changes are:

1. [tex]\( \text{Fe}_2\text{O}_3(\text{s}) + 3 \text{CO}(\text{g}) \rightarrow 2 \text{Fe}(\text{s}) + 3 \text{CO}_2(\text{g}) \)[/tex]; [tex]\( \Delta H_1^{\circ} = -26.8 \ \text{kJ} \)[/tex]
2. [tex]\( \text{FeO}(\text{s}) + \text{CO}(\text{g}) \rightarrow \text{Fe}(\text{s}) + \text{CO}_2(\text{g}) \)[/tex]; [tex]\( \Delta H_2^{\circ} = -16.5 \ \text{kJ} \)[/tex]

We need to manipulate these reactions to get the target reaction:

[tex]\[ \text{Fe}_2\text{O}_3(\text{s}) + \text{CO}(\text{g}) \rightarrow 2 \text{FeO}(\text{s}) + \text{CO}_2(\text{g}) \][/tex]

### Step 1: Reverse the Second Reaction
First, reverse the second reaction to produce:
[tex]\[ \text{Fe}(\text{s}) + \text{CO}_2(\text{g}) \rightarrow \text{FeO}(\text{s}) + \text{CO}(\text{g}) \][/tex]
Reversing the reaction changes the sign of ΔH:
[tex]\[ \Delta H_{\text{rev2}}^{\circ} = +16.5 \ \text{kJ} \][/tex]

### Step 2: Scale the Reversed Second Reaction
Next, we need to match the stoichiometry. Multiply the reversed reaction by 2:
[tex]\[ 2 \text{Fe}(\text{s}) + 2 \text{CO}_2(\text{g}) \rightarrow 2 \text{FeO}(\text{s}) + 2 \text{CO}(\text{g}) \][/tex]
This doubles the enthalpy change:
[tex]\[ \Delta H_{\text{rev2 and scaled}}^{\circ} = 2 \times 16.5 \ \text{kJ} = 33.0 \ \text{kJ} \][/tex]

### Step 3: Add the Adjusted Reactions
Finally, add the first reaction and the scaled reversed reaction:
1. [tex]\( \text{Fe}_2\text{O}_3(\text{s}) + 3 \text{CO}(\text{g}) \rightarrow 2 \text{Fe}(\text{s}) + 3 \text{CO}_2(\text{g}) \)[/tex]; [tex]\( \Delta H_1^{\circ} = -26.8 \ \text{kJ} \)[/tex]
2. [tex]\( 2 \text{Fe}(\text{s}) + 2 \text{CO}_2(\text{g}) \rightarrow 2 \text{FeO}(\text{s}) + 2 \text{CO}(\text{g}) \)[/tex]; [tex]\( \Delta H_{\text{rev2 and scaled}}^{\circ} = 33.0 \ \text{kJ} \)[/tex]

Adding these reactions:
[tex]\[ \text{Fe}_2\text{O}_3(\text{s}) + 3 \text{CO}(\text{g}) + 2 \text{Fe}(\text{s}) + 2 \text{CO}_2(\text{g}) \rightarrow 2 \text{Fe}(\text{s}) + 3 \text{CO}_2(\text{g}) + 2 \text{FeO}(\text{s}) + 2 \text{CO}(\text{g}) \][/tex]

### Simplifying
Cancel out the common terms on both sides:
[tex]\[ \text{Fe}_2\text{O}_3(\text{s}) + \text{CO}(\text{g}) \rightarrow 2 \text{FeO}(\text{s}) + \text{CO}_2(\text{g}) \][/tex]

### Calculate ΔH for the Target Reaction
[tex]\[ \Delta H_{\text{target}}^{\circ} = -26.8 \ \text{kJ} + 33.0 \ \text{kJ} = 6.2 \ \text{kJ} \][/tex]

So, the enthalpy change for the reaction [tex]\( \text{Fe}_2\text{O}_3(\text{s}) + \text{CO}(\text{g}) \rightarrow 2 \text{FeO}(\text{s}) + \text{CO}_2(\text{g}) \)[/tex] is [tex]\( 6.2 \ \text{kJ} \)[/tex].