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].
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].
