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Chapter 6: Problem 112

Given the following data \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-23 \mathrm{~kJ}\) \(3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-39 \mathrm{~kJ}\) \(\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=+18 \mathrm{~kJ}\) calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

Short Answer

Expert verified
The enthalpy change, ΔH, for the given reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ is approximately -25.67 kJ.

Step by step solution

01

Identify the common species in the reactions

Observe that CO is a reactant in all the given reactions and CO₂ is a product in all the given reactions.
02

Calculate the moles of the target reaction

The goal is to manipulate the given reactions to match the target reaction. So the target reaction is: $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$
03

Manipulate the given reactions

In order for the target reaction to be obtained, multiply the first given reaction by \(\frac{1}{3}\) and reverse the last given reaction.
04

Writing the modified reactions and their enthalpy changes

The modified reactions are: $$ \frac{1}{3}(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)) \quad \Delta H^{\circ}=-\frac{1}{3} \times 23 \, \mathrm{kJ}=-\frac{23}{3} \, \mathrm{kJ} $$ $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-18 \, \mathrm{kJ} $$ Notice that when we reverse the third given reaction, its enthalpy change becomes negative. Now add the first modified reaction and the second modified reaction together to get the target reaction.
05

Combine the modified reactions and find the ΔH for the target reaction

Summing both modified reactions, we get the target reaction: $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ Now, according to Hess's law, the enthalpy change for the target reaction is equal to the sum of the enthalpy changes for the modified reactions: $$ \Delta H^{\circ}_{\text{target}}=\Delta H^{\circ}_{1}+\Delta H^{\circ}_{2}= -\frac{23}{3} \, \mathrm{kJ} - 18 \, \mathrm{kJ} $$ Calculate the final enthalpy change for the target reaction: $$ \Delta H^{\circ}_{\text{target}}=-\frac{23}{3} \, \mathrm{kJ} - 18 \, \mathrm{kJ} = -\frac{77}{3} \, \mathrm{kJ} $$
06

Final answer

The enthalpy change, ΔH, for the given reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ is -25.67 kJ (approximately).

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