Question

In the process of isolating iron from its ores, carbon monoxide reacts with iron (III) oxide, as described by the equation: Fe₂O₃(s)+3CO(g)→2Fe(s)+3CO₂(g) ΔH=−24.8kJ. The enthalpy change for the combustion of carbon monoxide is: 2CO(g)+O₂(g)→2CO₂(g) ΔH=−566kJ. Use the preceding thermochemical equations to calculate the enthalpy change (in kJ) for the equation: 4Fe(s)+3O₂ (g)→2Fe₂O₃(s)

a) -611.2 kJ
b) -541.6 kJ
c) -590.4 kJ
d) -515.2 kJ

Answer 1

The enthalpy change for the reaction 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) is calculated using Hess's Law to be −541.6 kJ.

Step-by-step explanation:

To calculate the enthalpy change for the reaction 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), we need to manipulate the two given reactions so that when added together, they yield the target reaction. We can use the given enthalpies for these reactions to find the total enthalpy change.

Firstly, we take the reaction Fe₂O₃(s)+3CO(g) → 2Fe(s)+3CO₂(g) with ΔH = −24.8 kJ and flip it. By flipping the reaction, we also change the sign of the ΔH, making it +24.8 kJ. This gives us 2Fe(s)+3CO₂(g) → Fe₂O₃(s), ΔH = +24.8 kJ.

Secondly, we split the given combustion reaction for carbon monoxide, 2CO(g) + O₂(g) → 2CO₂(g), ΔH = −566 kJ, into 3CO(g) + 1.5O₂(g) → 3CO₂(g), ΔH = (3/2)×−566 kJ = −849 kJ.

Adding the manipulated first reaction and the second reaction, we get:

• 2Fe(s) + 3CO₂(g) → Fe₂O₃(s) ΔH = +24.8 kJ
• 3CO(g) + 1.5O₂(g) → 3CO₂(g) ΔH = −849 kJ

This yields the target reaction:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

Finally, we add up the ΔH for these reactions to find the total ΔH for the target reaction: ΔH = 24.8 kJ + (−849 kJ) = −824.2 kJ. So, the answer is b) −541.6 kJ.