Question

Use Hess's Law to calculate the enthalpy change for the following reaction:

3Fe₂O₃(s) + CO(g) → CO₂(g) + 2 Fe₃O₄(s) Delta H?
given the following data: Fe₂O₃ (s) + 3CO (s) → 2Fe(s) + 3CO₂(g) Delta H= -28.0kJ
3Fe(s) + 4CO₂(s) → 4CO(g) +Fe3₃O₄(s) delta H = +12.5kJ

Answer 1

Using Hess's law, the enthalpy change for the reaction 3Fe₂O₃(s) + CO(g) → CO₂(g) + 2Fe₃O₄(s) is calculated by manipulating and summing the given reactions, resulting in a total ΔH of -96.5 kJ.

Step-by-step explanation:

To calculate the enthalpy change for the reaction 3Fe₂O₃(s) + CO(g) → CO₂(g) + 2Fe₃O₄(s), we must manipulate the provided reactions to match this target equation using Hess's law.

We are given:

1. Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g), ΔH = -28.0 kJ
2. 3Fe(s) + 4CO₂(g) → 4CO(g) + Fe₃O₄(s), ΔH = +12.5 kJ

First, multiply Equation 1 by 3 to match the amount of Fe₂O₃ and CO on the reactant's side of our target equation.

3[Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)]

Then, reverse Equation 2 so that Fe₃O₄ is on the product side:

Fe₃O₄(s) + 4CO(g) → 3Fe(s) + 4CO₂(g), ΔH = -12.5 kJ

Adding the multiplied first reaction and reversed second reaction gives:

3Fe₂O₃(s) + CO(g) → CO₂(g) + 2Fe₃O₄(s), ΔH = 3(-28.0 kJ) + (-12.5 kJ)

The total ΔH is -96.5kJ.