Using standard thermodynamic data at 298K, calculate the free energy change when 1.91 moles of Fe2O3(s) react at standard conditions.

Question

Consider the reaction

Fe₂O₃(s) + 3H₂(g)2Fe(s) + 3H₂O(g)

Using standard thermodynamic data at 298K, calculate the free energy change when 1.91 moles of Fe₂O₃(s) react at standard conditions.

    G°_rxn =  ___ kJ

Answer 1

Hey Ahmad!

So for this problem our first step is going to be to look up the standard 298K thermodynamic values of ΔG°. I'm just using a chart I found on chem.wisc.edu however, if you have your textbook available, I'm sure you can find them in there and they may be a little different than mine. This is what I found:

1. ΔG_fº Fe₂O_3(s) = -742.200 kJ/mol
2. ΔG_fº H_2(g) = 0 kJ/mol
3. ΔG_fº Fe_(s) = 0 kJ/mol
4. ΔG_fº H₂O_(g) = -228.572 kJ/mol

After I found these values, I knew that I would have to use the summation formula of:

ΔG_rxn°= ∑nΔG_f°(products)-∑mΔG_f°(reactants)

Remember, "∑" is the sum of, and in this reaction, n and m are referring to moles.

So now, we need to find the values of n and m. We know that 1.91 moles of Fe₂O₃ are reacting, and we are given then balanced chemical equation, so all we have to do is look at our mole ratio.

Fe₂O_3(s) + 3H_2(g) → 2Fe_(s) + 3H₂O_(g)

For every 1 mol of Fe₂O_3(s), there are 3 moles of H_2(g)

1.91 mol Fe₂O_3(s) 3 mol H_2(g)

———————— x ———————— = 5.73 mol H_2(g)

1 1 mol Fe₂O_3(s)

For every 1 mol of Fe₂O_3(s), there are 2 moles of Fe_(s)

1.91 mol Fe₂O_3(s) 2 mol Fe_(s)

———————— x ———————— = 3.82 mol Fe_(s)

1 1 mol Fe₂O_3(s)

For every 1 mol of Fe₂O_3(s), there are 3 moles of H₂O_(g)

1.91 mol Fe₂O_3(s) 3 mol H₂O_(g)

———————— x ———————— = 5.73 mol H₂O_(g)

1 1 mol Fe₂O_3(s)

Now we can solve for ΔGº

ΔG_rxn°= ∑(5.73mol H2O(g) x -228.572 kJ/mol) + (3.82 mol Fe(s) x 0 kJ/mol) - ∑(1.91mol Fe2O3(s)x -742.2 kJ/mol) + (5.73 mol H2(g) x 0 kJ/mol)

Simplify ↓

ΔG_rxn°= ∑(5.73mol H2O(g) x -228.572 kJ/mol) - ∑(1.91mol Fe2O3(s)x -742.2 kJ/mol)

ΔG_rxn° ≈ 108 kJ/mol

I hope this helps!

note: to solve this problem, you could also use the summation formulas for ΔH° and ΔS° and then use the equation: ΔG° = ΔH° - TΔS° however, I find this way to be more time efficient.