Question

Dry chemical hand warmers utilize the oxidation of iron to form iron oxide according to the following reaction: 4Fe(s)+3O2(g)→2Fe2O3(s) Standard thermodynamic quantities for selected substances at 25 ∘C Reactant or product ΔH∘f(kJ/mol) Fe(s) 0.0 O2(g) 0.0 Fe2O3(s) −824.2 Calculate ΔH∘rxn for this reaction.

Answer 1

-1648.4 kJ

Step-by-step explanation:

The product has the only nonzero heat of formation, so it is the only value needed to calculate the enthalpy of this reaction. Normally, you would want to express the enthalpy of a reaction with respect to one mole of a chemical species, whether it is a reactant or product. However, since the balanced chemical equation contains only coefficients greater than 1, you should consider how the enthalpy relates to one mole of each substance according to the coefficients. In other words, − 1648.4 kJ of heat is released when 4 mol of Fe reacts with 3 mol of O2 to produce 2 mol of Fe2O3 .

Answer 2

-1648.4 kJ/mol

Step-by-step explanation:

Based on Hess's law:

ΔHr = ∑n×ΔH°f(products) - ∑n×ΔH°f(reactants)

In the reaction:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

ΔHr = 2 ΔH°f {Fe₂O₃} - (4ΔH°f {Fe(s)} + 3ΔH°f{O₂(g)}

As:

ΔH°f {Fe₂O₃} = -824.2kJ/mol

ΔH°f {Fe(s)} = 0.0kJ/mol

ΔH°f{O₂(g)} = 0.0kJ/mol.

Thus,

ΔHr = 2 ₓ -824.2kJ/mol =

-1648.4 kJ/mol