Question

using the data below, calculate the enthalpy for the formation of rust, iron(ii) oxide.
2fe(s) + 3h₂o(g) → fe₂o₃(s) + 3h₂(g)
compound | s° (j/mol k) | δh°f (kj/mol)
fe(s) | 15 | 21
h₂o(g) | 189 | -242
fe₂o₃(s) | 87 | -824
h₂(g) | 131 | 0
δhᵣₓₙ = ? kj
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for enthalpy of reaction

The formula for the enthalpy of a reaction (\(\Delta H_{rxn}\)) is \(\Delta H_{rxn}=\sum\Delta H_{f}^{\circ}(\text{products})-\sum\Delta H_{f}^{\circ}(\text{reactants})\).

Step2: Identify products and reactants with their stoichiometric coefficients

• Products: \(\ce{Fe2O3(s)}\) (coefficient = 1) and \(\ce{H2(g)}\) (coefficient = 3)
• Reactants: \(\ce{Fe(s)}\) (coefficient = 2) and \(\ce{H2O(g)}\) (coefficient = 3)

Step3: Calculate \(\sum\Delta H_{f}^{\circ}(\text{products})\)

For \(\ce{Fe2O3(s)}\): \(\Delta H_{f}^{\circ}=- 824\space kJ/mol\), so contribution is \(1\times(-824)\space kJ\)
For \(\ce{H2(g)}\): \(\Delta H_{f}^{\circ}=0\space kJ/mol\), so contribution is \(3\times0 = 0\space kJ\)
Sum of products: \(-824+0=-824\space kJ\)

Step4: Calculate \(\sum\Delta H_{f}^{\circ}(\text{reactants})\)

For \(\ce{Fe(s)}\): \(\Delta H_{f}^{\circ}=21\space kJ/mol\), contribution is \(2\times21 = 42\space kJ\)
For \(\ce{H2O(g)}\): \(\Delta H_{f}^{\circ}=-242\space kJ/mol\), contribution is \(3\times(-242)=-726\space kJ\)
Sum of reactants: \(42+(-726)=42 - 726=-684\space kJ\)

Step5: Calculate \(\Delta H_{rxn}\)

Using \(\Delta H_{rxn}=\sum\Delta H_{f}^{\circ}(\text{products})-\sum\Delta H_{f}^{\circ}(\text{reactants})\)
\(\Delta H_{rxn}=(-824)-(-684)=-824 + 684=-140\space kJ\)

Answer:

-140