Question

Consider the reaction:

2Al(s) + Fe2O3(s) - Al2O3(s) + 2Fe(s)
The AH, for Fe2O3(s) = -824.3 kJ/mole. The AH, for Al2O3(s) = -1675.7 kJ/mole.
Finish the equation.
AHxn = [(1)
C
Y
kJ/mole) + (2)
Y kJ/mole)] - [(1)
kJ/mole) + (2) (
kJ/mole)]

Answer 1

`\large \boxed{\text{-851.4 kJ/mol}}`

Step-by-step explanation:

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

`\Delta_{\text{r}}H^(\circ) = \sum \Delta_{\text{f}} H^(\circ) (\text{products}) - \sum\Delta_{\text{f}}H^(\circ) (\text{reactants})`

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)

ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0

`\begin{array}{rcl}\Delta_{\text{r}}H^(\circ) & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}`