contestada

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)]

Respuesta :

znk

Answer:

[tex]\large \boxed{\text{-851.4 kJ/mol}}[/tex]

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

[tex]\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})[/tex]

                            2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)

ΔfH°/kJ·mol⁻¹:         0         -824.3         -1675.7         0

[tex]\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}}[/tex]