Question

Calculate the $\Delta H^{\ominus }$ for the reduction of $F{e}_2{O}_3\left(s\right)$ by $Al\left(s\right)$ at ${25}^{\circ }C$. The enthalpies of formation of $F{e}_2{O}_3\left(s\right)$ and $A{l}_2{O}_3$ are $-825.5$ and $-1675.7kJmo{l}^{-1}$ respectively.

• A. $\Delta H=-850.2kJmo{l}^{-1}$
• B. $\Delta H=+850.2kJmo{l}^{-1}$
• C. $\Delta H=-2500kJmo{l}^{-1}$
• D. None of these

Answer 1

The correct option is A $\Delta H=-850.2kJmo{l}^{-1}$

The reduction of $F{e}_2{O}_3$ by Al is represented by following chemical reaction:
$F{e}_2{O}_3\left(s\right)+2Al\left(s\right)⟶2Fe\left(s\right)+A{l}_2{O}_3;\Delta H=?$ ......(1)

The balanced chemical reaction for the formation of $F{e}_2{O}_3$ is as shown below.
$3Fe\left(s\right)+\frac{3}{2}{O}_2\left(g\right)⟶F{e}_2{O}_3;\Delta H_1=-825.5kJmo{l}^{-1}$ ......(ii)

The balanced chemical reaction for the formation of $A{l}_2{O}_3$ is as shown below.
$2Al\left(s\right)+\frac{3}{2}{O}_2\left(g\right)⟶A{l}_2{O}_3;\Delta H_2=-1675.7kJmo{l}^{-1}$......(iii)

Subtract the enthalpy change for the third reaction from the enthalpy change for the second reaction to obtain the enthalpy change for the first reaction.

$\Delta H=\Delta H_1-\Delta H_2$ $=-1675.7-(-825.5)$ $=-850.2kJmo{l}^{-1}$

Hence, option A is correct.