Question

Calculate the standard enthalpy of the reaction.

\[\ce{2Fe_{(s)} + \frac{3}{2} O_{2(g)} -> Fe2O_{3(s)}}\]

Given:

1.	\[\ce{2Al_{(s)} + Fe2O_{3(s)} -> 2Fe_{(s)} + Al_2O_{3(s)}}\],	∆rH° = –847.6 kJ
2.	\[\ce{2Al_{(s)} + \frac{3}{2} O_{2(g)} -> Al2O_{3(s)}}\],	∆rH° = –1670 kJ

Answer 1

Given: Given equations are,

\[\ce{2Al_{(s)} + Fe2O_{3(s)} -> 2Fe_{(s)} + Al_2O_{3(s)}}\],  ∆_rH° = –847.6 kJ    .....(1)

\[\ce{2Al_{(s)} + \frac{3}{2} O_{2(g)} -> Al2O_{3(s)}}\],    ∆_rH° = –1670 kJ    .......(2)

To find: Standard enthalpy of the given reaction

Calculation: Reverse equation (1),

\[\ce{2Fe_{(s)} + Al_2O_{3(s)} -> 2Al_{(s)} + Fe2O_{3(s)}}\],     Δ_rH° = 847.6 kJ …. (3)

Add equation (2) to equation (3),

\[\ce{2Fe_{(s)} + Al_2O_{3(s)} -> 2Al_{(s)} + Fe2O_{3(s)}}\],	ΔrH° = 847.6 kJ
\[\ce{2Al_{(s)} + \frac{3}{2} O_{2(g)} -> Al2O_{3(s)}}\],	∆rH° = –1670 kJ
\[\ce{2Fe_{(s)} + \frac{3}{2} O_{2(g)} -> Fe2O_{2(s)}}\],

Δ_rH° = 847.6 + (−1670) = −822.4 kJ

The standard enthalpy of the given reaction is −822.4 kJ.