Question

a chemist measures the enthalpy change δh during the following reaction:
4fe(s) + 3o₂(g)→2fe₂o₃(s) δh = -320. kj
use this information to complete the table below. round each of your answers to the nearest kj.

reaction	δh
fe₂o₃(s) → 2fe(s) + 3/2 o₂(g)	kj
fe(s) + 3/4 o₂(g) → 1/2 fe₂o₃(s)	kj

Explanation:

Step1: Reverse the original reaction

Original: $4Fe(s)+3O_2(g)→2Fe_2O_3(s)$ $\Delta H=-320$ kJ. Reverse: $2Fe_2O_3(s)→4Fe(s)+3O_2(g)$, so $\Delta H=+320$ kJ.

Step2: Halve the reversed reaction

Reversed reaction halved: $Fe_2O_3(s)→2Fe(s)+\frac{3}{2}O_2(g)$. $\Delta H=\frac{320}{2}=160$ kJ.

Step3: Adjust original to target reaction

Original divided by 4: $Fe(s)+\frac{3}{4}O_2(g)→\frac{1}{2}Fe_2O_3(s)$. $\Delta H=\frac{-320}{4}=-80$ kJ.

Answer:

320
160
-80