The enthalpy change (\( \Delta H_2 \)) for the reverse reaction, forming 2Al₂O₃ from 4Al and 3O₂, is the negative of the enthalpy change (\( \Delta H_1 \)) for the forward reaction, which is the formation of Al₂O₃ from 2Al and 3O₂.
To express the enthalpy of the second reaction, \( \Delta H_2 \), in terms of \( \Delta H_1 \), we need to consider the reverse of the first reaction since the second reaction is the reverse of the first.
The given reaction for the formation of aluminum oxide is:
\[ 4\text{Al}(s) + 3\text{O}_2(g) \rightarrow 2\text{Al}_2\text{O}_3(s) \quad \Delta H_1 \]
Now, the second reaction is the reverse of the first:
\[ 2\text{Al}_2\text{O}_3(s) \rightarrow 4\text{Al}(s) + 3\text{O}_2(g) \]
The enthalpy change for the reverse reaction (\( \Delta H_2 \)) is the negative of the enthalpy change for the forward reaction (\( \Delta H_1 \)):
\[ \Delta H_2 = -\Delta H_1 \]
Therefore, \( \Delta H_2 = -\Delta H_1 \).
The probable question may be:
Consider the reaction for the formation of aluminum oxide from aluminum and oxygen.
4Al(s)+3O2(g)⟶2Al2O3(s) Δ1
1. Express the enthalpy of the following reaction, Δ2, in terms of Δ1.
2Al2O3(s)⟶4Al(s)+3O2(g) Δ2
Δ2=