Consider the reaction for the formation of aluminum oxide from aluminum and oxygen: 4Al(s) + 3O2(g) -> 2Al2O3(s). Express the enthalpy of the following reaction, Î”H, in terms of Î”H. 2Al2O3(s) -> 4Al(s) + 3O2(g) Î”H.

The reaction given is:

\[
4Al(s) + 3O_2(g) \rightarrow 2Al_2O_3(s)
\]

This is the formation reaction of aluminum oxide. The enthalpy change for this reaction is \(\Delta H\).

Now, you're asked to express the enthalpy for the reverse reaction:

\[
2Al_2O_3(s) \rightarrow 4Al(s) + 3O_2(g)
\]

In thermodynamics, the enthalpy change for a reverse reaction is the negative of the enthalpy change for the forward reaction. Therefore, the enthalpy change for the reverse reaction is:

\[
\Delta H_{\text{reverse}} = -\Delta H
\]

Thus, the enthalpy of the reaction \( 2Al_2O_3(s) \rightarrow 4Al(s) + 3O_2(g) \) is \( -\Delta H \).

Would you like further clarification on this concept or additional details?

Here are 5 related questions you might explore:
1. What is the enthalpy change for the formation of aluminum oxide per mole of aluminum?
2. How does Hess's Law relate to the enthalpy changes in these reactions?
3. What is the physical significance of a negative enthalpy change?
4. How does the stoichiometry of a reaction affect the calculation of enthalpy changes?
5. What is the difference between standard enthalpy of formation and the enthalpy change of a reaction?

**Tip:** The reverse of an exothermic reaction is endothermic, and vice versa.

Consider the reaction for the formation of aluminum oxide from aluminum and oxygen: 4Al(s) + 3O2(g) -> 2Al2O3(s). Express the enthalpy of the following reaction, ΔH, in terms of ΔH: 2Al2O3(s) -> 4Al(s) + 3O2(g) ΔH.

Explore the thermochemical concepts behind the formation and decomposition of aluminum oxide, focusing on the calculation of enthalpy change (ΔH) for reversible reactions. Learn about key principles like Hess's Law and the Law of Conservation of Energy. Ideal for high school students in Grades 10-12.

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