Given 2Al(s) + 3/2O2(g) → Al2O3(s), ΔH°f = -1,670 kJ/mol for Al2O3 (s). Determine ΔH° for the reaction 2Al2O3(s) → 4Al(s) + 3O2(g).

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To find the enthalpy change (ΔH°) for the reaction 2Al2O3(s) → 4Al(s) + 3O2(g), we can use the given ΔH°f value for Al2O3(s) and the stoichiometric coefficients of the balanced equation.

The enthalpy change for a reaction is equal to the sum of the enthalpy changes for the formation of the products minus the sum of the enthalpy changes for the formation of the reactants.

Given:

ΔH°f = -1,670 kJ/mol for Al2O3(s)

We need to use the ΔH°f values to calculate the enthalpy change for the formation of the compounds involved in the reaction.

Reactants:
- 2 moles of Al2O3(s)

Products:
- 4 moles of Al(s)
- 3 moles of O2(g)

Calculations:

For reactants, the enthalpy change will be:
ΔH°reactants = 2 * ΔH°f (Al2O3)

For products, the enthalpy change will be:
ΔH°products = 4 * ΔH°f (Al) + 3 * ΔH°f (O2)

Substituting the given values:
ΔH°reactants = 2 * (-1,670 kJ/mol)
ΔH°products = 4 * ΔH°f (Al) + 3 * 0 (O2)

Now, we need to find ΔH°f for Al, which can be found in a thermochemical table. Let's assume it is ΔH°f (Al) = X kJ/mol.

ΔH°products = 4 * X + 3 * 0
ΔH°products = 4X

Since ΔH°reactants and ΔH°products are for the same reaction, they should be equal.

Therefore:
2 * (-1,670) = 4X

Solving for X:
-3,340 = 4X
X = -3,340 / 4
X = -835 kJ/mol

So, the enthalpy change (ΔH°) for the reaction 2Al2O3(s) → 4Al(s) + 3O2(g) is -835 kJ/mol.

To determine ΔH° for the given reaction: 2Al2O3(s) → 4Al(s) + 3O2(g), you can use the concept of Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is the sum of the individual enthalpy changes of a series of reactions that make up the overall reaction.

In this case, you need to find a combination of reactions that adds up to the given equation: 2Al2O3(s) → 4Al(s) + 3O2(g).

First, let's reverse the given equation: Al2O3(s) → 2Al(s) + 3/2O2(g).

Secondly, let’s double the reversed equation to match the coefficients of the reaction we are trying to find: 2Al2O3(s) → 4Al(s) + 3O2(g).

Now, when adding these two reactions together, you'll notice that the Al(s) and O2(g) on both sides of the reaction will cancel out:

Al2O3(s) + Al2O3(s) → 2Al(s) + 2Al(s) + 3/2O2(g) + 3/2O2(g)

This simplifies to: 2Al2O3(s) → 4Al(s) + 3O2(g).

Since we have reversed the first reaction and doubled it, the ΔH° for the second reaction will also be the opposite and double of the ΔH° of the first reaction.

Given that ΔH°f = -1,670 kJ/mol for Al2O3(s), the ΔH° for the reaction 2Al2O3(s) → 4Al(s) + 3O2(g) will be -1,670 kJ/mol × 2 = -3,340 kJ/mol.

Therefore, the ΔH° for the reaction 2Al2O3(s) → 4Al(s) + 3O2(g) is -3,340 kJ/mol.