Given ∆Hºrxn = -1670 kJ/mol for 2Al(s) + (3/2)O2(g) --> Al2O3(s), determine ∆Hº for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g).
I flipped Al(s) + (3/2)O2(g) --> Al2O3(s) to
Al2O3(s) --> A(g) + (3/2)O2(s) and multiplied the molar values by 2.
I ended up with ∆Hº = 3340 kJ/mol
Is this correct?
No, your calculation is not correct. To determine the ΔHº for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g), you need to use the given ΔHºrxn for 2Al(s) + (3/2)O2(g) --> Al2O3(s).
The given reaction is: 2Al(s) + (3/2)O2(g) --> Al2O3(s) with ΔHºrxn = -1670 kJ/mol.
To find the ΔHº for the reverse reaction, you just need to change the sign: Al2O3(s) --> 2Al(s) + (3/2)O2(g) with ΔHºrxn for the reverse reaction = +1670 kJ/mol.
Now, to determine the ΔHº for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g), you need to use the stoichiometric coefficients of the given reaction. Multiply each ΔHº value by the corresponding stoichiometric coefficient:
2 x [ΔHºrxn for Al2O3(s) --> 2Al(s) + (3/2)O2(g)] = 2 x (+1670 kJ/mol) = +3340 kJ/mol.
So, the correct value for ΔHº for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g) is ΔHº = +3340 kJ/mol.
To determine ΔHº for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g), you can use the given ΔHº for the reaction 2Al(s) + (3/2)O2(g) --> Al2O3(s).
The equation you provided for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g) is balanced correctly. However, to find ΔHº for this reaction, you need to consider the stoichiometric coefficients and manipulate the given equation accordingly.
The given equation 2Al(s) + (3/2)O2(g) --> Al2O3(s) has a ΔHº of -1670 kJ/mol. This means that the reaction releases 1670 kJ of energy per mole of Al2O3 formed.
To calculate ΔHº for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g), we need to consider the stoichiometric relationships between the two equations. Multiplying the given equation by 2 does not correctly represent the relationship between the two reactions.
Instead, you can reverse the given equation to match the reaction we want, which gives Al2O3(s) --> 2Al(s) + (3/2)O2(g). When you reverse a reaction, the sign of ΔHº also flips. Thus, the new equation has a ΔHº of +1670 kJ/mol.
Now, to obtain the desired reaction 2Al2O3(s) --> 4Al(s) + 3O2(g), we can double the coefficients of the new equation as follows:
2 (Al2O3(s) --> 2Al(s) + (3/2)O2(g))
This gives us:
2Al2O3(s) --> 4Al(s) + 3O2(g)
Since we have multiplied the entire equation by 2, ΔHº is also multiplied by 2. Therefore, the final ΔHº for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g) is:
2 * 1670 kJ/mol = 3340 kJ/mol.
Therefore, your final answer of ΔHº = 3340 kJ/mol is correct.