Question

Consider the reaction 2Al(OH)3(s)→Al2O3(s)+3H2O(l) with enthalpy of reaction ΔHrxn∘=21.0kJ/mol What is the enthalpy of formation of Al2O3(s)?

Answer 1

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Answer 2

Answer: The enthalpy of the formation of $Al_2O_3(s)$ is coming out to be -399.5 kJ/mol.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$2Al(OH)_3(s)\rightarrow Al_2O_3(s)+3H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(Al_2O_3(s))})+(3\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(Al(OH)_3(s))})]$

We are given:

$\Delta H^o_f_{(H_2O(l))}=-285.5kJ/mol\\\Delta H^o_f_{(Al(OH)_3(s))}=-1277kJ/mol\\\Delta H^o_{rxn}=21.0kJ$

Putting values in above equation, we get:

$21.0=[(1\times \Delta H^o_f_{(Al_2O_3(s))})+(3\times (-285.8))]-[(1\times (-1277))]\\\\\Delta H^o_f_{(Al_2O_3(s))}=-399.5kJ/mol$

Hence, the enthalpy of the formation of $Al_2O_3(s)$ is coming out to be -399.5 kJ/mol.