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Artyom0805 [142]
3 years ago
6

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)?
Chemistry
2 answers:
bekas [8.4K]3 years ago
8 0
Google knows your answer
adell [148]3 years ago
7 0

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

<u>Explanation:</u>

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.

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