Respuesta :
Answer: The enthalpy of the reaction is coming out to be -361.86 kJ.
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 [tex]\Delta H[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}][/tex]
For the given chemical reaction:
[tex]2Al^{3+}(aq.)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(AlCl_3(s))})]-[(2\times \Delta H^o_f_{(Al^{3+}(aq.))})+(3\times \Delta H^o_f_{(Cl_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(AlCl_3(s))}=-705.63kJ/mol\\\Delta H^o_f_{(Al^{3+}(aq.))}=-524.7kJ/mol\\\Delta H^o_f_{(Cl_2)}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(2\times (-705.63))]-[(2\times (-524.7))+(3\times (0))]\\\\\Delta H^o_{rxn}=-361.86kJ[/tex]
Hence, the enthalpy of the reaction is coming out to be -361.86 kJ.
