Question

Using standard heats of formation found in the appendix in your textbook, what is ΔH∘rxn for the following process. 2Al3+(aq)+3Cl2(g)→2AlCl3(s)

Answer 1

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 $\Delta H$

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:

$2Al^{3+}(aq.)+3Cl_2(g)\rightarrow 2AlCl_3(s)$

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

$\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(AlCl_3(s))})]-[(2\times \Delta H^o_f_{(Al^{3+}(aq.))})+(3\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(AlCl_3(s))}=-705.63kJ/mol\\\Delta H^o_f_{(Al^{3+}(aq.))}=-524.7kJ/mol\\\Delta H^o_f_{(Cl_2)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times (-705.63))]-[(2\times (-524.7))+(3\times (0))]\\\\\Delta H^o_{rxn}=-361.86kJ$

Hence, the enthalpy of the reaction is coming out to be -361.86 kJ.