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.

Step-by-step 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* \Delta H^o_f_((product))]-\sum [n* \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* \Delta H^o_f_((AlCl_3(s))))]-[(2* \Delta H^o_f_((Al^(3+)(aq.))))+(3* \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* (-705.63))]-[(2* (-524.7))+(3* (0))]\\\\\Delta H^o_(rxn)=-361.86kJ`

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