Question

Aluminum chloride can be formed from its elements:

(i)\({\bf{2Al(s) + 3C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(s) \Delta H^\circ = ?}}\)

Use the reactions here to determine the ΔH° for reaction(i):

\(\begin{array}{*{20}{l}}{\left( {{\bf{ii}}} \right){\rm{ }}{\bf{HCl(g)}} \to {\bf{HCl(aq) \Delta H^\circ (ii) = - 74}}{\bf{.8 kJ}}}\\{\left( {{\bf{iii}}} \right){\rm{ }}{{\bf{H}}_{\bf{2}}}{\bf{(g) + C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2HCl(g) \Delta H^\circ (iii) = - 185 kJ}}}\\{\left( {{\bf{iv}}} \right){\rm{ }}{\bf{AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(aq)}} \to {\bf{AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(s) \Delta H^\circ (iv) = + 323 kJ}}}\\{\left( {\bf{v}} \right){\rm{ }}{\bf{2Al(s) + 6HCl(aq)}} \to {\bf{2AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(aq) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g) \Delta H^\circ (v) = - 1049 kJ}}}\end{array}\)

Short answer

ΔH° = -1430.8 kJ.

Step-by-step solution

Different reactions

The given reaction is:

\({\rm{2Al(s) + 3C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2AlC}}{{\rm{l}}_{\rm{3}}}{\rm{(s) }}.........\left( {\rm{i}} \right)\)

We will get the above reaction (i) by adding the reactions (ii), (iii), (iv), and (v).

First, two reactions (i) and (ii) give a HCl by subtraction. We have to multiply the reaction (ii) by 2 for balancing it.

\(\begin{array}{*{20}{l}}{{{\rm{H}}_{\rm{2}}}{\rm{(g) + C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)\;}} \to {\rm{2HCl(g) ; \Delta H^\circ (iii) = - 185 kJ}}}\\{{\rm{ 2HCl(g)}} \to {\rm{2HCl(aq); \Delta H^\circ (ii) = - 74}}{\rm{.8 kJ}} \times {\rm{2}}}\end{array}\)

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\({{\bf{H}}_{\bf{2}}}{\bf{(g)\; + C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)\;}} \to {\bf{2HCl(aq)\;\;\Delta H^\circ = \; - 334}}{\bf{.6 kJ\;}}..................{\bf{(vi)}}\)

Determination of enthalpy change

We have to multiply reaction (iv) by 2 for balancing. Then we have to subtract the values of enthalpy change of reactions (iv) and (v).

\(\begin{array}{c}\left( {{\rm{iv}}} \right){\rm{ 2AlC}}{{\rm{l}}_{\rm{3}}}{\rm{(aq)}} \to {\rm{2AlC}}{{\rm{l}}_{\rm{3}}}{\rm{(s); \Delta H^\circ (iv) = + 323 kJ}} \times {\rm{2\;}}\\\left( {\rm{v}} \right)\;{\rm{2Al(s) + 6HCl(aq)}} \to {\rm{\;2AlC}}{{\rm{l}}_{\rm{3}}}{\rm{(aq) + 3}}{{\rm{H}}_{\rm{2}}}{\rm{(g); \Delta H^\circ (v) = - 1049 kJ}}\end{array}\)

\({\bf{2Al(s) + \;6HCl(aq)\;}} \to {\bf{2AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(s)\;\;\;\Delta H^\circ = - 403 kJ\;}}...................{\bf{(vii)}}\)

Then, reactions (vi) and (vii) are taken.

For balancing, we have to multiply reaction (vi) by 3.

\(\begin{array}{l}{\rm{ 2}}{{\rm{H}}_{\rm{2}}}{\rm{(g)\; + 3C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)\;}} \to 6{\rm{HCl(aq);\;\;\Delta H^\circ = \; - 342}}{\rm{.6 kJ\; \times 3\;}}.........{\rm{ (vi)}}\\{\rm{2Al(s) + \;6HCl(aq)\;}} \to {\rm{2AlC}}{{\rm{l}}_{\rm{3}}}{\rm{(s);\;\;\Delta H^\circ = - 403 kJ\;}}..........{\rm{ (vii)}}\end{array}\)

\({\bf{2Al(s)\; + 2}}{{\bf{H}}_{\bf{2}}}{\bf{(g)\; + 3C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)\;}} \to {\bf{2AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(s)\;\;\;\;\;\;\Delta H^\circ \; = \; - 1430}}{\bf{.8 kJ}}.\)