Question
Calculate the enthalpy of formation (in kcal/mol) of anhydrous Al2Cl6 from the following data.
2Al(s) + 6HCl(aq) → Al2Cl6(aq) + 3H2(g): ΔH = −239.760 kcal
H2(g) + Cl2(g) → 2HCl(g): ΔH = −44 kcal
HCl(g) + aq → HCl(aq): ΔH = −17.315 kcal
Al2Cl6(s) + aq → Al2Cl6(aq): ΔH = −153.690 kcal
Options
Solution
$(a)2Al(s) + 6HCl(aq) \rightarrow Al_{2}Cl_{6}(aq) + 3H_{2}(g);\Delta H = - 259.76\text{ kcal}$
$(b)H_{2}(g) + Cl_{2}(g) \rightarrow 2HCl(g);\Delta H = - 44\text{ kcal} $$${(c)HCl(g) + aq \rightarrow HCl(aq);\Delta H = - 17.315\text{ Kcal} }{(d)Al_{2}Cl_{6}(s) + aq \rightarrow Al_{2}Cl_{6}(aq);\Delta H = - 153.69\text{ Kcal} }{\text{Required T.C.E. is} }{2Al(s) + 3Cl_{2}(g) \rightarrow Al_{2}Cl_{6}(s);\Delta H = ?}$$
$\text{From }(a) + 3(b) + 6(c) - (d); $$$\Delta H = ( - 239.76) + 3( - 44) + 6( - 17.315) - ( - 153.69) = - 321.96\text{ kcal}$$
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