Question

Calculate the heat formation of anhydrous ${Al}_2{Cl}_6$ from
$2Al\left(s\right)+6HCl\left(aq\right)⟶{Al}_2{Cl}_6\left(aq\right)+3H_2\left(g\right),\Delta H=-239.76kcal$
$H_2\left(g\right)+{Cl}_2\left(g\right)⟶2HCl\left(g\right);\Delta H=-44.0kcal$
$HCl\left(g\right)+Aq⟶HCl\left(aq\right);\Delta H=-17.32kcal$
${Al}_2{Cl}_6\left(s\right)+Aq⟶{Al}_2{Cl}_6\left(aq\right);\Delta H=-153.69kcal$

Answer 1

$2Al\left(s\right)+{3Cl}_2\left(s\right)⟶{Al}_2{Cl}_6\left(s\right)\Delta H=?$

(i) $2Al\left(s\right)+6HCl\left(aq\right)\to {Al}_2{Cl}_6\left(aq\right)+3H_2\left(or\right),{\Delta H}_1=-239.76kCal$

(ii) $H_2\left(g\right)+{Cl}_2\left(g\right)\to 2HCl\left(g\right),{\Delta H}_2=-44.0kCal$

(iii) $HCl\left(g\right)+Av\to HCl\left(aq\right),{\Delta H}_3=-17.32kCal$

(iv) ${Al}_2{Cl}_6\left(s\right)+Aq\to {Al}_2{Cl}_6\left(aq\right),{\Delta H}_4=-153.6kCal$

${Al}_2{Cl}_6$ formation can be given by :

$\left(i\right)+3\times \left(ii\right)-6\times \left(iii\right)-\left(iv\right)\Rightarrow$

${\Delta H}_{{Al}_2{Cl}_6}={\Delta H}_1+3\times {\Delta H}_2-6\times {\Delta H}_3-{\Delta H}_4$

$=\{-239.76+3\times (-44.0)-6\times (-17.32)-(-153.6)\}kCal$

$=-114.24kCal$