Question

The bond dissociation energy of gaseous $H_2,C{l}_2$ and HCl are 104, 58 and 103 $kcalmo{l}^{-1}$ respectively. The enthalpy of formation for $HCl$ gas will be

• A. $-44.0kcal$
• B. $-22.0kcal$
• C. $22.0kcal$
• D. $44.0kcal$

Answer 1

The correct option is B $-22.0kcal$

$$\begin{array}{cc}{H}_2\left(g\right)\to 2H\left(g\right);& △H=104kcal....\left(1\right)\\ C{l}_2\left(g\right)\to 2Cl\left(g\right)& △H=58kcal....\left(2\right)\\ HCl\left(g\right)\to H\left(g\right)+Cl\left(g\right)& △H=103kcal....\left(3\right)\end{array}$$
Heat of formation for $HCl$
$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}C{l}_2\left(g\right)\to HCl\left(g\right)△H=?$
Divide equations (1) and (2) by 2, and then add
$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}C{l}_2\left(g\right)\to H\left(g\right)+Cl\left(g\right);△H=81kcal....\left(4\right)$

$HCl\left(g\right)\to H\left(g\right)+Cl\left(g\right);△H=103kcal$
Subtracting equation (3) from equation (4) gives enthalpy of formation of HCl
$\therefore$ Enthalpy of formation of HCl gas = $-22.0kcal$