Question

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

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

Answer 1

Answer: (B)

$-22.0$ kcal

$\frac{1}{2}{H}_2+\frac{1}{2}C{l}_2\to HCl$

${\Delta }_f{H}_{\left(HCl\right)}=\Delta H\left(\text{Rea}\right)-\Delta H\left(\text{Pro}\right)$

Implies that

$=\frac{1}{2}\left(104\right)+\frac{1}{2}\left(58\right)-103=-22kcal$.