Question

The heat of formation HCl(g) from the reaction
$H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$; $\Delta H$ =$-44$ kcal is:

• A. $+44kcal$
• B. $-44kcal$
• C. $+22kcal$
• D. $-22kcal$

Answer 1

Answer: (D)

$-22kcal$

${H_2}_{\left(g\right)}+{{Cl}_2}_{\left(g\right)}⟶2{HCl}_{\left(g\right)}\Delta H=-44kcal$

$\frac{1}{2}\times [{H_2}_{\left(g\right)}+{{Cl}_2}_{\left(g\right)}⟶2{HCl}_{\left(g\right)}\Delta H=-44kcal]$

$\frac{1}{2}{H_2}_{\left(g\right)}+\frac{1}{2}{{Cl}_2}_{\left(g\right)}⟶{HCl}_{\left(g\right)}\Delta H=-22kcal$

As we know that enthalpy of formation is the same as the enthalpy of reaction when $1$ mole of product is formed.

Hence the enthalpy of formation of ${HCl}_{\left(g\right)}$ is $-22kcal$.

Hence, the correct option is $\text{D}$