Question

$H_2\left(g\right)+\left(\begin{array}{c}\frac{1}{2}\end{array}\right)O_2\left(g\right)=H_{2}O\left(l\right);\Delta H_{298K}=-68.00Kcal$.

Heat of vaporisation of water at $1atm$ and ${25}^{\circ }C$ is $10.00Kcal$.

The standard heat of formation (in Kcal) of a $1mol$ vapour at ${25}^{\circ }C$ is:

• A. $-78.00$
• B. $-58.00$
• C. $+58.00$
• D. $78.00$

Answer 1

The correct option is B $-58.00$

$H_2\left(g\right)+\frac{1}{2}{O}_2⟶H_{2}O\left(l\right);\Delta H_1=-68.00Kcal$

$H_{2}O\left(l\right)⟶H_{2}O\left(g\right);\Delta H_2=10.0Kcal$

The above two reactions are added to obtained:

$H_2\left(g\right)+\frac{1}{2}{O}_2⟶H_{2}O\left(g\right);$

$\Delta H=\Delta H_1+\Delta H_2=-68.00+10.0=-58.0Kcal$

Hence, the standard heat of formation of 1 mole of water vapour at ${25}^{o}C$ is $-58.0kcal$.