Question

(i) $CaO\left(s\right)+H_{2}O\left(l\right)=Ca\left(OH{)}_2\right(s)$, $\Delta H_{{180}^{o}C}=-15.26kcal$
(ii) $H_{2}O\left(l\right)=H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$, $\Delta H_{{180}^{o}C}=68.37kcal$
(iii) $Ca\left(s\right)+\frac{1}{2}{O}_2\left(g\right)=CaO\left(s\right)$, $\Delta H_{{180}^{o}C}=-151.80kcal$

From the following data, the heat of formation of $Ca\left(OH{)}_2\right(s)$ at ${18}^{\circ }C$ is:

• A. $-98.69$ kcal
• B. $-235.43$ kcal
• C. $194.91$ kcal
• D. $98.69$ kcal

Answer 1

The correct option is B $-235.43$ kcal

The thermochemical reactions are as given below

$CaO\left(s\right)+H_{2}O\left(l\right)=Ca\left(OH{)}_2\right(s),\Delta H_{{180}^{o}C}=-15.26Kcal$ ......(1)

$H_{2}O\left(l\right)=H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right),\Delta H_{{180}^o}C=68.37Kcal$ ......(2)

$Ca\left(s\right)+\frac{1}{2}{O}_2\left(g\right)=CaO\left(s\right),\Delta H_{{180}^{o}C}=-151.80Kcal$ ......(3)

The reaction (2) is reversed and added to the reactions (1) and (3)

Hence, the heat of formation of calcium hydroxide is $-15.26-68.37-151.80=-235.43$ Kcal