Question

Following are the thermochemical reactions:

$H_2+\frac{1}{2}{O}_2\to H_{2}O;\Delta H=-68.39$ kcal/mol
$K+H_{2}O\to KOH\left(aq\right)+\frac{1}{2}{H}_2;\Delta H=-48.0$ kcal/mol

$KOH+H_{2}O\to KOH\left(aq\right);\Delta H=-14.0$ kcal/mol

The heat of formation (in kcal/mol) of $KOH$ is:

• A. $-68.39+48-14.0$
• B. $-68.39-48.0+14.0$
• C. $+68.39-48.0+14.0$
• D. $+68.39+48.0-14.0$

Answer 1

The correct option is B $-68.39-48.0+14.0$

The thermochemical equations are given below.

$H_2+\frac{1}{2}{O}_2\to H_{2}O;\Delta H=-68.39$ kcal/mol $......\left(1\right)$

$K+H_{2}O\to KOH\left(aq\right)+\frac{1}{2}{H}_2;\Delta H=-48.0$ kcal/mol $......\left(2\right)$

$KOH+H_{2}O\to KOH\left(aq\right);\Delta H=-14.0$ kcal/mol $......\left(3\right)$

The reaction $\left(3\right)$ is reversed and added to reactions $\left(1\right)$ and $\left(2\right)$ to obtain the reaction for the formation of $KOH$.

$K+\frac{1}{2}{H}_2+\frac{1}{2}{O}_2\to KOH$

Hence, the heat of formation of $KOH$ is $-68.39-48.0+14.0$ kcal/mol.