Question

If $H_2+\frac{1}{2}{O}_2\to H_{2}O;\Delta H=-68.39Kcal$
$K+H_{2}O+water\to KOH\left(aq\right)+\frac{1}{2}{H}_2$;
$\Delta H=-48.0Kcal$
$KOH+water\to KOH\left(aq\right)\Delta H=-14.0Kcal$ the heat of formation 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

As given,

$H_2+\frac{1}{2}{O}_2\to H_{2}O;\Delta H=-68.39Kcal$
$K+H_{2}O+water\to KOH\left(aq\right)+\frac{1}{2}{H}_2;\Delta H=-48.0Kcal$
$KOH+water\to KOH\left(aq\right)\Delta H=-14.0Kcal$

and required is

$K+1/2H_2+1/2O_2\to KOH;\Delta H=?$

So by calculating using above equations

$K+1/2H_2+1/2O_2\to KOH;\Delta H=-68.39-48.0+14.0$