Question

Calculate the heat of formation of $KOH$ from the following data in K.Cal
$K_{\left(s\right)}+H_{2}O+aq\to {KOH}_{\left(aq\right)}+\frac{1}{2}{H}_2;\Delta H=-48.4K.Cal$
$H_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to H_2{O}_{\left(l\right)};\Delta H=-68.44K.Cal$
${KOH}_{\left(s\right)}+aq\to {KOH}_{\left(aq\right)};\Delta H=-14.0K.Cal$

• A. $+102.83$
• B. $+130.85$
• C. $-102.83$
• D. $-130.85$

Answer 1

Answer: (C)

$-102.83$

Reaction of formation of $KOH$ is
$K\left(s\right)+\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to KOH\left(s\right)$ ............. (i)
$K\left(s\right)+H_{2}O+aq\to KOH\left(aq\right)+\frac{1}{2}{H}_2$ ; $\Delta H_1=-48.4$ $Kcal$
$H_2+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)$............... (ii) ; $\Delta H_2=-68.44$ $Kcal$.
$KOH\left(s\right)+aq\to KOH\left(aq\right)$.............(iii) ; $\Delta H_3=-14.0$ $Kcal$
On adding equation $\left(i\right)$& $\left(ii\right)$& subtarcting $\left(iii\right)$ from them we get equation $\left(i\right)$
$\Delta H_f=\Delta H_1+\Delta H_2-\Delta H_3$
$=(-48.4-68.44+4.0)$ $Kcal$
$\Delta H_f=-102.83$ $Kcal$