Calculate the heat of formation of KOH from the following data in K.Cal.
K_(s) + H₂O + aq → KOH_(aq)+1/2 H₂; ΔH = -48.4K.Cal
H_2(g) + 1/2O_2(g) → H₂O_(l); ΔH = -68.44K.Cal
KOH_(s) + aq → KOH_(aq); ΔH = -14.01K.Cal

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– 102.83

– 130.85

+130.85

+102.83

answer is C.

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Detailed Solution

$\begin{array}{l} eq.(1) \Rightarrow {K_{(s)}} + {\rm{ }}{H_2}O{\rm{ }} + {\rm{ }}aq{\rm{ }} \to {\rm{ }}KO{H_{(aq)}} + 1/2{\rm{ }}{H_2};{\rm{ }}\Delta {H_1}{\rm{ }} = {\rm{ }} - 48.4K.Cal\\ eq.(2) \Rightarrow {H_{2(g)}} + {\rm{ }}1/2{O_{2(g)}} \to {\rm{ }}{H_2}{O_{(l)}};{\rm{ }}\Delta {H_2}{\rm{ }} = {\rm{ }} - 68.44K.Cal\\ \underline {\frac{1}{{eq.(3)}} \Rightarrow KO{H_{(aq)}} \to KO{H_{(s)}} + {\rm{ }}aq;\Delta {H_3}{\rm{ }} = {\rm{ + }}14.01K.Cal\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \,\\ On\,addition\,\,\,{K_{(s)}} + 1/2{O_{2(g)}} + 1/2{\rm{ }}{H_2} \to KO{H_{(s)}};\,\Delta H = - 48.4 - 68.44 + 14.01\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 102.83Kcal \end{array}$