Question

Reaction $ {{\text{H}}_2}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_2} \to {{\text{H}}_2}{{O;\Delta }}{{\text{H}}^{\text{0}}}{\text{ = }} - 68 $ Kcal
 $ {\text{K + }}{{\text{H}}_{\text{2}}}{\text{O + aq}} \to {\text{KOH(aq) + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{H}}_{\text{2}}}{{;\Delta }}{{\text{H}}^{\text{0}}}{\text{ = }} - 48 $ Kcal
 $ {\text{KOH + aq}} \to {{KOH(aq);\Delta }}{{\text{H}}^{\text{0}}}{\text{ = }} - 14 $ Kcal
From the above data, what is the standard heat of formation of $ {\text{KOH}} $ in Kcal?
(A) $ - 68 + 48 - 14 $
(B) $ - 68 - 48 + 14 $
(C) $ 68 - 48 + 14 $
(D) $ 68 + 48 + 14 $

Answer 1

Hint: In the above question, it is asked about standard heat of formation of KOH in kcal. We are provided with 3 different equations and their heat of formation. We will first write the required and compare it with the 3 equations given as the options given are a combination of the three equations of heat of formation.

Complete step by step solution:
Let us write the given equations:
 $ {{\text{H}}_2}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_2} \to {{\text{H}}_2}{{O;\Delta }}{{\text{H}}^{\text{0}}}{\text{ = }} - 68 $ Kcal
 $ {\text{K + }}{{\text{H}}_{\text{2}}}{\text{O + aq}} \to {\text{KOH(aq) + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{H}}_{\text{2}}}{{;\Delta }}{{\text{H}}^{\text{0}}}{\text{ = }} - 48 $ Kcal
 $ {\text{KOH + aq}} \to {{KOH(aq);\Delta }}{{\text{H}}^{\text{0}}}{\text{ = }} - 14 $ Kcal
The required equation is:
 $ {\text{K + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{H}}_{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}} \to {\text{KOH}} $
Now, let us calculate the heat of formation, for this let us first add the three equations:
 $ {{\text{H}}_2}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_2} + {\text{K + }}{{\text{H}}_{\text{2}}}{\text{O + KOH}} \to {{\text{H}}_2}{\text{O + KOH(aq) + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{H}}_{\text{2}}} + {\text{KOH(aq)}} $
We can see that we 1 molecule of water on both the side of reactant and product, so we can simplify it as:
 $ {{\text{H}}_2}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_2} + {\text{K + KOH}} \to {\text{KOH(aq) + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{H}}_{\text{2}}} + {\text{KOH(aq)}} $
Since , we have 1 molecule of KOH on reactant side and 2 molecules of KOH on product side and half molecule of oxygen on reactant side and 1 molecule of oxygen on product side and hence, by rearranging, we get:
 $ \dfrac{1}{2}{{\text{H}}_2}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_2} + {\text{K}} \to {\text{KOH(aq)}} $
This is the desired equation which is obtained by adding the above three equations and hence, we can find the heat of formation by adding all the heat enthalpies.
Hence, we get $ {{\Delta }}{{\text{H}}^{\text{o}}}{\text{ = }} - 68 - 48 + 14 $
So, the correct option is option B.

Note:
Heat of formation can be defined as the amount of heat which is absorbed or evolved when one mole of a compound is formed from its constituent elements with each substance being in its normal physical state.