Calculate heat of formation of `KOH(s)` using the following equations
`K(s) +H_(2)O(l) +aq rarr KOH(aq)+1//2 H_(2)O(g),DeltaH =- 48.0 kcal …(i)`
`H_(2)(g) +2O_(2)(g) rarr H_(2)O(l),DeltaH =- 68.4 kcal …(ii)`
`KOH(s) +(aq) rarr KOH(aq),DeltaH =- 14.0 kcal ....(iii)`

Explanation

We have to evaluate `DeltaH` for
`K(s) +1//2O_(2)(g) +1//2H_(2)(g) rarr KOH(s), DeltaH = ?`
Adding equations (i) and (ii),
`K(s)+H_(2)O(l) +aq rarrKOH(aq) +1//2H_(2)(g)`
`DeltaH =- 48.0 kcal`
`H_(2)(g)+1//2O_(2)(g) rarr H_(2)O(l)`,
`DeltaH =- 68.4 kcal`
`{:ulbar(K(s)+1//2H_(2)(g)+1//2O_(2)(g)+aqunderset(DeltaH=-116.4Kcal...(iv))(rarrKOH(aq)):}`
Subtracting equation (iii) from equation (iv) `{:(KOH(s)+aqrarrunderset(DeltaH =- 102.4 kcal ..(v))(KOH(aq),)),("- - - +" ),(ulbar(K(s)+1//2H_(2)(g)+1//2O_(2)(g)rarrunderset(DeltaH =- 102.4 kcal)(KOH(s),))):}`