The heat of formation of KOH is ( in kcal) (a) −68.39 + 48 − 14 (b) −68.39 − 48 + 14

Question

H₂ + 1/2O₂ → H₂O, ΔH = − 68.39 kcal

K + H₂O + water → KOH (aq) + 1/2 H₂ ΔH = −48 kcal

KOH + water → KOH (aq) ΔH = −14 kcal

The heat of formation of KOH is ( in kcal) 

(a) −68.39 + 48 − 14 

(b) −68.39 − 48 + 14 

(c) −68.39 + 48 + 14 

(d) 68.39 + 48 + 14

Answer 1

Correct option (b) −68.39 − 48 + 14

Explanation:

The desired therochemical equation is :

Now, on adding (i), (ii) and (iv), we will get the desired equation