Heat of neutralization of HCl by NaOH is 13.7 kcal per equivalent and by `NH_4OH` is 12.27 kcal. The heat of dissociation of `NH_4OH` is

Options

A. `-25.97` kcal
B. 25.97 kcal
C. `-1.43` kcal
D. 1.43 kcal

Answer

D

Explanation

Here,
`H^(+) + NH_4OH to NH_4^(+) + H_2O , DeltaH`=-12.27 kcal
The neutalization may be regarded to proceed to two steps :
i. `NH_4OH to NH_4^(+) + OH^(-) , DeltaH_1`=?
ii.`H^(+)+ OH^(-) to H_2O, DeltaH_2`=-13.7 kcal
So, `DeltaH=DeltaH_1+DeltaH_2`
or, `DeltaH_1=DeltaH-DeltaH_2`
=-12.27-(-13.7)=1.43 kcal