The enthalpy of neutralization of `NH_(4)OH` and `CH_(3)COOH` is – 10.5 kcal/mole and enthalpy of neutralization of strong base and `CH_(3)COOH` is – 12.5 kcal/mole. Calculate the enthalpy of dissociation of `NH_(4) OH`-

Options

A. `4.0k"cal mol"^(-1)`
B. `3.0 k"cal mol"^(-1)`
C. `2.0 k "cal mol"^(-1)`
D. `3.2 k"cal mol"^(-1)`

Answer

C

Explanation

`DeltaH_(N)` for strong acid with strong base `=-13.7k` cal `"equiv"^(-1)`
`DeltaH_("ion")(CH_(3)COOH)=-12.5-(-13.7)=+1.2k"cal mol"^(-1)`
`DeltaH_("ion")(NH_(4)OH)=-10.5-(-13.7)-DeltaH_("ion")(CH_(3)COOH)`
`=13.7-10.5-1.2=2 k "cal mol"^(-1)`