Question

Enthalpy of neutralization of the reaction between $C{H}_{3}COOH\left(aq\right)$ and $NaOH\left(aq\right)$ is -13.2 $kcale{q}^{-1}$ and that of the reaction between $H_{2}S{O}_4\left(aq\right)$ and $KOH\left(aq\right)$ is -13.7 $kcale{q}^{-1}$. The enthalpy of dissociation of $C{H}_{3}COOH\left(aq\right)$ is:

• A. -0.5 $kcale{q}^{-1}$
• B. +0.5 $kcale{q}^{-1}$
• C. -26.9 $kcale{q}^{-1}$
• D. +0.5 $kcale{q}^{-1}$

Answer 1

The correct option is B +0.5 $kcale{q}^{-1}$

Dissociation enthalpy of $C{H}_{3}COOH$ = 13.7 - 13.2 = 0.5 $kcale{q}^{-1}$

Therefore, 0.5 $kcale{q}^{-1}$ heat will be used to dissociate $C{H}_{3}COOH$ completely