Question

If heat of dissociation of $CHC{l}_{2}COOH$ is 0.7 kcal/mole then $\Delta H$ for the reaction:
$CHC{l}_{2}COOH+KOH\to CHC{l}_{2}COOK+H_{2}O$

• A. -13 kcal/mol
• B. + 13 kcal/mol
• C. -14.4 kcal/mol
• D. -13.7 kcal/mol

Answer 1

The correct option is A -13 kcal/mol

Heat of neutralization of strong acid by a strong base,
$H^{+}+O{H}^{-}\to H_{2}O\left(l\right)$
$\Delta H=-13.7kcal.......\left(i\right)$
For neutralization of weak acid by a strong base some of the heat is consumed in dissociation of the acid.
$CHC{l}_{2}COOH\to CHC{l}_{2}CO{O}^{-}+H^{+}\Delta H$
= 0.7 kcal ..........(ii)
Hence, heat of neutralization of $CHC{l}_{2}COOH$ by a strong base is obtained by adding (i) and (iii)
$CHC{l}_{2}COOH+O{H}^{-}\to$
$CHC{l}_{2}CO{O}^{-}+H_{2}O\left(l\right)\Delta H=?$
Adding equation (i) and $\left(ii\right)\Delta H$ = -13 kcal.