Question

$CHC{l}_{2}COOH+KOH\to CHC{l}_{2}COOK+H_{2}O$

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

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

Answer 1

The correct option is A $-13kcal$

The average enthalpy of neutralization of any strong acid by a strong base is found to be $–57.7$ This is because strong acids and strong bases are completely ionized in aqueous solutions. The aqueous solution of one gram equivalent of all strong acids contains the same number of $H^{+}$ ions. Similarly, the aqueous solution of one gram equivalent of all strong bases also contains the same number of $O{H}^{–}$. The neutralization reactions between strong acids and strong bases in aqueous solutions involve simply the combination of $H^{+}$ ions (from an acid) and $O{H}^{–}$ ions (from a base) to form unionized water molecules.

When strong acid and a weak base or weak acid and a strong base or weak acid and weak base are mixed in equivalent amounts, the heat evolved or change in enthalpy is less than $-57.7$

so here dissociation energy is $0.7kcal/mol$

so the heat of reaction for weak acid

$=-13.7+0.7$

$=-13kcal$

Hence, the correct option is $\text{A}$