Consider the equations:<br><br><br><br><br> for the neutralization of HCl and NaOH is –13.75 kcal/eq — Thermodynamics and Thermochemistry Chemistry Question
Question
Consider the equations:<br>$\text{CH}_3\text{CH(OH)CHClCOOH} + 2\text{KOH} \to \text{CH}_3\text{CHCHCOOK} + \text{KCl} + 2\text{H}_2\text{O};$<br>$\Delta H = -14.7\text{ kcal}$<br>$\text{CH}_3\text{CH(OH)CHClCOOK} + \text{KOH} \to \text{CH}_3\text{CHCHCOOK} + \text{KCl} + \text{H}_2\text{O};$<br>$\Delta H = -0.522\text{ kcal}$<br>$\Delta H$ for the neutralization of HCl and NaOH is –13.75 kcal/eq. The enthalpy of ionization of $\text{CH}_3\text{CH(OH)CHClCOOH}$ is
💡 Solution & Explanation
Subtracting Reaction 2 from Reaction 1 isolates the pure neutralization of the carboxylic acid group: $\Delta H_{neut} = -14.7 - (-0.522) = -14.178\text{ kcal/mol}$. Comparing this to the strong acid standard gives the ionization enthalpy: $\Delta H_{ion} = \Delta H_{neut} - \Delta H_{strong} = -14.178 - (-13.75) = -0.428\text{ kcal/mol}$, closest to -0.429.