Question

While performing a thermodynamics experiment, a student made the following observations, 

$ HCl + NaOH \rightarrow NaCl + H _2 O$

$ \Delta H =-573 \,kJ \,mol ^{-1}$ 

$ CH _3 COOH + NaOH \rightarrow CH_3COONa + H _2 O$ 

$ \Delta H =-553 \,kJ\, mol ^{-1} $ 

The enthalpy of ionization of $CH _3 COOH$ as calculated by the student is _____ $kJ \,mol ^{-1}$ (nearest integer)

Answer 1

Correct Answer: 2

To determine the enthalpy of ionization of $CH_3COOH$, we need to consider the two given reactions: 

$HCl + NaOH \rightarrow NaCl + H_2O$	ΔH = -573 kJ/mol
$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$	ΔH = -553 kJ/mol

In the reaction of $HCl$ with $NaOH$, both are strong electrolytes and dissociate completely in water. Thus, the reaction represents the net ionic equation:

$H^+ + OH^- \rightarrow H_2O$

For this equation, the standard enthalpy change (ΔH) is -573 kJ/mol.

In the case of $CH_3COOH$, it is a weak acid and partially ionizes. When $CH_3COOH$ reacts with $NaOH$, the net ionic equation is effectively:

$CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$
ΔHionisation ​ of CH3​COOH=∣−57.3−(−55.3)∣ 
=2KJ/mol