Question

While performing a thermodynamics experiment, a student made the following observations.
HCl + NaOH \(\rightarrow\) NaCl + H₂O; ΔH = – 57.3 kJ mol^–1
CH₃COOH + NaOH \(\rightarrow\) CH₃COONa + H₂O; ΔH = –55.3 kJ mol^–1
The enthalpy of ionization of CH₃COOH, as calculated by the student, is ______ kJ mol^–1. [nearest integer]

Answer 1

Correct Answer: 2

To find the enthalpy of ionization of CH₃COOH, we use the principle of Hess's Law. We know:
1. HCl + NaOH → NaCl + H₂O; ΔH = –57.3 kJ/mol
2. CH₃COOH + NaOH → CH₃COONa + H₂O; ΔH = –55.3 kJ/mol
The ionization of CH₃COOH can be expressed as:
CH₃COOH → CH₃COO⁻ + H⁺
Using Hess's Law: Enthalpy change in a reaction is equal to the sum of enthalpy changes of individual steps.
Thus, Ionization Enthalpy (ΔH_ionization) = ΔH(CH₃COOH + NaOH) - ΔH(HCl + NaOH)
= (–55.3 kJ/mol) - (–57.3 kJ/mol)
= 2.0 kJ/mol
Therefore, the enthalpy of ionization of CH₃COOH is 2 kJ mol^–1, which fits within the expected range of 2,2.