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

(I) HCl + NaOH $\rightarrow$ NaCl + H2O
ΔH1 = – 57.3 KJ mol–1
(II) CH3COOH + NaOH $\rightarrow$ CH3COONa + H2O
ΔH2 = – 55.3 KJ mol–1
Reaction (I) can be written as
(III) NaCl + H2O $\rightarrow$ HCl + NaOH
ΔH3 = 57.3 KJ mol–1
By adding (II) and (III)
CH3COOH + NaCl $\rightarrow$ CH3COONa + HCl
ΔHr = ΔH3 + ΔH2 = 57.3 – 55.3
= 2 kJ mol–1