While performing a thermodynamics experiment, a student made the following observations.
HCl + NaOH $\to$ NaCl + H2O $\Delta$H = $-$57.3 kJ mol$-$1
CH3COOH + NaOH $\to$ CH3COONa + H2O $\Delta$H = $-$55.3 kJ mol$-$1
The enthalpy of ionization of CH3COOH as calculated by the student is _____________ kJ mol$-$1. (nearest integer)
Answer (integer)
2
Solution
(I) $\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$
<br/><br/>
$\Delta \mathrm{H}_{1}=-57.3 \,\mathrm{KJ} \mathrm{mol}^{-1}$
<br/><br/>
(II) $$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$$
<br/><br/>
$\Delta \mathrm{H}_{2}=-55.3 \,\mathrm{KJ} \mathrm{mol}^{-1}$
<br/><br/>
Reaction (I) can be written as
<br/><br/>
$$
\text { (III) } \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\mathrm{NaOH}
$$
<br/><br/>
$\Delta \mathrm{H}_{3}=57.3 \,\mathrm{KJ} \mathrm{mol}^{-1}$
<br/><br/>
By adding (II) and (III)
<br/><br/>
$$
\begin{aligned}
&\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaCl} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{HCl} \quad \Delta \mathrm{H}_{\mathrm{r}} \\
&\begin{aligned}
\Delta \mathrm{H}_{\mathrm{r}}=\Delta \mathrm{H}_{3}+\Delta \mathrm{H}_{2} &=57.3-55.3 \\
&=2 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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