JEE MAIN - Chemistry (2022 - 25th July Evening Shift - No. 16)
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
2
Explanation
(I) $$\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$$
$$ \Delta \mathrm{H}_{1}=-57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $$
(II) $$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$$
$$ \Delta \mathrm{H}_{2}=-55.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $$
Reaction (I) can be written as
$$ \text { (III) } \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\mathrm{NaOH} $$
$$ \Delta \mathrm{H}_{3}=57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $$
By adding (II) and (III)
$$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaCl} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{HCl} \quad \Delta \mathrm{H}_{\mathrm{r}} \\ &\begin{aligned} \Delta \mathrm{H}_{\mathrm{r}}=\Delta \mathrm{H}_{3}+\Delta \mathrm{H}_{2} &=57.3-55.3 \\ &=2 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \end{aligned} $$
$$ \Delta \mathrm{H}_{1}=-57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $$
(II) $$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$$
$$ \Delta \mathrm{H}_{2}=-55.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $$
Reaction (I) can be written as
$$ \text { (III) } \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\mathrm{NaOH} $$
$$ \Delta \mathrm{H}_{3}=57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $$
By adding (II) and (III)
$$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaCl} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{HCl} \quad \Delta \mathrm{H}_{\mathrm{r}} \\ &\begin{aligned} \Delta \mathrm{H}_{\mathrm{r}}=\Delta \mathrm{H}_{3}+\Delta \mathrm{H}_{2} &=57.3-55.3 \\ &=2 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \end{aligned} $$
Comments (0)
