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  • While performing a thermodynamics experiment, a student made the following observations. <img alt="\begin{aligned} &\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2}

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

\begin{aligned} &\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \Delta \mathrm{H}=-57.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O} \Delta \mathrm{H}=-55.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

 The enthalpy of ionization of  \mathrm{CH}_{3} \mathrm{COOH}  as calculated by the student is ____________\mathrm{\mathrm{kJ}\; \mathrm{mol}^{-1}}. (nearest integer)

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The ionization of \mathrm{CH_{3}COOH} -

\mathrm{CH}_3 \mathrm{COOH} \longrightarrow \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{H}^{+}\; \; \mathrm{\Delta H}^{-}=\text {? }

Given,

\text { (1) } \mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}, \Delta \mathrm{H}=-57.3 \mathrm{KJ} / \mathrm{mol}

(2) \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}, \mathrm{\Delta H}=-55.3 \mathrm{KJ} / \mathrm{mol}

By doing (2) - (1), will become ionisation of \mathrm{CH_{3}COOH}

\mathrm{CH_{3}COOH+NaCl\rightarrow CH_{3}COO^{\circleddash}Na^{\oplus}+HCl}

So, \mathrm{\Delta H=-55.3-(-57.3)}

       \mathrm{\Delta H=2 \; kJ/mol}

Answer = 2

Posted by

Gaurav

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