AP EAMCET · Chemistry · Thermodynamics (C)
For strong acid and strong base neutralisation net chemical change is
\(\begin{aligned} \mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow & \mathrm{H}_2 \mathrm{O}(l) ; \\ \Delta_r H^{\circ} & =-55.84 \mathrm{kJmol}^{-1} .\end{aligned}\)
If enthalpy of neutralisation of \(\mathrm{CH}_3 \mathrm{COOH}\) by \(\mathrm{NaOH}\) is \(49.86 \mathrm{~kJ} \mathrm{~mol}^{-1}\), then enthalpy of ionisation of \(\mathrm{CH}_3 \mathrm{COOH}\) is
- A \(5.98 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- B \(-5.98 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C \(105.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D \(-59.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(5.98 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\because \mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O} ; \Delta_r H^{\circ}=-55.84 \mathrm{~kJ} / \mathrm{mol}\)…
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