Enthalpy of neutralisation of CH3COOH by NaOH is – 50.6 kJ/mol and the heat of neutralisation of a strong acid with NaOH is – 55.9 kJ/mol. The value of ΔH for the ionisation of CH3COOH is :

Option 1 - <p>3.5 kJ / mol</p>
Option 2 - <p>4.6 kJ / mol</p>
Option 3 - <p>5.3 kJ / mol</p>
Option 4 - <p>6.4 kJ / mol</p>
1 Views|Posted 4 months ago
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1 Answer
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4 months ago
Correct Option - 3
Detailed Solution:

CH3COOH + NaOH → CH3COONa + H2O

ΔH = –50.6 kJ/mol

NaOH + SA [HCl] → NaCl + H2O

ΔH = –55.9 kJ/mol

the value of ΔH for ionisation of CH3COOH

⇒ ΔH = +55.9 – 50.6

5.3 kJ/mol

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Chemistry Organic Chemistry 2025

Chemistry Organic Chemistry 2025

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