While performing a thermodynamics experiment, a student made the following observations. HCl + NaOH → NaCl + H₂O Δ H = - 57.3 kJ mol⁻¹ CH₃COOH + NaOH → CH₃COONa + H₂O Δ H = - 55.3 kJ mol⁻¹ The enthalpy of ionization of CH₃COOH as calculated by the student is _ _ _ _ _ _ k J mol⁻¹. (nearest integer)

Chemical Thermodynamics: JEE Main 2022 Chemistry question

JEE Main 2022 · 25 Jul, Evening shift · Q23

JEE Main 202225 Jul, Evening shiftChemistryMediumChemical Thermodynamics84× since 2002

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)

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2022 Chemical Thermodynamics Question with Answer (25 Jul Evening shift, Q23) | JEE Arc