For strong acid and strong base neutralization net chemical change is {H}^{+}+{OH}^{-} \\to {H}_{2} {O} (l) \\DeltaH_{r}^{0}=-55.84 {kJ} { mol}^{-1}. If enthalpy of neutralization of {CH}_{3} {COOH} by {Na} {OH} is 49.86 { kJ} { mol}^{-1} then enthalpy of ionization of {CH}_{3} {COOH} is _______

Correct Answer is : 5.98 kJ mol−1

AP EAPCET 24-Aug-2021 Shift 2 Paper

Section: Chemistry

Question : 134 of 160

Marks: +1, -0

For strong acid and strong base neutralization net chemical change is
H++OH−→H2O (l) ΔHr0=−55.84kJ mol−1. If enthalpy of neutralization of CH3 COOH by Na OH is 49.86 kJ mol−1 then enthalpy of ionization of CH3 COOH is _______

5.98 kJ mol−1
−5.98 kJ mol−1
105.7 kJ mol−1
−59.8 kJ mol−1

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