Enthalpy of neutralization of acetic acid and sodium hydroxide is -55.4 kJ/mol. The enthalpy of ionization of acetic acid is

- 5.54 kJ/mol

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+ 5.54 kJ/mol

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- 1.7 kJ/mol

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+ 1.7 kJ/mol

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Solution

The correct option is D + 1.7 kJ/mol
Maximum $Δ H_{i}$ value possible (for strong acid and strong base)=57.1 kJ/mol
$∴ Δ H_{i} o f a b o v e c o m b i n a t i o n i s = 57.1 - 55.4 = 1.7 k J / m o l$

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Enthalpy of neutralization of acetic acid and sodium hydroxide is -55.4 kJ/mol. The enthalpy of ionization of acetic acid is

The correct option is D + 1.7 kJ/molMaximum ΔHi value possible (for strong acid and strong base)=57.1 kJ/mol ∴ΔHiofabovecombinationis=57.1−55.4=1.7kJ/mol

The correct option is D + 1.7 kJ/mol
Maximum $Δ H_{i}$ value possible (for strong acid and strong base)=57.1 kJ/mol
$∴ Δ H_{i} o f a b o v e c o m b i n a t i o n i s = 57.1 - 55.4 = 1.7 k J / m o l$