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Question

The heat of neutralization is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt. Heat of neutralization of strong acid with strong base is 13.7 kcal mol1.

List - IList - II(|Enthalpy change in kcal|)(Neutralisation)(I) < 13.7 kcal(P) HCl1 mol+NaOH1 mol(II) = 13.7 kcal(Q) HF1 mol+NaOH1 mol(III) > 13.7 kcal(R) HCl1 mol+NH4OH1 mol(IV) = 27.4 kcal(S) NaOH2 mol+H2SO41 mol(T) NaOH1 mol+CH3COOH1 mol

Which of the following options has the correct combination considering List-I and List-II?

A
IIIS,T, IVQ
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B
IIIQ,S, IVS
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C
IIIP,S, IVR
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D
IIIR,S, IVS
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Solution

The correct option is B IIIQ,S, IVS
(P) NaOHNaCl+H2O
Heat of neutralization for strong acid with strong base is 13.7 kcal mol1.
(Q) |ΔHnu|>13.7 kcal, for HF, heat is evolved due to hydration of F.
(R) NH4OH is a weak base, so |ΔHnu|<13.7 kcal because some amount of heat will be used up in ionisation of NH4OH.
(S) For 2 moles of NaOH,
|ΔHnu|=2×13.7=27.4 kcal
(T) CH3COOH is a weak acid, so |ΔHnu|<13.7 kcal because some amount of heat will be used up in ionisation of CH3COOH.

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