Question

# Match the following (Assume heat of neutralisation of strong acid with strong base is 13.7 kcal): Column - IColumn - II(Enthalpy change in kcal)(Neutralisation)(A)<13.7kcal(p)HCl(1mol)+NaOH(1mol)(B)=13.7kcal(q)HF(1mol)+NaOH(1mol)(C)>13.7kcal(r)NH4OH(1mol)+HCl(1mol)(D)=27.4kcal(s)NaOH(2mol)+H2SO4(1mol)(t)NaOH(1mol)+CH3COOH(1mol)a - r, t b - p c - q, s d - sa - r, t,s b - p c - q,  d - sa - r, t,s b - p,s c - q, s d - sa - r, t b - p,s c - q, s d - p

Solution

## The correct option is A a - r, t b - p c - q, s d - sPoint to remember: ⇒For Strong Acid+Strong Base, heat of neutralisation is always equal to –13.7kcal/mole or –57.1kJ/mole. ⇒ For any other combination it must be less than –13.7kcal/mole or –57.1kJ/mole. ⇒ Rest of energy utilised for ionisation of weak acid or base  For HCl+NaOH Both are strong so ΔH=13.7 kcal For HF1 mole+NaOH1 mole HF weak acid thus ΔH≤13.7 kcal For NH4OH1+HCl1 weak base and strong acid so ΔH≤13.7 kcal For NaOH2+H2SO41 two mole of [H+] reacting with two mole of [OH−] so ΔH=27.4 kcal For CH3COOH1+NH4OH1 both are weak acid  so ΔH≤13.7 kcal

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