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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)
  1. a - r, t
    b - p
    c - q, s
    d - s
  2. a - r, t,s
    b - p
    c - q, 
    d - s
  3. a - r, t,s
    b - p,s
    c - q, s
    d - s
  4. a - 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 - s
Point 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 ΔH13.7 kcal

For
NH4OH1+HCl1
weak base and strong acid so ΔH13.7 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 ΔH13.7 kcal

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