-->

Thermodynamics

Question
CBSEENCH11006195

Explain why enthalpy of neutralization of a strong acid and strong base remains the same and the value changes if one of them is weak.

Solution

Enthalpy of neutralisation for a strong acid and a strong base is always constant: This is because in dilute solution all strong acids and strong bases are completely ionised. The neutralisation of a strong acid and strong base simply involves the combination of H+ions (from acid) and OH ions (from base) to form unionised water molecules with the evolution of 57.1 kJ heat.
stack straight H to the power of plus left parenthesis aq right parenthesis with From space acid below space plus space stack OH to the power of minus left parenthesis aq right parenthesis with From space base below space rightwards arrow stack straight H subscript 2 straight O left parenthesis straight l right parenthesis semicolon space with increment subscript straight n straight H to the power of 0 space equals space minus 57.1 space kJ below
Since the same reaction takes place during neutralisation of all strong acids and strong bases, the value of enthalpy of neutralisation is constant.
The neutralisation of HCl and NaOH can be represented as:
straight H to the power of plus left parenthesis aq right parenthesis plus Cl to the power of minus left parenthesis aq right parenthesis plus Na to the power of plus left parenthesis aq right parenthesis plus OH to the power of minus left parenthesis aq right parenthesis
space space space space space space space space space space space space space space space space space space space space space space rightwards arrow space Na to the power of plus plus Cl to the power of minus left parenthesis aq right parenthesis plus straight H subscript 2 straight O left parenthesis straight l right parenthesis
                                                                        increment subscript straight n straight H to the power of 0 space equals space minus 57.1 space kJ
Cancelling the common ions,
straight H to the power of plus left parenthesis aq right parenthesis space plus space OH to the power of minus left parenthesis aq right parenthesis rightwards arrow space straight H subscript 2 straight O left parenthesis straight l right parenthesis semicolon space

increment subscript straight n straight H to the power of 0 space equals space minus 57.1 space kJ

Enthalpy of neutralisation if either acid or base is weak : If one of the acid or bases is weak, then its ionisation is not complete in solution. Therefore, a part of the energy liberated is utilised for the ionisation of a weak acid (or base). Consequently, the value of enthalpy of neutralisation of the weak acid-strong base or strong acid-weak base is numerically less than 57.1 kJ. For example, neutralisation of acetic acid and sodium hydroxide can be represented as:
CH subscript 3 COOH space rightwards harpoon over leftwards harpoon space CH subscript 3 COO to the power of minus space plus space straight H to the power of plus semicolon
space increment subscript straight i straight H to the power of 0 space equals space 1.9 space kJ

straight H to the power of plus plus space Na to the power of plus space plus space OH to the power of minus space space space rightwards arrow space space space space Na to the power of plus space plus space straight H subscript 2 straight O semicolon
space increment subscript straight n straight H to the power of 0 space equals space minus 57.1 space kJ
CH subscript 3 COOH space plus space Na to the power of plus space plus space OH to the power of minus space space space rightwards arrow space CH subscript 3 COO to the power of minus plus space Na to the power of plus space plus space straight H subscript 2 straight O
semicolon space increment straight H space equals space minus 55.2 space kJ
Thus, enthalpy of neutralisation of acetic acid and sodium hydroxide is –55.2 kJ.
Similarly, in the neutralisation of NH4OH and HCl, 5.6 kJ of heat is used up for the dissociation of weak base i.e. NH4OH. Hence enthalpy of neutralisation, in this case, is –57.1 + 5.6 = –51.5 kJ.