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Question: Given that 1\. \(C{{H}_{3}}COOH+NaOH\to C{{H}_{3}}COONa+{{H}_{2}}O,\,\Delta H=-55.0kJmo{{l}^{-1}}\...

Given that
1. CH3COOH+NaOH→CH3COONa+H2O, ΔH=−55.0kJmol−1C{{H}_{3}}COOH+NaOH\to C{{H}_{3}}COONa+{{H}_{2}}O,\,\Delta H=-55.0kJmo{{l}^{-1}}
2. HCl+NaOH→NaCl+H2O, ΔH=−57.0kJmol−1HCl+NaOH\to NaCl+{{H}_{2}}O,\,\Delta H=-57.0kJmo{{l}^{-1}}
The heat of ionisation CH3COOHC{{H}_{3}}COOHis
(a) 2.0 kJ mol−12.0\,kJ\,mo{{l}^{-1}}
(b) 0.2 kJ mol−10.2\,kJ\,mo{{l}^{-1}}
(c) 20.0 kJ mol−120.0\,kJ\,mo{{l}^{-1}}
(d) 2.2 kJ mol−12.2\,kJ\,mo{{l}^{-1}}

Explanation

Solution

The heat of ionisation of a reaction is the amount of heat required to ionise the substance into corresponding ion and specifically that amount of heat required to ionise one mole of the substance.
We can calculate the heat of ionisation acetic acid. We have to find the difference between the heat of reaction given for the two reactions above.

Complete answer:
In the question two reactions are given, in the first equation given acetic acid, a weak acid is reacting with NaOH, a strong base to give the corresponding salt and water and the enthalpy is also given.
In the second equation the neutralisation reaction between a strong acid HCl and strong base NaOH, is given and the heat of neutralisation is also given.
The chemical reaction between the strong acid HCl and strong base NaOH can be represented as,H+(aq)+OH−(aq)→H2O(l){{H}^{+}}_{\left( aq \right)}+O{{H}^{-}}_{\left( aq \right)}\to {{H}_{2}}{{O}_{\left( l \right)}}
The chemical reaction involving acetic acid can be written as,
CH3COOH+OH−→CH3COO−+H2OC{{H}_{3}}COOH+O{{H}^{-}}\to C{{H}_{3}}CO{{O}^{-}}+{{H}_{2}}O
We want the equation for ionization of acetic acid and it can be written as,
CH3COOH→CH3COO−+H+C{{H}_{3}}COOH\to C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}
The heat of reaction associated with equation (1.1) is ΔH1=−57kJ/mol\Delta {{H}_{1}}=-57kJ/mol and with equation (1.2) is ΔH1=−55kJ/mol\Delta {{H}_{1}}=-55kJ/mol
And we get equation (1.3) by subtracting the equation (1.2) and equation (1.1).
So we can calculate the heat of ionization of CH3COOHC{{H}_{3}}COOH
By Hess's law, heat of ionization of acetic acid is,
ΔH=ΔH2−ΔH1\Delta H=\Delta {{H}_{2}}-\Delta {{H}_{1}}
ΔH=−55−(−57)=2kJ/mol\Delta H=-55-(-57)=2kJ/mol

So the correct answer for the given question is option (a).

Note:
Since the neutralization reaction between an acid and an alkali involves the evolution of heat it is an exothermic reaction and as the evolution of heat takes place, the enthalpy of neutralization will have negative sign only.
Hess's law is the law that states the heat change of a reaction involved in a reaction with a single step or multiple steps will be equal to the total sum of all the changes.