
The enthalpy of neutralisation of oxalic acid by a strong base is \[-25.4kcalmo{{l}^{-1}}\] . The enthalpy of neutralisation of strong acid and strong base is\[-13.7kcale{{q}^{-1}}\]. The enthalpy of dissociation of oxalic acid is:
Given: Heat of neutralisation will be \[2\times -13.7kcalmo{{l}^{-1}}\]
(a) 1 \[kcalmo{{l}^{-1}}\]
(b) 2 \[kcalmo{{l}^{-1}}\]
(c) 18.55 \[kcalmo{{l}^{-1}}\]
(d) 11.7 \[kcalmo{{l}^{-1}}\]
Answer
585.3k+ views
Hint: When one equivalent of an acid and a base undergo neutralisation reaction in order to form salt and water, the change in the enthalpy of the reaction is called the Enthalpy of neutralisation.
Complete step by step solution:
Strong acids and strong bases undergo complete neutralisation. Their salts remain fully dissociated in the solution. Therefore, the neutralisation reaction can be represented as
\[{{H}^{+}}+{{A}^{-}}+O{{H}^{-}}\to {{H}_{2}}O+{{B}^{+}}+{{A}^{-}}\]
From such a neutralisation reaction we get 1 mole of water. So as given the heat change accompanying the reaction is \[-13.7kcale{{q}^{-1}}\].
Oxalic acid is a weak acid and it will not dissociate completely in water. The solution will contain a mixture of dissociated and undissociated molecules of oxalic acid. So, there is some amount of heat used up to ionise the undissociated molecules. As only a little amount of energy is consumed, less amount of the energy is released.
It is given that enthalpy neutralisation of oxalic acid for 1 mole is\[-25.4kcalmo{{l}^{-1}}\]
Oxalic acid is dibasic so heat of neutralisation = \[(2\times -13.7kcal)\]
Therefore, using the equation
Enthalpy of dissociation= Enthalpy of neutralisation of weak acid – Enthalpy of neutralisation of strong acid
Enthalpy of dissociation\[=(2\times -13.7kcal)-(-25.4kcalmo{{l}^{-1}})=-27.4+25.4\]
\[=2kcalmo{{l}^{-1}}\]
So, the correct answer is “Option B”.
Additional Information:
Standard enthalpy of neutralisation is the enthalpy of neutralisation reaction when carried out in standard condition (Temperature 298K, pressure 1atm). Heat of neutralisation for weak acids and strong bases are pH dependent. If no mineral acid or base is present, then some heat is required for complete dissociation.
Note: For such types of questions we need to always check whether the acid is strong or weak acid. Another point to keep in mind is the basicity of the acid.
Complete step by step solution:
Strong acids and strong bases undergo complete neutralisation. Their salts remain fully dissociated in the solution. Therefore, the neutralisation reaction can be represented as
\[{{H}^{+}}+{{A}^{-}}+O{{H}^{-}}\to {{H}_{2}}O+{{B}^{+}}+{{A}^{-}}\]
From such a neutralisation reaction we get 1 mole of water. So as given the heat change accompanying the reaction is \[-13.7kcale{{q}^{-1}}\].
Oxalic acid is a weak acid and it will not dissociate completely in water. The solution will contain a mixture of dissociated and undissociated molecules of oxalic acid. So, there is some amount of heat used up to ionise the undissociated molecules. As only a little amount of energy is consumed, less amount of the energy is released.
It is given that enthalpy neutralisation of oxalic acid for 1 mole is\[-25.4kcalmo{{l}^{-1}}\]
Oxalic acid is dibasic so heat of neutralisation = \[(2\times -13.7kcal)\]
Therefore, using the equation
Enthalpy of dissociation= Enthalpy of neutralisation of weak acid – Enthalpy of neutralisation of strong acid
Enthalpy of dissociation\[=(2\times -13.7kcal)-(-25.4kcalmo{{l}^{-1}})=-27.4+25.4\]
\[=2kcalmo{{l}^{-1}}\]
So, the correct answer is “Option B”.
Additional Information:
Standard enthalpy of neutralisation is the enthalpy of neutralisation reaction when carried out in standard condition (Temperature 298K, pressure 1atm). Heat of neutralisation for weak acids and strong bases are pH dependent. If no mineral acid or base is present, then some heat is required for complete dissociation.
Note: For such types of questions we need to always check whether the acid is strong or weak acid. Another point to keep in mind is the basicity of the acid.
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