Question

When 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous solution 108 kJ of heat is liberated then the enthalpy of ionisation of the oxalic acid is (Given H+(aq) + OH-(aq)--->H2O (l)
Δneut H = -57.3kJ
plz explain with complete derivation

Answer 1

Answer : The enthalpy of ionisation of oxalic acid is 6.6 kJ

Explanation :

The reaction of oxalic acid with NaOH can be written as,

$C_{2}H_{2}O_{4}(aq)+2NaOH(aq) \rightarrow Na_{2}C_{2}O_{4}(aq)+ 2H_{2}O(l)$

The net ionic equation for the above reaction is,

$C_{2}H_{2}O_{4}(aq)+2OH^{-}(aq)\rightarrow C_{2}O_{4}^{2-}(aq)+ 2H_{2}O(l)$.......Equation (1)

The enthalpy for above reaction is -108kJ.

The neutralisation reaction of H+ is written as,

$H^{+}(aq)+OH^{-}(aq)\rightarrow H_{2}O(l)$......... equation (2)

The enthalpy of neutralisation for above reaction is -57.3 kJ

The given question can be solved by using Hess's law.

Let us reverse equation 2 and multiply it by 2.

The new equation that we get is,

$2H_{2}O(l)\rightarrow 2H^{+}(aq)+ 2OH^{-}(aq)$......... equation (3)

Enthalpy change for above equation is 2*57.3 kJ= 114.6kJ [ Enthalpy becomes positive because the reaction is reversed]

Let us add equation 1 and 3.

$C_{2}H_{2}O_{4}(aq)+2OH^{-}(aq)\rightarrow C_{2}O_{4}^{2-}(aq)+ 2H_{2}O(l)$ ........ΔH = -108 kJ

$2H_{2}O(l)\rightarrow 2H^{+}(aq)+ 2OH^{-}(aq)$ ........ΔH = 114.6 kJ

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$C_{2}H_{2}O_{4}(aq)\rightarrow 2H^{+}(aq)+C_{2}O_{4}^{2-}(aq)$..........ΔH = 6.6 kJ

The above reaction represents ionisation of C₂H₂O₄ .

Therefore the enthalpy of ionisation of oxalic acid is 6.6 kJ

Answer 2

The ionization energy is qualitatively explained as the minimum amount of energy required to remove the most freely bound electron, the valence electron, of an isolated neutral gaseous atom to form a cation.

Ionization Enthalpy is explained as the amount of energy required by an isolated gaseous atom to lose an electron in its initial state.