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Salt of Strong Acid and Weak Base

The heat of n...

Question

The heat of neutralisation of oxalic acid is $- 25.4 K c a l m o l^{- 1}$ using strong base, $N a O H$ . Hence, the enthalpy change of the process is
$H_{2} C_{2} O_{4} \to\leftarrow 2 H^{\oplus} + C_{2} O_{4}^{2 -}$ iS

2.0 Kcal

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-11.8 Kcal

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1.0 Kcal

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-1.0 Kcal

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Solution

The correct option is A 2.0 Kcal
oxalic acid has two ionisable $H^{\oplus}$ . Hence, expected heat of neutralisation, if it behaves as a strong acid would have been = $- 13.7 \times 2 = - 27.4 K c a l m o l^{- 1}$
But experimental value = $- 25.4 K c a l m o l^{- 1}$
$∴$ Heat of ionisation = $2.0 K c a l m o l^{- 1}$

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(H_{2} C_{2} O_{4})

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H_{2} C_{2} O_{4} \to 2 H^{+} + C_{2} O_{4}^{2 -}

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- 13.7 k c a l m o l^{- 1}

\begin{matrix} List - I & List - II (|Enthalpy change in kcal|) & (Neutralisation) (I) < 13.7 kcal & (P) H C l 1 mol + N a O H 1 mol (II) = 13.7 kcal & (Q) H F 1 mol + N a O H 1 mol (III) > 13.7 kcal & (R) H C l 1 mol + N H_{4} O H 1 mol (IV) = 27.4 kcal & (S) N a O H 2 mol + H_{2} S O_{4} 1 mol (T) N a O H 1 mol + C H_{3} C O O H 1 mol \end{matrix}

\begin{matrix} Column - I & Column - II (Enthalpy change in kcal) & (Neutralisation) (A) & < 13.7 kcal & (p) & H C l (1 m o l) + N a O H (1 m o l) (B) & = 13.7 kcal & (q) & H F (1 m o l) + N a O H (1 m o l) (C) & > 13.7 kcal & (r) & N H_{4} O H (1 m o l) + H C l (1 m o l) (D) & = 27.4 kcal & (s) & N a O H (2 m o l) + H_{2} S O_{4} (1 m o l) (t) & N a O H (1 m o l) + C H_{3} C O O H (1 m o l) \end{matrix}

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