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Formation of Enamines

The enthalpy ...

Question

The enthalpy of neutralisation of oxalic acid by NaOH is –x kcal mol–1. What will be the dissociation energy of oxalic acid if enthalpy of neutralization of strong acid and strong base is –y kcal eq–1? Options: a-(2x + y) kcal mol–1 b- (2y – x) kcal mol–1 c-(–2y – x) kcal mol–1 d (2x – y) kcal mol–1

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

(Δ H)

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

H_{2} C_{2} O_{4}

\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}

- 25.4 K c a l m o l^{- 1}

N a O H

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

N H_{4} O H

C H_{3} C O O H

m o l^{- 1}

C H_{3} C O O H

m o l^{- 1}

N H_{4} O H

- 25.4 k

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

- 13.7 k c a l

{e q}^{- 1}

(2 \times - 13.7 k c a l)

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