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Heat of Reaction

The enthalpy ...

Question

The enthalpy of neutralisation of $N H_{4} O H$ and $C H_{3} C O O H$ is -10.5 kcal $m o l^{- 1}$ and enthalpy of neutralisation of $C H_{3} C O O H$ with strong base is -12.5 kcal $m o l^{- 1}$ . The enthalpy of ionisation of $N H_{4} O H$ will be:

4.0 kcal

m o l^{- 1}

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3.0 kcal

m o l^{- 1}

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2.0 kcal

m o l^{- 1}

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3.2 kcal

m o l^{- 1}

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Solution

The correct option is C 2.0 kcal $m o l^{- 1}$
Answer:-
$N H_{4} O H + C H_{3} C O O H ⟶ C H_{3} C O O N H_{4} + H_{2} O \Rightarrow {Δ H}_{1} = - 10.5 k c a l {m o l}^{- 1}$
$C H_{3} C O O H + N a O H Strong base ⟶ C H_{3} C O O N a + H_{2} O \Rightarrow {Δ H}_{2} = - 12.5 k c a l {m o l}^{- 1}$
$\Rightarrow Enthalpy of ionisation of N H_{4} O H = {Δ H}_{1} - {Δ H}_{2} = (- 10.5 - (- 12.5)) k c a l {m o l}^{- 1} = 2.0 k c a l {m o l}^{- 1}$

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Similar questions

\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}

{C H}_{3} C O O H

{Δ H}

{C H}_{3} C O O H

- 186

+

^{-}

- 97

- 85

C s O H

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

C s O H

H F

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

Δ H^{⊖}

H F

X e F_{4} ⟶ X e^{+} + F^{-} + F_{2} + F; Δ H = ?

X e - - F

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

{I E}_{1}

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

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

e_{F - - F} = 38 k c a l m o l^{- 1}

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