If $H^{+} + {O H}^{-} = H_{2} O + 13.7$ Kcal,then heat of complete neutralization of one gram mole of $H_{2} {S O}_{4}$ with strong base will be:

13.7

Kcal

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27.4

Kcal

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6.85

Kcal

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3.425

Kcal

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Solution

The correct option is A $13.7$ Kcal
Given, $H^{+} + O H^{-} \to H_{2} O$ $△ H = 13.7 K c a l$
One mole of $H_{2} S O_{4}$ give $2$ mole of $H^{+}$ ions.
One mole of $(K O H)$ strong base give one mole of $O H^{-}$ ions.
Since, one mole of $O H^{-}$ of strong base react with one mole of $H^{+}$ of strong acid.
$H^{+} + O H^{-} \to H_{2} O$
$△ H = 13.7 K C a l$

Suggest Corrections

Similar questions

If heat of dissociation of $C H C l_{2} C O O H$ is $0.7 k c a l / m o l e$ then $Δ H$ for the reaction:

$C H C l_{2} C O O H + K O H \to C H C l_{2} C O O K + H_{2} O$

C H C l_{2} C O O H

Δ H

C H C l_{2} C O O H + K O H \to C H C l_{2} C O O K + H_{2} O

H^{+} + {O H}^{-} ⇋ H_{2} O + 13.7 k c a l

H_{2} {S 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}

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