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Enthalpy of Neutralisation

The enthalpy ...

Question

The enthalpy of neutralisation of strong acid by a strong base is $- 57.32 k J m o l^{- 1}$ . The enthalpy of formation of water is $- 285.84 k J m o l^{- 1}$ . The enthalpy of formation of hydroxyl ion is :

+ 228.52 k J m o l^{- 1}

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- 114.26 k J m o l^{- 1}

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- 228.52 k J m o l^{- 1}

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+ 114.2 k J m o l^{- 1}

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Solution

The correct option is C $- 228.52 k J m o l^{- 1}$
Neutralisation of strong acid and strong base-
$H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l) Δ H = - 57.32 k J m o l^{- 1}$
reversing the reaction we get-
$H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q) Δ H = 57.32 k J m o l^{- 1} . . . . .1$
Formation of water-
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - 285.84 k J m o l^{- 1} . . . . .2$

add equation 1 & 2 to get the enthalpy of formation of $O H^{-}$ .
therefore,

$H_{2} O (l) + H_{2} (g) + \frac{1}{2} O_{2} (g) \to H^{+} (a q) + O H^{-} (a q) + H_{2} O (l) Δ H = 57.32 - 285.84 k J m o l^{- 1}$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H^{+} (a q) + O H^{-} (a q) Δ H = - 228.52 k J m o l^{- 1}$

hence, the enthalpy of formation of $O H^{-}$ is $- 228.52 k J m o l^{- 1}$

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