The absolute enthalpy of neutralisation of the reaction will be:
$M g O (s) + 2 H C l (a q) \to M g C l_{2} (a q) + H_{2} O (l)$

+ 57.33 k J m o l^{- 1}

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

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greater than

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

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less than

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

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Solution

The correct option is C greater than $- 57.33 k J m o l^{- 1}$
The absolute value of enthalpy of neutralization of this reaction is greater than $- 57.33 k J m o l^{- 1}$ .
This is due to the very high enthalpy of hydration of $M g^{2 +}$ ion. The change in enthalpy of this reaction is nearly -146 kJ.

Thus enthalpy of neutralization per mole of $H^{+}$ ion is nearly -73 kJ, which is greater than the enthalpy of neutralization strong acid against strong base i.e. -57.33 kJ.

Hence, the correct answer is an option (C).

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Similar questions

M g O (s) + 2 H C l (a q .) \to M g {C l}_{2} (a q .) + H_{2} O (l)

M g O (s) + 2 H C l (a q) \to M g C l_{2} (a q) + H_{2} O (l)

27^{\circ} C

15 k J m o l^{- 1}

M g (s) + 2 H C l (a q) \to M g C l 2 (a q) + H 2 (g)

△ H = - 143 k J

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

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