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

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

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Solution

The correct option is A greater than $- 57.33 k J {m o l}^{- 1}$
The enthalpy of hydration of ${M g}^{2 +}$ ion is very high. Due to this, the absolute enthalpy of neutralization of the reaction is greater than $- 57.33 K J {m o l}^{- 1}$ . The $△ H$ for the above reaction is found to be $- 146 K J$ which states that the enthalpy of neutralization of $H^{+}$ is $- 73 K J / m o l$ which is greater than the enthalpy of neutralization i.e. $57.33 K J / m o l$ .

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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)

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

△ H = - 143 k J

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

△ H = - 216 k J

H_{2} (g) + 1 / 2 O_{2} \to H_{2} O (I)

△ H = - 285 k J

M g (s) + 1 / 2 O_{2} \to M g O (s)

27^{\circ} C

15 k J m o l^{- 1}

H_{2} O (l)

- 285.7

m o l^{- 1}

- 57.6

m o l^{- 1}

O H^{- 1}

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