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Enthalpy

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Question

Calculate the hydration enthalpy of ions for $N a C l (s)$ from the following data.
$Δ_{l a t t i c e} H = + 788 k J m o l^{- 1}$
$Δ_{s o l} H = + 4 k J m o l^{- 1}$

A

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

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B

- 1084 k J m o l^{- 1}

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C

+ 784 k J m o l^{- 1}

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D

+ 1084 k J m o l^{- 1}

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Solution

The correct option is A $- 784 k J m o l^{- 1}$
We know that,
$Δ_{s o l n} H = Δ_{h y d} H + Δ_{l a t t i c e} H$
but given that,
$Δ_{l a t t i c e} H = + 788 k J m o l^{- 1}$
$Δ_{s o l n} H = + 4 k J m o l^{- 1}$

putting the above values in the formula we get,
$+ 4 k J m o l^{- 1} = Δ_{h y d} H + 788 k J m o l^{- 1}$
$Δ_{h y d} H = - 784 k J m o l^{- 1}$

Theory:
Lattice enthalpy $Δ_{L a t t i c e} H$ :
Energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions.

Standard lattice enthalpy $Δ_{L a t t i c e} H^{o}$ :
Energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions at unit pressure $(1 b a r)$ and temperature of $298 K$ .

Standard enthalpy of hydration $Δ_{h y d} H^{o}$ Conditions specified = $1 b a r$ and $298 K$
Enthalpy change when one mole of anhydrous or partially hydrated salt combines with the requisite amount of water to form a new hydrated stable salt at unit pressure $(1 b a r)$ and $298 K$ .

Integral enthalpy of solution/enthalpy of dissolution:
Enthalpy change when one mole of the solute is dissolved in a definite amount of the solvent to make a solution.
$Δ_{s o l n} H = Δ_{h y d} H + Δ_{l a t t i c e} H$

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