Lattice enthalpy for NaCl is +788 kJ mol^{- | vedclass

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(B) The enthalpy of solution $(\Delta H_{	ext{sol}})$ is given by the sum of the lattice enthalpy and the enthalpy of hydration: $\Delta H_{	ext{sol

Vedclass · Accepted Answer

$+4 \ kJ \ mol^{-1}$

Vedclass · Answer

(B) The enthalpy of solution $(\Delta H_{	ext{sol}})$ is given by the sum of the lattice enthalpy and the enthalpy of hydration: $\Delta H_{	ext{sol}} = \Delta H_{	ext{lattice}} + \Delta H_{	ext{hyd}}$ Given: $\Delta H_{	ext{lattice}} = +788 \ kJ \ mol^{-1}$ and $\Delta H_{	ext{hyd}} = -784 \ kJ \ mol^{-1}$ $\Delta H_{	ext{sol}} = 788 \ kJ \ mol^{-1} + (-784 \ kJ \ mol^{-1}) = +4 \ kJ \ mol^{-1}$

Lattice enthalpy for $NaCl$ is $+788 \ kJ \ mol^{-1}$ and $\Delta H_{\text{hyd}}^{\circ} = -784 \ kJ \ mol^{-1}$. The enthalpy of solution of $NaCl$ is:

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