If lattice enthalpy and hydration enthalpy of | vedclass

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(D) The enthalpy of solution $(\Delta_{	ext{soln}} H)$ is given by the sum of the lattice enthalpy $(\Delta_{L} H)$ and the hydration enthalpy $(\Del

Vedclass · Accepted Answer

$17.20 \ kJ \ mol^{-1}$

Vedclass · Answer

(D) The enthalpy of solution $(\Delta_{	ext{soln}} H)$ is given by the sum of the lattice enthalpy $(\Delta_{L} H)$ and the hydration enthalpy $(\Delta_{	ext{hyd}} H)$.$\Delta_{	ext{soln}} H = \Delta_{L} H + \Delta_{	ext{hyd}} H$Given: $\Delta_{L} H = 699 \ kJ \ mol^{-1}$ and $\Delta_{	ext{hyd}} H = -681.8 \ kJ \ mol^{-1}$.Substituting the values:$\Delta_{	ext{soln}} H = 699 \ kJ \ mol^{-1} + (-681.8 \ kJ \ mol^{-1})$$\Delta_{	ext{soln}} H = 17.2 \ kJ \ mol^{-1}$.

If lattice enthalpy and hydration enthalpy of $KCl$ are $699 \ kJ \ mol^{-1}$ and $-681.8 \ kJ \ mol^{-1}$ respectively,what is the enthalpy of solution of $KCl$?

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