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 The lattice energy of solid NaCl is 150\: \mathrm{kcal} / \mathrm{mol}. The dissolution of the solid in water in the form of ions is endothermic to the extent of 1\: \mathrm{kcal} / \mathrm{mol} . If the hydration energies of \mathrm{Na^{+}}and \mathrm{Cl^{-}} are in the ratio 5 : 4, what is the enthalpy of hydration of the ion?

Option: 1

-82.8 \mathrm{kcal}


Option: 2

-72.8 \mathrm{kcal}


Option: 3

-62.8 \mathrm{kcal}


Option: 4

-52.8 \mathrm{kcal}


Answers (1)

best_answer

The lattice enthalpy and the enthalpy change for hydration are added to determine the enthalpy change for dissolution.

\Delta \mathrm{H}_{\mathrm{sol}}=\Delta \mathrm{H}_{\text {lattice }}+\Delta \mathrm{H}_{\mathrm{hyd}}
Substitute values in the above expression,

\begin{aligned} & 1=150+\Delta \mathrm{H}_{\mathrm{h}} \\ & \Delta \mathrm{H}_{\mathrm{h}}=-149 \mathrm{kcal} \end{aligned}

The sum of the enthalpies of hydration for sodium ions and chloride ions is the enthalpy of hydration for sodium chloride.

\Delta \mathrm{H}_{\mathrm{Na}^{+}(\mathrm{h})}+\Delta \mathrm{H}_{\mathrm{Cl}^{-}(\mathrm{h})}=-149 \mathrm{kcal}

The hydration energies of sodium ions and chloride ions are in the ratio 5: 4.

\begin{aligned} & \therefore \Delta \mathrm{H}_{\mathrm{Na}^{+}(\mathrm{h})}=\frac{5 \times(-149)}{9} \\ & =-82.77 \\ & \approx-82.8 \end{aligned}

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Divya Prakash Singh

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