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Question: The lattice energy of \[CsI(s)\] is \[ - 604Kjmo{l^{ - 1}}\] and the enthalpy of solution is \[33Kjm...

The lattice energy of CsI(s)CsI(s) is −604Kjmol−1 - 604Kjmo{l^{ - 1}} and the enthalpy of solution is 33Kjmol−133Kjmo{l^{ - 1}} . How would you calculate the enthalpy of hydration of 0.650.65 moles of CsICsI ? Enter a numeric value only, do not enter units in your answer.

Explanation

Solution

Lattice enthalpy is the heat released due to the formation of all ionic bonds in a compound and the enthalpy of solution is the heat released upon dissolution of the compound. Hydration enthalpy is the heat needed to separate the ions of a lattice and then dissolve them in water.

Complete answer:
Enthalpy of hydration is the amount of heat released or evolved when the ions of a particular ionic compound get dissolved in a water and get stabilized by the ion-dipole interaction in it. This stabilization causes the heat to release and we expect hydration enthalpy to be an exothermic process.
The first step of hydrating ions of CsICsI is by breaking apart its lattice to get individual cesium and iodide ions. The ionic bonds consist of strong electrostatic forces of attraction which we need to get rid of by supplying heat energy equal to the lattice energy of the compound.
The second step is the dissolution, which is an exothermic step and the energy evolved should be taken as it is. Thus, the formula of hydration enthalpy in terms of lattice enthalpy and enthalpy of solution can be written as follows:
ΔhydH=(−ΔsolH)+(−ΔLEH){\Delta _{hyd}}H = ( - {\Delta _{sol}}H) + ( - {\Delta _{LE}}H)
Inserting the values of lattice and solution enthalpy with appropriate sign convention in the above formula we get,
ΔhydH=−(−604Kjmol−1)+(−33Kjmol−1)=571Kjmol−1{\Delta _{hyd}}H = - ( - 604Kjmo{l^{ - 1}}) + ( - 33Kjmo{l^{ - 1}}) = 571Kjmo{l^{ - 1}}
Thus enthalpy is a measure of heat released by the hydration of one mole of cesium iodide. This value needs to be multiplied by the number of moles to get the final enthalpy of hydration.
ΔhydH=571Kjmol−1×0.65mol=371.15Kj{\Delta _{hyd}}H = 571Kjmo{l^{ - 1}} \times 0.65mol = 371.15Kj
⇒\Rightarrow Thus, the numerical value of enthalpy of hydration of cesium iodide is 371.15371.15.

Note:
The positive sign of hydration enthalpy indicates the fact that the cesium iodide is not soluble in water and its hydration is an endothermic process rather than an exothermic process. The large size of the ions is responsible for low hydration.