Find the enthalpy of solution when lattice energy and enthalpy of hydration is given

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Lattice energy of NaI is -686kj/mol enthalpy of hydration is -694kj/mol what is the enthalp of solution per mol NaI

1) Lattice energy refers to Na+(g) + I-(g) ----> Na+I-(s) DeltaH = -686kJ/mol

2) Hydration enthalpy refers to Na+(g) + I-(g) + (aq)  ----> Na+(aq) + I-(aq) DeltaH = -694kJ/mol

If you reverse the first one and add to the second one, you will get the enthalpy of solution of Na+I-(s) which is Na+I-(s) + aq ---> Na+(aq) + I-(aq)

Reversing 1) Na+I-(s) ---->  Na+(g) + I-(g) DeltaH = +686kJ/mol

Na+(g) + I-(g) + (aq)  ----> Na+(aq) + I-(aq) DeltaH = -694kJ/mol

Adding gives Na+I-(s) + Na+(g) + I-(g) + (aq)  ----> Na+(g) + I-(g) + Na+(aq) + I-(aq) and simplifying gives

Na+I-(s)    ----> Na+(aq) + I-(aq) So DeltaH = -8 kJ/mol


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