Question

The heat of solution of calcium chloride is -121 kJ/mol. Given that the lattice energy of calcium chloride is -2258 kJ/mol and the heat of hydration of a chloride ion is -338 kJ/mol calculate the heat of hydration of a calcium ion.

Answer 1

Heat of hydration of calcium ion is -1703 kJ/mol

Step-by-step explanation:

Dissolution equilibrium of calcium chloride:

`CaCl_(2)(s)\rightleftharpoons Ca^(2+)(aq.)+2Cl^(-)(aq.)`

`\Delta H_(sol)=[1mol* \Delta H_(hyd)(Ca^(2+))_(aq.)]+[2mol* \Delta H_(hyd)(Cl^(-))_(aq.)]-[1mol* U(CaCl_(2))_(s)]`

Where
`\Delta H_(sol)` is heat of solution,
`\Delta H_(hyd)` is heat of hydration and U represents lattice energy.

Here,
`\Delta H_(sol)` = -121 kJ/mol,
`\Delta H_(hyd)(Cl^(-))_(aq.)` = -338 kJ/mol and
`U(CaCl_(2))(s)` = -2258 kJ/mol

So, plug-in all the given values in the above equation-

`-121=(1* \Delta H_(hyd)(Ca^(2+))_(aq))+(2* -338)-(1* -2258)`

So,
`\Delta H_(hyd)(Ca^(2+))_(aq)=-1703 kJ/mol`