Respuesta :
Answer:
Heat of hydration of calcium ion is -1703 kJ/mol
Explanation:
Dissolution equilibrium of calcium chloride:
[tex]CaCl_{2}(s)\rightleftharpoons Ca^{2+}(aq.)+2Cl^{-}(aq.)[/tex]
[tex]\Delta H_{sol}=[1mol\times \Delta H_{hyd}(Ca^{2+})_{aq.}]+[2mol\times \Delta H_{hyd}(Cl^{-})_{aq.}]-[1mol\times U(CaCl_{2})_{s}][/tex]
Where [tex]\Delta H_{sol}[/tex] is heat of solution, [tex]\Delta H_{hyd}[/tex] is heat of hydration and U represents lattice energy.
Here, [tex]\Delta H_{sol}[/tex] = -121 kJ/mol, [tex]\Delta H_{hyd}(Cl^{-})_{aq.}[/tex] = -338 kJ/mol and [tex]U(CaCl_{2})(s)[/tex] = -2258 kJ/mol
So, plug-in all the given values in the above equation-
[tex]-121=(1\times \Delta H_{hyd}(Ca^{2+})_{aq})+(2\times -338)-(1\times -2258)[/tex]
So, [tex]\Delta H_{hyd}(Ca^{2+})_{aq}=-1703 kJ/mol[/tex]
