First slide
Ionic bond
Question

Ionization enthalpy of sodium = 492 kJ/mole;

Electron gain enthalpy of chlorine = -349 kJ/mole;

Sublimation enthalpy of sodium = +108 kJ/mole ;

Bond dissociation enthalpy of chlorine = +243 kJ/mole ;

Change in enthalpy for the process, Nas+12Cl2gNa+g+Cl-g will  be

Moderate
Solution

NasNag,          H=+108 kJ...............1 NagNa+g+e, H=+492 kJ................2 12Cl2gClg,      H=+121.5 kJ/mole....3 Clg+eCl-g,    H=-349 kJ................4 Adding 1 ,2, 3and 4 Nas+12Cl2gNa+g+Cl-g, H=+372.5 kJ ;

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