Calculate the lattice enthalpy of `CaCl_(2)` given that the enthalpy of :
i) Sublimation of Ca in `"121.8 kJ mol"^(-1)`
ii) Dissociation of `Cl_(2)` to `2Cl` is `"242.8 kJ mol"^(-1)`
iii) Ionisation of `Ca" to "Ca^(2+)" is 2422 kJ mol"^(-1)`
iv) Electron gain for `Cl" to "Cl^(-)" is - 355 kJ mol"^(-1)`
v) `DeltaH_(f)^((o))" overall is "-"795 kJ mol"^(-1)`

Calculate the lattice energy of CaCl_2 from the given data. Ca_((s)) + Cl_(2(g)) to CaCl_(2(s)) DeltaH_f^0 =-795 "kJ mol"^(-1) Sublimation : Ca_((s)) to Ca_((g)) " " DeltaH_1^0=+121 "kJ mol"^(-1) Ionisation : Ca_((g)) to Ca_((g))^(2+) +2e^(-) " " DeltaH_2^0=+2422 "kJ mol"^(-1) Dissociation : Cl_(2(g)) to 2Cl_((g)) " " DeltaH_3^0=+242.8 "kJ mol"^(-1) Electron affinity : Cl_((g)) + e^(-) to Cl_((g))^(-) " " DeltaH_4^0=-355 "kJ mol"^(-1)

Calculate the lattice energy of formation of NaCl from the following data : Na_((s)) +1//2Cl_(2(g)) to NaCl_((s)) Delta H_(f) -411.3 KJ.mol^(-1) Heat of sublimation of Na_((s)) = 108 . 7 kJ mol ^(-1) Ionisation energy of Na_((g)) = 49.5.0 kJ mol^(-1) Dissociation energy of Cl_(2(g)) = 244 kJ mol^(-1) Electron affinity of Cl_((g)) = - 349 . 0 kJ mol^(-1)

When a mole of magnesium bromide is prepared from 1 mole of magnesium and 1 mole of liquid bromine, 524 kJ of energy is released. The heat of sublimation of Mg metal is 148 "kJ mol"^(-1) . The heat of dissociation of bromine gas into atoms is 193 "kJ mol"^(-1) . The heat of vaporization of liquid bromine is 31 kJ "mol"^(-1) . The ionisation energy of magnesium is 2187 kJ mol^(-1) and the electron affinity of bromine is - 662 "kJ mol"^(-1) . Calculate the lattice energy of magnesium bromide.

Calculate the entropy change when 1 mole of ethanol is evaporated at 351 K. The molar heat of vaporization of ethanol is 39.84 kJ "mol"^(-1)