For `Br_2(l)` Enthalpy of atomisation = x kJ/mol, Bond dissociation enthalpy of bromine = y kJ/mole, then

Use the following data to calculate Delta_("lattice") H^(Θ) " for " NaBr. Delta_("Sub")H^(Θ) for sodium metal = 108.4 kJ mol^(-1) , ionisation enthalpy of sodium = 496 kJ mol^(-1) , electron gain enthalpy of bromine = -325 kJ mol^(-1) , bond dissociation enthalpy of bromine = 192 kJ mol^(-1), Delta_(f) H^(Θ) " for " NaBr(s) = - 360.1 kJ mol^(-1)

Use the following data to calculate Delta_("lattice")H^(@) for NaBr. Delta_("sub")H^(@) for sodium metal = 108.4 kJ "mol"^(-1) .Ionization enthalpy of sodium = 496 kJ "mol"^(-1) Electron gain enthalpy of bromine = -325 kJ "mol"^(-1) .Bond dissociation enthalpy of bromine = 192 kJ "mol"^(-1) . Delta_(f)^(H^(@)) for NaBr (s) = -360.1 kJ "mol"^(-1) .

Use the following data to calculate Delta_("lattice") H^(@) for NaBr. Delta_("sub")H^(@) for sodium metal =108.4 kJ mol^(-1) , ionization enthalpy of sodium =496 kJ mol^(-1) ., electron gain enthalpy of bromine =-325 kJ mol^(-1) bond dissociation enthalpy of bromine =192 kJ mol^(-1) , Delta _f H^(@) for NaBr(s) - 360 kJ mol^(-1) .

Calculate the lattice enthalpy of KCl from the following data by Born- Haber's Cycle. Enthalpy of sublimation of K = 89 kJ mol^(–1) Enthalpy of dissociation of Cl = 244 kJ mol^(–1) Ionization enthalpy of potassium = 425 kJ mol^(–1) Electron gain enthalpy of chlorine = –355 kJ mol^(–1) Enthalpy of formation of KCl = –438 kJ mol^(-1)

Bond enthalpy of bromine is 194 kJ mol^(-) . If enthalpy of vapourisation of Br_2 is +30 kJ mol^(-) , electron gain enthalpy of Br is -325 kJ mol^(-1) and hydration enthalpy of bromide is -339 kJ mol^(-1) calculate the change in enthalpy for the reaction, 1/2Br_(2)(l) + e^(-) overset(aq)(rarr) Br^(-)(aq) .

If enthalpy of atomisation for Br_((2(l))) is xKJ//mol bond enthalpy for Br_2 is Y KJ//mol , the relation between them :