With the help of given following information find `deltaH_f(NaBr)^@`
Ionisation enthalpy of Na(g) to `Na^+` is 495.8KJ/mole, electron given enthalpy of Br is -325KJ/mole & Lattice enthalpy of NaBr(s) -747KJ/mole.

The correct order of electron gain enthalpy (in kJ/mole) is

The sublimation energy of I_(2) (solid) is 57.3 KJ/mole and enthalpy of fusion is 15.5 KJ/mole. The enthalpy of vapourisation of I_(2) is

The ionization enthalpy of Na^+ formation from Nad is 495.8 kJ mol^(-1) , while the electron gain enthalpy of Bris -325.0 kJ mol^(-1) .Given the lattice enthalpy of NaBr is -728.4 kJ mol^(-1) The energy for the formation of NaBr ionic solid is (-)"______" xx 10^(-1)kJ "mol"^(-1)

The electron gain enthalpy (in kJ/mole) of F, O, N and C respectively are :

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^(@) 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) .

Correct order of electron gain enthalpy (Kj/mole) of F, Cl, Br, I

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)