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For Br2(l) Enthalpy of atomisation = x (...

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

A

is `x lty`

B

is `x gty`

C

does not exist

D

is `x = 7y`

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The correct Answer is:
B
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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) = -360.1 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)

The enthalpy of dissociation (in kJ/mol) of fluorine, chlorine, bromine and iodine are :

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