JEE Main2013ChemistryThermodynamics (C)Actual
Given Reaction Energy Change array ll & (in kJ) Li ( s ) Li ( g ) & 161 Li ( g ) Li ⁺( g ) & 520 1 2 ~F ₂( ~g ) F ( g ) & 77 ~F ( ~g )+ e ⁻ F ⁻( g ) & array l (Electron gain enthalpy) array Li ⁺( g )+ F ⁻( g ) LiF ( s ) & -1047 Li ( s )+ 1 2 ~F ₂( ~g ) LiF ( s ) & -617 array Based on data provided, the value of electron gain enthalpy of fluorine would be :
Options
- A-300 ~kJ ~mol ⁻¹
- B-350 ~kJ ~mol ⁻¹
- C-328 ~kJ ~mol ⁻¹
- D-228 ~kJ ~mol ⁻¹
Correct answer
C. -328 ~kJ ~mol ⁻¹
Step-by-step solution
Applying Hess's Law aligned _ f H ^ = & _ sub H + 1 2 _ diss H + I.E. + E.A + _ lattice H & -617=161+520+77+ E.A. +(-1047) & E.A. =-617+289=-328 ~kJ ~mol ⁻¹ & electron affinity of fluorine & =-328 ~kJ ~mol ⁻¹ aligned