Given{:("Reaction","Energy Change (inKJ)",),(Li(s) rarr Li(g),161,),(Li(g) rarr Li^(+) (g),520,),((1)/(2) F_(2)(g) rarr F(g),77,),(F(g) + e^(-) rarr F(g),("Electron gain enthalpy"),),(Li^(+) (g) + F^(-)(g) rarr LiF(s),-1047,),(Li (s) + (1)/(2) f_(2) (g) rarr LiF(s),-617,):} Based on data provided, the value of electron gain enthalpy of fluorine would be :

Given{:("Reaction","Energy Change (inKJ)",),(Li(s) rarr Li(g),161,),(L

1.	Given{:("Reaction","Energy Change (inKJ)",),(Li(s) rarr Li(g),161,),(Li(g) rarr Li^(+) (g),520,),((1)/(2) F_(2)(g) rarr F(g),77,),(F(g) + e^(-) rarr F(g),("Electron gain enthalpy"),),(Li^(+) (g) + F^(-)(g) rarr LiF(s),-1047,),(Li (s) + (1)/(2) f_(2) (g) rarr LiF(s),-617,):} Based on data provided, the value of electron gain enthalpy of fluorine would be :
Answer» `-300 kJ mol^(-1)` `-228 kJ mol^(-1)` `-328 kJ mol^(-1)` `-350 kJ mol^(-1)` Solution : `Li (s) + (1)/(2)F_(2) (g) RARR LIF(s) = - 617` `DeltaH = -617 = 161 + 520 + 77 + x - 1047` `x = -328 kJ mol^(-1)`