Given: The heat of sublimation of K(s) is 89 kJ mol^(-1). K(g) rarrK^(o+)(g)+e^(-), DeltaH^(Theta) = 419 kJ F_(2)(g) rarr 2F(g),DeltaH^(Theta) = 155 kJ The lattice enegry of KF(s) is -813kJ mol^(-1), the heat of formation of KF(s) is -563 kJ mol^(-1). the E_(A) of F(g) is

Given: The heat of sublimation of K(s) is 89 kJ mol^(-1). K(g) rarrK^

1.	Given: The heat of sublimation of K(s) is 89 kJ mol^(-1). K(g) rarrK^(o+)(g)+e^(-), DeltaH^(Theta) = 419 kJ F_(2)(g) rarr 2F(g),DeltaH^(Theta) = 155 kJ The lattice enegry of KF(s) is -813kJ mol^(-1), the heat of formation of KF(s) is -563 kJ mol^(-1). the E_(A) of F(g) is
Answer» `-413` `-336` `-1149` `+413` Solution :Given that: Heat of sulmiation of `K(s)` `K(s) rarr K(v),Delta_("sub")H^(Theta) = 89 kJ mol^(-1)` Heat of ionisation `K(g) rarr K^(o+) (g) +e^(-),Delta_(i)H^(Theta) = 419 kJ mol^(-1)` Heat of atomisation of `F_(2)` `F_(2)(g) rarr 2F(g),Delta_(a)H^(Theta) = 155 kJ mol^(-1)` LATTICE enegry `Delta_(U)H^(Theta) =- 813 kJ mol^(-1)` Heat of formation `Delta_(f)H^(Theta) =- 563 kJ mol^(-1)` [Heat of electron affinify of `F(g) = ?]` `:. Delta_(f)H = Delta_("sub")H^(Theta) +Delta_(lie)H^(Theta) +(1)/(2)Delta_(a)H^(Theta) +Delta_(E_(A))H^(Theta) +Delta_(i)H^(Theta)` or `Delta_(E_(A))H^(Theta) = Delta_(f)H - Delta_("sub")H^(Theta) - Delta_(i)H^(Theta) - (1)/(2)Delta_(a)H^(Theta) - Delta_(i)H^(Theta)` Substituting all the VALUES: `Delta_(E_(A))H^(Theta) =- 563 - 89 - 419 - (155)/(2)+813.0` `~~-336 kJ mol^(-1)`