The following reaction can be written as the sum of two reactions, one of which

Question

The following reaction can be written as the sum of two reactions, one of which relates to ionization energy and one of which relates to electron affinity: Rb(g) + I(g)-->Rb+(g) + I-(g)

Explanation / Answer

2 Rb (s) + Cl2 (g) ----> 2 RbCl (s) Enthalpy of Formation of RbCl is -430.5 kJ / mole Enthalpy of Formation of 2 Mole of RbCl is -430.5 kJ / mole x 2 = -861 kJ ____ ( per 2 moles ) 2 Rb (s) ----> 2 Rb (g) ____ Sublimation Enthalpy of Sublimation for Rubidium is 86 kJ / mole Enthalpy of Sublimation for 2 Mole of Rubidium is 86 kJ / mole x 2 = 172 kJ ____ ( per 2 moles ) = H1 2 Rb (g) ----> 2 Rb+ (g) + 2 e- ____ Ionization Ionization Energy of Rubidium is 402 kJ / mole Enthalpy of Ionization of 2 Mole of Rubidium is 402 kJ / mole x 2 = 804 kJ ____ ( per 2 moles ) = H2 Cl2 (g) ----> 2 Cl (g) ____ Bond Dissociation Bond Dissociation Energy of Cl2 (g) is 243 kJ / mole Enthalpy of this Dissociation ( 1 Mole of Cl2 gas ) = 243 kJ ____ ( per 1 mole of Cl2 ; per 2 moles of Cl atoms ) = H3 2 Cl (g) + 2 e- ----> 2 Cl- (g) Electron Affinity of Chlorine is -349 kJ / mole Enthalpy of this reaction ( of 2 Mole of Chlorine ) is -349 kJ / mole x 2 = -698 kJ ____ ( per 2 moles of Cl- ) = H4 2 Rb+ (g) + 2 Cl- (g) ----> 2 RbCl (s) ____ Formation of Crystal Lattice from Ions Enthalpy of this reaction ( which is reverse of Crystal Lattice Breakdown ) = - ( Lattice Energy of RbCl / mole ) x 2 mole = - 2 x ( Lattice Energy of RbCl ) = H5 Enthalpy of Formation of 2 Mole of RbCl = H1 + H2 + H3 + H4 + H5 ( -861 ) = ( 172 ) + ( 804 ) + ( 243 ) + ( -698 ) + ( H5 ) -1382 = H5 Lattice Energy = ( -1382 ) / ( -2 ) = ( +691 )

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