What is the change in enthalpy for this gas-phase reaction between potassium and bromine?
The ionization energy for potassium is 419 kJ/mol. The electron affinity for bromine is -325 kJ/mol. Use these values and Hess's law to calculate the change in enthalpy for the following reaction per mole of reagent:
K(g)+Br(g)→K+(g)+Br−(g), ΔH=?
The ionization energy for potassium is 419 kJ/mol. The electron affinity for bromine is -325 kJ/mol. Use these values and Hess's law to calculate the change in enthalpy for the following reaction per mole of reagent:
K(g)+Br(g)→K+(g)+Br−(g), ΔH=?
1 Answer
You should write out the reaction definitions of electron affinity and ionization energy:
#"X"(g) + e^(-) -> "X"^(-)(g)# ,
Electron affinity, electron added in
#"M"(g) -> "M"^(+)(g) + e^(-)# ,
Ionization energy, electron removed
Hence, one can deconstruct this into two half-reactions.
#"K"(g) -> "K"^(+)(g) + cancel(e^(-))# ,#" "DeltaH_(IE) = "419 kJ/mol"#
#ul("Br"(g) + cancel(e^(-)) -> "Br"^(-)(g))# ,#" "DeltaH_(EA) = -"325 kJ/mol"#
#"K"(g) + "Br"(g) -> "K"^(+)(g) + "Br"^(-)(g)#
And the sum is the net change in enthalpy, as per Hess's law...
#color(blue)(DeltaH_(rxn)) = DeltaH_1 + DeltaH_2 + . . . #
#= 419 - 325# #"kJ/mol"#
#=# #color(blue)("94 kJ/mol")#
Is this endothermic or exothermic?
