What is #Delta"E"# for the formation of gaseous sodium and chloride ions from gaseous sodium and chlorine atoms?

Question

Ellie Andrews · Accepted Answer

The #Delta"E"# is #"147.0 kJ/mol"#.

#Delta"E"# for the formation of gaseous sodium and chloride ions from gaseous sodium and chlorine atoms is the sum of the first ionization energy (I.E.) for gaseous Na, and the electron affinity (E.A.) of a chlorine atom in the gaseous state for an electron.

#Delta"E" = "I.E"#. + #"E.A."#

The following data comes from https://tutor.hix.ai

1st Ionization Energy [#"Na"("g")# #rarr# #"Na"^+("g")# + #"e"^-#]: #"495.8 kJ/mol"#

Electron Affinity [#"Cl(g)"# + #"e"^-# #rarr# #"Cl"^-""("g")#]: #-348.8# #"kJ/mol"#

#Delta"E"# = #"495.8 kJ/mol"# + #(-348.8)##"kJ/mol"# = #"147.0 kJ/mol"#

Note: I included only the data that started with the sodium and chlorine atoms in the gaseous state, and ended with the formation of the gaseous sodium and chloride ions.

Serenity Allen · Answer

undefined The standard molar enthalpy change ($ \Delta H^\circ $) for the formation of gaseous sodium chloride ($NaCl$) from its elements is $ -411 \, \text{kJ/mol} $ and the standard entropy change ($ \Delta S^\circ $) for the reaction is $ -3.41 \, \text{J/(mol K)} $. Therefore, the standard Gibbs free energy change ($ \Delta G^\circ $) for the reaction can be calculated using the equation:

$$ \Delta G^\circ = \Delta H^\circ - T \cdot \Delta S^\circ $$

Where $ \Delta G^\circ $ is the standard Gibbs free energy change, $ \Delta H^\circ $ is the standard enthalpy change, $ \Delta S^\circ $ is the standard entropy change, and $ T $ is the temperature in Kelvin. The value of $ \Delta G^\circ $ for this reaction can be calculated at various temperatures using the provided data and the equation.