*By definition, the gaseous cation and anion creating the matching ionic link release power termed the lattice energy, the energy had within the lattice structure.You are watching: Use the born-haber cycle and the data shown to calculate the lattice energy of cao.*

For an alternate explanation, see here.

The **Born-Haber cycle** takes benefit of the state role property that the change in enthalpy to indirectly recognize the lattice power of ionic link through procedures that utilize known thermodynamic quantities prefer ionization energy and electron affinity.

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Let"s take #"NaCl"# together an example. We begin by writing the development reaction, i beg your pardon is *by definition* from the elemental states at #25^

"C"# and #"1 atm"#:

#"Na"(s) + 1/2"Cl"_2(g) -> "NaCl"(s)#

**Our score is to change the reactants right into their ionic gases, as that is the reaction that describes the procedure for which "lattice energy" is defined.**

Put this every together, v some data, and also we get, because that #"1 mol"# the #"NaCl"(s)#:

#"NaCl"(s) -> "Na"(s) + 1/2"Cl"_2(g)#, #-DeltaH_(f,"NaCl"(s)) = +"411 kJ"#

#"Na"(s) -> "Na"(g)#, #DeltaH_("sub","Na") = "107 kJ"#

#"Na"(g) -> "Na"^(+)(g) + e^(-)#, #"IE"_(1,"Na"(g)) = "502 kJ"#

#1/2"Cl"_2(g) -> "Cl"(g)#, #1/2DeltaH_("bond","Cl"_2(g)) = 1/2xx"242 kJ"#

#"Cl"(g) + e^(-) -> "Cl"^(-)(g)#, #"EA"_(1,"Cl"(g)) = -"355 kJ"#

#"Na"^(+)(g) + "Cl"^(-)(g) -> "NaCl"(s)#, #DeltaH_"lattice" = ???#

#"-----------------------------------------------------------------------------"#

#"These release out totally upon adding, proving"#

#"we have a complete cycle."#

And now if we wish, the lattice energy can be calculated.

Take the #DeltaH_f^

# step as gift upwards to create a finish cycle, for which #DeltaH_"cycle" = 0# (since #H_f = H_i# because that a complete cycle). Therefore, we have actually this equation:

#0 = DeltaH_"cycle" = DeltaH_(f,"NaCl"(s))^

+ DeltaH_("sub","Na") + "IE"_(1,"Na"(g)) + 1/2DeltaH_("bond","Cl"_2(g)) - "EA"_(1,"Cl"(g)) - DeltaH_"lattice"#

Solving because that #DeltaH_"lattice"# *usually* gives a confident answer, so we take the negative of the prize by convention to get:

#color(blue)(DeltaH_"lattice" -= -|DeltaH_"lattice"|)#

#= color(blue)(-