How do you use the following five values to calculate a lattice energy (in kilojoules per mole) for sodium hydride, NaH?

Question

Eea for H = -72.8 kJ/mol
Ei1 for Na = +495.8 kJ/mol
Heat of sublimation for Na= +107.3 kJ/mol
Bond dissociation energy for H2 = 435.9 kJ/mol
Net energy charge for the formation of NaH from its elements= -60kJ/mol

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Answer 1 · 2016-03-20T18:28:19

I like to do a Hess's Law calculation.

The lattice energy is the energy that is released when the ions in the gas phase come together to form the crystal lattice.

The target equation is

#"Na"^+("g") + "H"^("-")("g") → "NaH(s)"; Δ""_"latt"H = "?"#

The equations given are

#1. "H(g)" + e^"-" → "H"^"-"("g")#; #E_"EA" = "-72.8 kJ/mol"#

#2. "Na(g)" → "Na"^+("g") + "e"^"-"#; #E_"i₁" = "+495.8 kJ/mol"#

#3. "Na(s)" → "Na(g)"#; #Δ_"sub"H = "+107.3 kJ/mol"#

#4. "H"_2 "(g") → "2H(g)"#; #D = "+435.9 kJ/mol"#

#5. "Na(s)" + 1/2H_2("g") → "NaH(s)"#; #Δ_"f"H^@= "-60 kJ/mol"#

Now we set up the given equations to get the target equation.

#color(white)(mmmmmmmmmmmmmm)ΔH"/kJ·mol"^"-1"#

#"Na"^+("g") + color(red)(cancel(color(black)("e"^"-"))) →color(red)(cancel(color(black)("Na(g)")))# #color(white)(mmml)"-495.8"#

#color(red)(cancel(color(black)("Na(g)"))) → color(red)(cancel(color(black)("Na(s)"# #color(white)(mmmmmmll)"-107.3"#

#color(red)(cancel(color(black)("Na(s)"))) + color(red)(cancel(color(black)(1/2"H"_2("g")))) → "NaH(s)"# #color(white)(m)"-60"#

#color(red)(cancel(color(black)("H"("g"))) →color(red)(cancel(color(black)(1/2"H"_2 ("g"))))# #color(white)(mmmmmm)"-218.0"#

#"H"^"-"("g") + color(red)(cancel(color(black)("e"^"-"))) → color(red)(cancel(color(black)("H"("g"))))# #color(white)(mmmmm)"+72.8"#

#"Na"^+"(g)" + "H"^"-"("g") → "NaH(s)"# #color(white)(mll) "-808.3"#

The lattice energy of #"NaH"# is -808 kJ/mol.