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.