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

## 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

Mar 20, 2016

I like to do a Hess's Law calculation.

#### Explanation:

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

$\text{Na"^+("g") + "H"^("-")("g") → "NaH(s)"; Δ""_"latt"H = "?}$

The equations given are

1. "H(g)" + e^"-" → "H"^"-"("g"); ${E}_{\text{EA" = "-72.8 kJ/mol}}$

$2. \text{Na(g)" → "Na"^+("g") + "e"^"-}$; ${E}_{\text{i₁" = "+495.8 kJ/mol}}$

$3. \text{Na(s)" → "Na(g)}$; Δ_"sub"H = "+107.3 kJ/mol"

$4. \text{H"_2 "(g") → "2H(g)}$; $D = \text{+435.9 kJ/mol}$

$5. \text{Na(s)" + 1/2H_2("g") → "NaH(s)}$; 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)"))) $\textcolor{w h i t e}{m m m l} \text{-495.8}$

color(red)(cancel(color(black)("Na(g)"))) → color(red)(cancel(color(black)("Na(s)" $\textcolor{w h i t e}{m m m m m m l l} \text{-107.3}$

color(red)(cancel(color(black)("Na(s)"))) + color(red)(cancel(color(black)(1/2"H"_2("g")))) → "NaH(s)" $\textcolor{w h i t e}{m} \text{-60}$

color(red)(cancel(color(black)("H"("g"))) →color(red)(cancel(color(black)(1/2"H"_2 ("g")))) $\textcolor{w h i t e}{m m m m m m} \text{-218.0}$

"H"^"-"("g") + color(red)(cancel(color(black)("e"^"-"))) → color(red)(cancel(color(black)("H"("g")))) $\textcolor{w h i t e}{m m m m m} \text{+72.8}$

$\text{Na"^+"(g)" + "H"^"-"("g") → "NaH(s)}$ $\textcolor{w h i t e}{m l l} \text{-808.3}$

The lattice energy of $\text{NaH}$ is -808 kJ/mol.