What is the lattice energy of NaH? Use the given information below:

Heat of formation for NaH = -56.442 kJ/mol
Head of sublimation for Na = 107.3 kJ/mol
Ionization energy for Na = 496.00 kJ/mol
Bond dissociation energy for H2 = 435.9kJ/mol
Electron affinity of H = -72.8

To calculate the lattice energy (ΔH lattice) of NaH, we can use the following formula:

ΔH lattice = Heat of formation of NaH - (Heat of sublimation of Na + Ionization energy of Na + 1/2 * Bond dissociation energy of H2 - Electron affinity of H)

Plugging in the given values:

ΔH lattice = (-56.442 kJ/mol) - (107.3 kJ/mol + 496.00 kJ/mol + 1/2 * 435.9 kJ/mol - (-72.8 kJ/mol))

Simplifying the expression:

ΔH lattice ≈ -56.442 kJ/mol - 107.3 kJ/mol - 496.00 kJ/mol - 1/2 * 435.9 kJ/mol + 72.8 kJ/mol

Calculating the value:

ΔH lattice ≈ -1037.942 kJ/mol

Therefore, the lattice energy of NaH is approximately -1037.942 kJ/mol.

To determine the lattice energy of NaH, we can use the Born-Haber cycle, which relates various energy changes involved in the formation of an ionic compound.

The lattice energy (U) of an ionic compound is the energy released when one mole of a solid ionic compound is formed from its constituent ions in the gas phase. It can be calculated using the following equation:

U = Hf(NaH) - [Hsub(Na) + IE(Na) + 0.5 * BE(H2) + EA(H)]

Where:
Hf(NaH) = Heat of formation of NaH
Hsub(Na) = Heat of sublimation of Na
IE(Na) = Ionization energy of Na
BE(H2) = Bond dissociation energy of H2
EA(H) = Electron affinity of H

Given information:
Hf(NaH) = -56.442 kJ/mol
Hsub(Na) = 107.3 kJ/mol
IE(Na) = 496.00 kJ/mol
BE(H2) = 435.9 kJ/mol
EA(H) = -72.8 kJ/mol

Substituting the values into the equation:

U = (-56.442 kJ/mol) - [107.3 kJ/mol + 496.00 kJ/mol + 0.5 * 435.9 kJ/mol + (-72.8 kJ/mol)]

Let's calculate the value:

U = -56.442 kJ/mol - 107.3 kJ/mol - 496.00 kJ/mol - 0.5 * 435.9 kJ/mol + 72.8 kJ/mol
U = - 265.942 kJ/mol

Therefore, the lattice energy of NaH is approximately -265.942 kJ/mol.