The ionization enthalpy of Na⁺ formation from Na (g) is 495.8 kJ mol⁻¹, while the electron gain enthalpy of Br is -325.0 kJ mol⁻¹. Given the lattice enthalpy of NaBr is -728.4 kJ mol⁻¹. The energy for the formation of NaBr ionic solid is (-) ____________ × 10⁻¹ kJ mol⁻¹.

Solution

Chemistry ma’am · 3 steps

Build NaBr from gaseous atoms in three steps, then add the three enthalpies with their signs.

1. Three steps. First remove an electron from sodium: Na(g) → Na⁺(g) + e⁻. Then hand that electron to bromine: Br(g) + e⁻ → Br⁻(g). Finally let the gaseous ions come together as the solid: Na⁺(g) + Br⁻(g) → NaBr(s).
2. Signs of each. Ionization costs energy, so it is +495.8 kJ mol⁻¹. Electron gain releases energy, -325.0 kJ mol⁻¹. Forming the lattice releases a lot more, -728.4 kJ mol⁻¹.

   Δ H = 495.8 + (-325.0) + (-728.4) = -557.6 kJ mol⁻¹

3. So. The energy for forming NaBr is -557.6 kJ mol⁻¹ = -5576 × 10⁻¹ kJ mol⁻¹, so the blank is 5576.

The common slip: 557.6. Writing the result in kJ mol⁻¹ gives 557.6. The blank is in units of 10⁻¹ kJ mol⁻¹, so multiply by 10 to get 5576.

The idea to keep. In a Born-Haber sum, energy you put in is positive and energy released is negative. Write each step with its sign, then add. The large lattice enthalpy is what makes an ionic solid stable.

Answer from the NTA final answer key. Solution by PrepWiser.