For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, Δ C H Θ = -601.70 kJ mol⁻¹, the magnitude of change in internal energy for the reaction is __________ kJ. (Nearest integer)
(Given : R = 8.3 J K⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

Only gases do expansion work: half a mole of O₂ is used up and no gas forms.

1. Write the reaction. For one mole of magnesium, Mg(s) + 1/2 O₂(g) → MgO(s). Solids hardly change volume, so only the gas counts: Δ n_g = 0 - 1/2 = -1/2.

   Δ H = Δ U + Δ n_g RT ⇒ Δ U = Δ H - Δ n_g RT

2. Find the RT term. RT = 8.3 × 300 = 2490 J = 2.49 kJ. So Δ n_g RT = -1/2 × 2.49 = -1.245 kJ.
3. Solve for ΔU. Δ U = -601.70 - (-1.245) = -600.455 kJ. Its magnitude is a little smaller than that of Δ H, because the surroundings push in on the system as the gas is used up.
4. So. The magnitude of Δ U is 600.455 kJ, which rounds to 600.

The common slip: 643. Leaving RT in joules gives Δ U = -601.70 + 1245 = 643.3, a magnitude of 643. Convert 1245 J to 1.245 kJ before you combine it with Δ H.

The idea to keep. To go between Δ H and Δ U, count only the gaseous moles: Δ n_g is gas products minus gas reactants. Keep its sign, and change RT to kJ before you add.

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