The reaction of cyanamide, NH₂CN₍ₛ₎ with oxygen was run in a bomb calorimeter and ΔU was found to be -742.24 kJ mol⁻¹. The magnitude of ΔH₂₉₈ for the reaction
NH₂CH_((S)) + 3/2O_(2(g)) → N_(2(g)) + O_(2(g)) + H₂O_((I))
is _________ kJ. (Rounded off to the nearest integer) [Assume ideal gases and R = 8.314 J mol⁻¹ K⁻¹]

Solution

Chemistry ma’am · 4 steps

The bomb gives you Δ U. Add Δ n_g RT, and count only the gases when you find Δ n_g.

1. Fix the equation. The printed equation has typos. Cyanamide is NH₂CN, and its carbon burns to CO₂, so the reaction is the one below.

   NH₂CN(s) + 3/2 O₂(g) → N₂(g) + CO₂(g) + H₂O(l)

2. Count gas moles. Leave out the solid and the liquid. The gaseous products are N₂ and CO₂, which is 2 mol. The gaseous reactant is 3/2 mol of O₂. So Δ n_g = 2 - 1.5 = 0.5.
3. Put in numbers. Δ n_g RT = 0.5 × 8.314 × 298 = 1238.8 J = 1.239 kJ. Then Δ H = Δ U + Δ n_g RT = -742.24 + 1.239 = -741.0 kJ.
4. So. The magnitude of Δ H₂₉₈ is 741 kJ.

The idea to keep. A bomb calorimeter works at constant volume, so it measures Δ U. Convert with Δ H = Δ U + Δ n_g RT, and count only gaseous species in Δ n_g.

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