For complete combustion of ethene.
C₂ H₄(g)+3 O₂(g) → 2 CO₂(g)+2 H₂ O(l)
the amount of heat produced as measured in bomb calorimeter is 1406 kJ mol⁻¹ at 300 K. The minimum value of T Δ S needed to reach equilibrium is (-) _________ kJ. (Nearest integer)
Given : R=8.3 J K⁻¹ mol⁻¹

Solution

Chemistry ma’am · 4 steps

A bomb calorimeter gives you Δ U, so convert it to Δ H first, then set Δ G = 0.

1. Read the calorimeter. A bomb calorimeter works at constant volume, so the heat it measures is Δ U. The heat is released, so Δ U = -1406 kJ mol⁻¹.
2. Count gas moles. Only gases count, and water here is a liquid. Gaseous products are 2 mol of CO₂ and gaseous reactants are 1 + 3 = 4 mol, so Δ n_g = 2 - 4 = -2.

   Δ H = Δ U + Δ n_g RT = -1406 + (-2)(8.3)(300)/1000 = -1406 - 4.98 = -1410.98 kJ

3. Equilibrium condition. At equilibrium Δ G = Δ H - TΔ S = 0, so TΔ S = Δ H = -1410.98 kJ.
4. So. TΔ S = -1411 kJ. The blank after the minus sign is 1411.

The common slip: 6386. Adding Δ n_g RT = -4980 J to -1406 kJ without converting to kJ gives 6386, not 1411. Divide RT by 1000 first.

The idea to keep. Bomb calorimeter means constant volume, so it gives Δ U, not Δ H. Use Δ H = Δ U + Δ n_g RT with only gas moles, and set Δ G = 0 at equilibrium.

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