The standard entropy change for the reaction
4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) is -550 J K⁻¹ at 298 K.
[Given : The standard enthalpy change for the reaction is -165 kJ mol⁻¹]. The temperature in K at which the reaction attains equilibrium is _____________. (Nearest Integer)

Solution

Chemistry ma’am · 4 steps

At equilibrium Δ G = 0, so just divide Δ H by Δ S, with both in joules.

1. Equilibrium condition. A reaction is at equilibrium when Δ G = 0. Put this into Δ G = Δ H - TΔ S and you get Δ H = TΔ S.

   T = (Δ H)/(Δ S)

2. Match the units. Δ H = -165 kJ mol⁻¹ = -165000 J mol⁻¹ and Δ S = -550 J K⁻¹ mol⁻¹. Both signs are negative, so T comes out positive.
3. Put in numbers. T = -165000/-550 = 300 K. Below this temperature the negative Δ H wins and the rusting reaction is spontaneous. Above it, the -TΔ S term wins and it is not.
4. So. The reaction attains equilibrium at 300 K.

The common slip: 0.3. Dividing 165 by 550 without converting kJ to J gives 0.3 K, not 300 K.

The idea to keep. Equilibrium means Δ G = 0, so T = Δ H / Δ S. Always bring Δ H into joules before dividing by an entropy in J per K.

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