When Δ Hᵥₐₚ=30 kJ / mol and Δ Sᵥₐₚ=75 J mol⁻¹ K⁻¹, then the temperature of vapour, at one atmosphere is _________ K.

Solution

Chemistry ma’am · 3 steps

At the boiling point liquid and vapour are in equilibrium, so set Δ G = 0 and solve for T.

1. Equilibrium condition. At 1 atm, the liquid and its vapour coexist at the normal boiling point. At equilibrium Δ G = Δ H - TΔ S = 0.

   T_b = (Δ Hᵥₐₚ)/(Δ Sᵥₐₚ)

2. Match the units. Δ Hᵥₐₚ is in kJ but Δ Sᵥₐₚ is in J, so convert first: 30 kJ mol⁻¹ = 30000 J mol⁻¹.
3. So. T_b = 30000/75 = 400 K. The answer is 400.

The common slip: 0.4. Dividing 30 by 75 without converting kJ to J gives 0.4 K, which is far too small. Change 30 kJ to 30000 J first.

The idea to keep. Any phase change at its transition temperature is an equilibrium, so Δ G = 0 and T = Δ H/Δ S. Always bring Δ H and Δ S to the same energy unit first.

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