Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process?

Solution

Chemistry ma’am · 3 steps

Haber's process is exothermic and loses gas moles, so Δ H and Δ S are both negative.

1. Fix the signs. For N₂ + 3H₂ → 2NH₃, heat is released, so Δ H° ≈ -92 kJ mol⁻¹. Four moles of gas become two, so Δ S° ≈ -198 J K⁻¹mol⁻¹. Both stay roughly constant with temperature, so their graphs are nearly flat lines below zero.
2. Link the curves. Divide Δ G = Δ H - TΔ S by -T. The -Δ G/T curve is the -Δ H/T curve shifted by Δ S.

   -(Δ G°)/T = -(Δ H°)/T + Δ S°

   Quantity	Sign	Behaviour as T rises
   Δ H°	negative	nearly constant
   Δ S°	negative	nearly constant
   -Δ H°/T	always positive	falls like 1/T, never crosses zero
   -Δ G°/T	positive at low T, then negative	same shape as -Δ H°/T but lower by |Δ S°|, so it crosses zero near 465 K

3. So. Only (a) shows all four correctly: -Δ H°/T positive and falling, -Δ G°/T below it by about 200 and crossing zero, and Δ H° and Δ S° as flat negative lines. The answer is (a).

The idea to keep. Write -Δ G/T = -Δ H/T + Δ S. Once you know the signs of Δ H and Δ S, every curve's sign, shape and relative position follows from this one line.

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