JEE Main · 2025 · Shift-IIhardTHERMO-142

Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process? (N2(g)…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process? (\ceN2(g)+3H2(g)<=>2NH3(g)\ce{N2(g) + 3H2(g) <=> 2NH3(g)})

Options
  1. a

    Graph (a): ΔHR/T-\Delta H_R^\circ/T decreases; ΔGR/T-\Delta G_R^\circ/T decreases; ΔSR-\Delta S_R^\circ constant positive

  2. b

    Graph (b): ΔHR-\Delta H_R^\circ constant; ΔGR/T-\Delta G_R^\circ/T and ΔSR/T-\Delta S_R^\circ/T vary

  3. c

    Graph (c): ΔSR-\Delta S_R^\circ positive large; ΔHR-\Delta H_R^\circ negative

  4. d

    Graph (d): ΔGR/T-\Delta G_R^\circ/T increases with T

Correct Answera

Graph (a): ΔHR/T-\Delta H_R^\circ/T decreases; ΔGR/T-\Delta G_R^\circ/T decreases; ΔSR-\Delta S_R^\circ constant positive

Detailed Solution

🧠 Haber's process: ΔH<0\Delta H < 0 and ΔS<0\Delta S < 0 — both properties decrease with temperature in a Ellingham-style plot For \ceN2+3H2>2NH3\ce{N2 + 3H2 -> 2NH3}: the reaction is exothermic (ΔHR<0\Delta H_R < 0) and entropy decreases (ΔSR<0\Delta S_R < 0) because gas moles decrease from 4 to 2.

Analyzing the thermodynamic plots:

  • ΔHR/T-\Delta H_R^\circ / T increases as TT increases (since ΔHR\Delta H_R is roughly constant, dividing by larger TT reduces the magnitude of the negative value, making ΔH/T-\Delta H / T less positive).
  • ΔSR-\Delta S_R^\circ is constant and positive (since ΔSR<0\Delta S_R < 0, its negation is >0> 0).
  • ΔGR/T=ΔHR/T+ΔSR-\Delta G_R^\circ / T = -\Delta H_R^\circ / T + \Delta S_R^\circ decreases with temperature (the positive entropy term gets subtracted increasingly by the diminishing ΔH/T-\Delta H/T contribution).

Graph (a) correctly shows ΔH/T-\Delta H/T decreasing, ΔS-\Delta S constant and positive, and ΔG/T-\Delta G/T decreasing.

Answer: (a)\boxed{\text{Answer: (a)}}

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