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GATE 2019 MT – Question 45

Metallurgical Thermodynamics · Reaction equilibria and Ellingham diagrams · 2 marks · Multiple choice

For C(graphite)+2H₂(g)→CH₄(g), ΔH°=−74900 J/mol and ΔS°=−80 J/(mol K) at 300 K. With R=8.314 J/(mol K), find the equilibrium constant at 300 K.

  1. 5.6×10⁶
  2. 3.6×10⁷
  3. 4.0×10⁸
  4. 7.3×10⁸

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Correct answer: (D) 7.3×10⁸

Explanation

$\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=-74900-300(-80)=-50900$ J/mol. Since ΔG°=−RT lnK, lnK=50900/(8.314×300)=20.407. Exponentiation gives K≈7.3×10⁸, D. The entropy term reduces the driving force relative to ΔH° alone.

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