MCQEasyJEE Main 2026 · 28 January, Shift 2Gibbs Free Energy & Equilibrium Constant

Chemistry Question from JEE Main 2026 · 28 January, Shift 2

The plot of log⁡10K\log_{10}K vs 1T\frac{1}{T} gives a straight line. The intercept and slope respectively are (where KK is equilibrium constant).

  • A

    2.303RΔH∘, 2.303RΔS∘\dfrac{2.303R}{\Delta H^\circ},\ \dfrac{2.303R}{\Delta S^\circ}

  • B

    −ΔS∘R2.303, ΔH∘R2.303-\dfrac{\Delta S^\circ R}{2.303},\ \dfrac{\Delta H^\circ R}{2.303}

  • C

    ΔS∘2.303R, −ΔH∘2.303R\dfrac{\Delta S^\circ}{2.303R},\ -\dfrac{\Delta H^\circ}{2.303R}

  • D

    −ΔH∘2.303R, ΔS∘2.303R-\dfrac{\Delta H^\circ}{2.303R},\ \dfrac{\Delta S^\circ}{2.303R}

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Sign in to see the solution

Practice more Gibbs Free Energy & Equilibrium Constant questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Start free trialMore on Gibbs Free Energy & Equilibrium Constant

Related questions