Jeeify
MCQEasyJEE Main 2024 · 27 January, Shift 1Gibbs Free Energy & Equilibrium Constant

Chemistry Question from JEE Main 2024 · 27 January, Shift 1

For a reaction, if the equilibrium constant at 500 K500 \, \text{K} is 44, then the value of the standard Gibbs free energy ΔG∘\Delta G^\circ at this temperature is:

  • A

    −1155 J-1155 \, \text{J}

  • B

    1386 J1386 \, \text{J}

  • C

    −1386 J-1386 \, \text{J}

  • D

    1155 J1155 \, \text{J}

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