Jeeify
MCQEasyJEE Main 2024 · 31 January, Shift 1Gibbs Free Energy & Equilibrium Constant

Chemistry Question from JEE Main 2024 · 31 January, Shift 1

Consider the following reaction at 298 K298 \, \text{K}:

32O2(g)→O3(g),Kp=2.47×10−29.\frac{3}{2} O_2(g) \rightarrow O_3(g), \quad K_p = 2.47 \times 10^{-29}.

ΔG∘\Delta G^\circ for the reaction is kJ\text{kJ}. (Given R=8.314 J K−1 mol−1R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1})

  • A

    150150

  • B

    163163

  • C

    175175

  • D

    190190

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