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JEE MainChemistryChemical Equilibrium

For the gaseous equilibrium A ₂( g ) 2 A ( g ) , the standard Gibbs free energy change at temperature T is G^ . Starting with 1 mole of pure A ₂ , the degree of dissociation x is very small ( x 1 ). The expression for x in terms of the total equilibrium pressure P , G^ , the universal gas constant R , and T is:

Options

  1. Ax = P^ -1/2 e^ - G^ / 2RT
  2. Bx = 1 2 P^ -1/2 e^ - G^ / RT
  3. Cx = 1 2 P^ -1/2 e^ - G^ / 2RT
  4. Dx = 1 4 P^ -1/2 e^ - G^ / 2RT

Correct answer

C. x = 1 2 P^ -1/2 e^ - G^ / 2RT

Step-by-step solution

The thermodynamic equilibrium constant K_p is related to standard Gibbs free energy change by: K_p = e^ - G^ / RT For the reaction: A ₂( g ) 2 A ( g ) Initial moles: 1 mole of A ₂ At equilibrium: 1 - x moles of A ₂ and 2x moles of A Total moles at equilibrium = 1 - x + 2x = 1 + x Since x is very small, 1 + x 1 and 1 - x 1 . Partial pressures at equilibrium: p_ A ₂ = 1 1 P = P p_ A = 2x 1 P = 2xP The equilibrium constant K_p is: K_p = (p_ A )^2 p_ A ₂ = (2xP)^2 P = 4x^2 P Equating the two expressions for K_p : 4x^2

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