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JEE Main Chemistry Chemical Equilibrium 2026 JEE Main 2026 (22 January Shift 1)

JEE Main Chemistry Question (2026) — Solution

Question

Dissociation of a gas A _ 2 takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K. A _ 2 ( ~g ) 2 ~A ( ~g ) The standard Gibbs energy of formation of the involved substances has been provided below: \( array |c|c| Substance & G _ f ^ / kJ ~mol ^ -1 \\ ~A _2 & -100.00 \\ ~A & -50.832 \\ array \) The degree of dissociation of A _ 2 ( ~g ) is given by (x 10^ -2 )^ 1 / 2 where x= \_\_\_\_. (Nearest integer). [Given: R =8 ~J ~mol ^ -1 ~K ^ -1 , 2=0.3010, 3=0.48 ] Assume degree of dissociation is not negligible.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

For the dissociation reaction A_2(g) 2A(g), first calculate the standard Gibbs free energy: G°_ rxn = 2(-50.832) - (-100.00) = -1.664 kJ/mol. Using G° = -RT K_p: -1664 = -8 300 K_p, giving K_p = e^ 0.6933 = 2. For dissociation with degree of dissociation , starting with 1 mole of A_2 at 1 bar total pressure: At equilibrium, moles are (1- ) for A_2 and 2 for A, with total (1+ ) moles. K_p = 4 ^2 1- ^2 = 2. Solving: 2(1- ^2) = 4 ^2 gives ^2 = 1/3 or = 0.577. Since degree of dissociation is (x 10^ -2 )^ 1/2 : (x 10^ -2 )^ 1/2 = 0.577 gives x 10^ -2 = 0.333, so x = 33.3 33.

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