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JEE Main Chemistry Chemical Equilibrium 2026 JEE Main 2026 (05 April Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

The reaction A(g) B(g) + C(g) was initiated with the amount 'a' of A(g). At equilibrium it is found that the amount of A(g) remaining is (a - x) at a total pressure of p. The equilibrium constant K_p of the reaction can be calculated from the expression :

Options

  1. A. x^2 a^2 + x^2 p
  2. B. x^2 a^2 - x^2 p
  3. C. a + x^2 x^2 p
  4. D. a^2 - x^2 x^2 p

Answer

B. x^2 a^2 - x^2 p

Step-by-step solution

The reaction is A(g) B(g) + C(g) Initial moles: a for A, 0 for B, 0 for C Moles at equilibrium: (a - x) for A, x for B, x for C Total moles at equilibrium = (a - x) + x + x = a + x Partial pressures at equilibrium: p_A = a - x a + x p p_B = x a + x p p_C = x a + x p The equilibrium constant K_p is given by: K_p = p_B p_C p_A K_p = ( x a + x p ) ( x a + x p ) a - x a + x p K_p = x^2 (a + x)(a - x) p K_p = x^2 a^2 - x^2 p Answer: x^2 a^2 - x^2 p

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Related: Chemistry — Chemical Equilibrium · All PYQ Banks