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JEE Main Chemistry Chemical Equilibrium 2026 JEE Main 2026 (28 January Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

Observe the following equilibrium in a 1 L flask. A ( ~g ) B ( ~g ) At T ( K ), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T ( K ) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively

Options

  1. A. 0.53,0.4.
  2. B. 0.742,0.557.
  3. C. 0.557,0.418.
  4. D. 0.367,0.275.

Answer

C. 0.557,0.418.

Step-by-step solution

Initial equilibrium: [A]=0.5 M, [B]=0.375 M; K_c= 0.375 0.5 =0.75 After adding 0.1 mol A: [A]=0.6 M, [B]=0.375 M; Q=0.625 Q Let x be A that decomposes: 0.375+x 0.6-x =0.75 0.375+x=0.45-0.75x 1.75x=0.075; x≈0.0429 New [A]=0.6-0.0429=0.557 M; [B]=0.375+0.0429=0.418 M

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Related: Chemistry — Chemical Equilibrium · All PYQ Banks