See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

 A s B D   B D P P 40    2P = 40 Or P = 20 atm    2 p K P 20 20 400     Similarly,    C s E g D g   E D P P 80    2P = 80 Or P = 40 atm  p K 40 40 1600    For simultaneous equilibrium  A s B D   1 1 2 P P P   C d E D   P2 + P2 + P1    1 1 2 400 P P P      2 1 2 1600 P P P   Or 1 2 1 2 P 1 or P 4P P 4   Also,   1 1 2 P P P 400   AITS-FT-XVII-PCM(Sol.)-JEE(Main)/22 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 12   1 1 1 P P 4P 400   2 1 5P 400  P1 = 8.94 atm Total pressure at eq. =   1 1 2 2 P P P P    =     1 2 1 1 2 P P 2 P 4P    = 1 10P 10