See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

     2 3 2 1 1 1 2 N O g NO g NO g Initial 4 Equi. 4 P P P P          2 2 4 1 1 2 2 2NO g N O g Initial P Equi. P P P / 2     Now,   1 1 1 2 P 1 P P P K 4 P    Since 1 P 1 K 1, P 2     2 2 2 2 P 1 P 1 4 2          2 2 P 2 2 1 2 P / 2 1/ 2 K