See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

    2 2 H I 2HI 0 60 42 0 60 42 2 volumeat t Volumeat equilibrium x x x       The volume of gas  mole for a gas at same P and T           2 2 2 2 2 .... 60 42 c HI x K i H I x x      For reactions 0, n  one can use mole in place of concentration (mol litre-1) since formula will not involve volume terms. Given, 2 28 14 x x    By eqs. (i) and (ii)      28 28 28 ..... 60 14 42 14 46 c K iii      Now for dissociation of HI       2 2 2HI H I 0 1 0 0 . 1 / 2 / 2 moleat t Moleat eq         Where  is degree of dissociation        1 2 2 2 / 2 / 2 1 4 1 c K              1 2 2 46 1 46 28 28 4 1 c c K K             