See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

   2 4 1 2 ; 17 C s H g CH g H kcal     ……………… (i)    2 2 6 2 2 3 ; 24 C s H g C H g H kcal     …………… (ii)    2 3 8 3 4 ; ? f C s H g C H g H     ……………….. (iii) From equations (i) and (ii), we get Let x kcal 1 mol is the energy of   C s C g  Let y kcal 1 mol is the BE energy of   H H  From equation (i), we get 17 2 4 99 x y     From equation (ii), we get   24 2 3 84 6 99 x y      Solved for x and y, AITS-FT-VI (RE NAME-FT-V)-PCM(Sol.)-JEE (Main)/2021 Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 8 171 , 104 x kcal y kcal   From equation (iii), we get   3 4 2 84 8 99 f H x y       1 31kcalmol 