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BITSAT Chemistry Thermodynamics (C) 2024 BITSAT 2024 (Memory Based Paper 3)

BITSAT Chemistry Question (2024) — Solution

Question

The bond dissociation energy of X_2, Y_2 and X Y are in the ratio of 1: 0.5: 1 . H for the formation of X Y is -200 ~kJ / mol . The bond dissociation energy of X_2 will be

Options

  1. A. 200 ~kJ / mol
  2. B. 100 ~kJ / mol
  3. C. 400 ~kJ / mol
  4. D. 800 ~kJ / mol

Answer

D. 800 ~kJ / mol

Step-by-step solution

Let the bond dissociation energy of X_2=a ~kJ / mol i.e. BE (X_2 )=a ~kJ / mol Then, BE ( Y _2 )=0.5 a and BE (X Y)=a ~kJ / mol Given, 1 2 X_2+ 1 2 Y_2 X Y, H=-200 ~kJ / mol aligned _r H & = BE ( Rreactants )- BE ( Product ) \\ _r H & = [ 1 2 BE (X_2 )+ 1 2 BE (Y_2 ) ]- BE (X Y) \\ -200 & = a 2 + 0.5 a 2 -a \\ a & = 200 0.25 =800 ~kJ / mol aligned

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