Let B.E. of x2,y2 and xy are xkJmol−1,0.5xkJmol−1 and xkJmol−1 respectively
21X2+21Y2→XY;ΔH=−200kJmol−1
ΔH=−200=Σ(B.E)Reactant−Σ(B.E)Product
−200=[21×(x)+21×(0.5x)]−[1×(x)]⇒x=800kJmol
B.E of X2=800kJmol−1
NEET UG 2018 — Chemistry Physical Chemistry
The bond dissociation energies of X2Y2 and XY are in the ratio of 1:0.5:1.ΔH for the formation of XY is −200kJmol−1. What will be the bond dissociation energy of X2?
Held on 30 Apr 2018 · Verified 9 Jul 2026.
800kJmol−1
100kJmol−1
200kJmol−1
400kJmol−1
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