The bond dissociation energies of X2, Y2 and XY are in the ratio of 1 : 0.5 : 1. H for the formation of XY is -200 kJ,mol-1. The bond dissociation energy of X2 will be

NEET 2018 · ChemistryPrevious Year Question

The bond dissociation energies of $X X_{2}$ , $Y X_{2}$ and $XY$ are in the ratio of $1 : 0.5 : 1$ . $Δ H$ for the formation of $XY$ is $- 200 k J m o l^{- 1}$ . The bond dissociation energy of $X X_{2}$ will be

Answer: (A) $800 k J m o l^{- 1}$ . \textbf{Answer:} (A) $800 k J m o l^{- 1}$ \textbf{Solution:} Formation reaction: $\frac{1}{2} X X_{2} + \frac{1}{2} Y X_{2} XY$ .

A. $800 k J m o l^{- 1}$ ✓
B. $100 k J m o l^{- 1}$
C. $200 k J m o l^{- 1}$
D. $400 k J m o l^{- 1}$

Correct Answer

(A) $800 k J m o l^{- 1}$

Solution & Explanation

\textbf{Answer:} (A) $800 k J m o l^{- 1}$ \textbf{Solution:} Formation reaction: $\frac{1}{2} X X_{2} + \frac{1}{2} Y X_{2} XY$ . Let the bond dissociation energy of $X X_{2}$ be $x$ ; then BDE $(Y X_{2}) = 0.5 x$ and BDE $(XY) = x$ . Enthalpy of formation $=$ (bonds broken) $-$ (bonds formed): $Δ_{f} H = (\frac{1}{2} x + \frac{1}{2} (0.5 x)) - x = - 200$ . This gives $\frac{x}{2} + \frac{x}{4} - x = - 200 \Rightarrow - \frac{x}{4} = - 200 \Rightarrow x = 800 k J m o l^{- 1}$ . Mapped: the source key gave numeric option "1" $= 800 k J m o l^{- 1}$ , which is option A here.

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The bond dissociation energies of XX2​, YX2​ and XY are in the ratio of 1:0.5:1. ΔH for the formation of XY is −200 kJmol−1. The bond dissociation energy of XX2​ will be

Solution & Explanation

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The bond dissociation energies of $X X_{2}$ , $Y X_{2}$ and $XY$ are in the ratio of $1 : 0.5 : 1$ . $Δ H$ for the formation of $XY$ is $- 200 k J m o l^{- 1}$ . The bond dissociation energy of $X X_{2}$ will be