Question

The bond dissociation energies of $X_2$, $Y_2$ and $XY$ are in the ratio of 1 : 0.5 : 1. $△H$ for the formation of $XY$ is –200 kJ $mo{l}^{–1}$. The bond dissociation energy of $X_2$ will be

• A. 800 kJ $mo{l}^{–1}$
• B. 100 kJ $mo{l}^{–1}$
• C. 200 kJ $mo{l}^{–1}$
• D. 400 kJ $mo{l}^{–1}$

Answer 1

The correct option is A 800 kJ $mo{l}^{–1}$

The reaction for ${△}_r{H}^o$$\left(XY\right)$
$\frac{1}{2}$ $X_2$(g) + $\frac{1}{2}$ $Y_2$(g) $⟶$ $XY$(g)

given bond energies of $X_2$, $Y_2$ and $XY$ are P, P/2 and P respctively, so enthalpy of reaction can be written as

$△H$ = bond energies of reactant - bond energies of product

$△H$= $(P/2+P/4)$-$P$= -200
P= 800 kJ/mol