The bond dissociation energies of X_2,Y_2 and | vedclass

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(C) Let the bond dissociation energies ($B$.$E$.) of $X_2$,$Y_2$,and $XY$ be $x \ kJ \ mol^{-1}$,$0.5x \ kJ \ mol^{-1}$,and $x \ kJ \ mol^{-1}$ respec

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(C) Let the bond dissociation energies ($B$.$E$.) of $X_2$,$Y_2$,and $XY$ be $x \ kJ \ mol^{-1}$,$0.5x \ kJ \ mol^{-1}$,and $x \ kJ \ mol^{-1}$ respectively.The chemical equation for the formation of $XY$ is: $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) ightarrow XY(g)$.The enthalpy of reaction is given by: $\Delta H = \Sigma(B.E.)_{	ext{reactants}} - \Sigma(B.E.)_{	ext{products}}$.Substituting the values: $-200 = [\frac{1}{2} 	imes (x) + \frac{1}{2} 	imes (0.5x)] - [1 	imes (x)]$.$-200 = [0.5x + 0.25x] - x$.$-200 = 0.75x - x$.$-200 = -0.25x$.$x = \frac{200}{0.25} = 800 \ kJ \ mol^{-1}$.Therefore,the bond dissociation energy of $X_2$ is $800 \ kJ \ mol^{-1}$.

The bond dissociation energies of $X_2$,$Y_2$ and $XY$ are in the ratio of $1 : 0.5 : 1$. $\Delta H$ for the formation of $XY$ is $-200 \ kJ \ mol^{-1}$. The bond dissociation energy of $X_2$ will be $...... \ kJ \ mol^{-1}$.

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