The bond dissociation energies of $X_2$, $Y_2$ and $XY$ are in the ratio of $1 : 0.5 : 1$. $\Delta H$ for the

Question

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 $-200 	ext{ kJ mol}^{-1}$. The bond dissociation energy of $X_2$ will be

Accepted Answer

The reaction for $\Delta_f H^\circ(XY)$ is $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) ightarrow XY(g)$. Given bond energies of $X_2, Y_2, XY$ are $X, 0.5X, X$. $\Delta H = [\frac{1}{2}(X) + \frac{1}{2}(0.5X)] - X = -200$. Solving this, $\frac{X}{2} + \frac{X}{4} - X = -200 \Rightarrow -\frac{X}{4} = -200 \Rightarrow X = 800 	ext{ kJ/mol}$.

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