The bond dissociation energies of , , and are observed to be in the ratio . If the standard enthalpy — Thermodynamics and Thermochemistry Chemistry Question
Question
The bond dissociation energies of $X_2$, $Y_2$, and $XY$ are observed to be in the ratio $1 : 0.5 : 1$. If the standard enthalpy of formation ($\Delta_f H^\circ$) for $XY$ is $-200 \text{ kJ mol}^{-1}$, what is the exact bond dissociation energy of $X_2$?
💡 Solution & Explanation
Let the bond dissociation energies ($BE$) of $X_2$, $Y_2$, and $XY$ be $x$, $0.5x$, and $x \text{ kJ mol}^{-1}$ respectively. The formation reaction is $\frac{1}{2}X_2 + \frac{1}{2}Y_2 \rightarrow XY$. The enthalpy of formation is evaluated by: $\Delta H_f = \sum BE_{reactants} - \sum BE_{products} = [\frac{1}{2}BE(X_2) + \frac{1}{2}BE(Y_2)] - BE(XY)$. Therefore, $-200 = [\frac{1}{2}(x) + \frac{1}{2}(0.5x)] - x = 0.75x - x = -0.25x$. Solving yields $x = \frac{-200}{-0.25} = 800 \text{ kJ mol}^{-1}$.