The experimentally determined bond dissociation energies of homonuclear molecules $A_2$, $B_2$, and

Question

The experimentally determined bond dissociation energies of homonuclear molecules $A_2$, $B_2$, and the heteronuclear molecule $AB$ exist in the ratio of $1 : 0.25 : 1$. If the standard enthalpy of formation ($\Delta_f H^\circ$) of the gas $AB$ is established as $-150 	ext{ kJ mol}^{-1}$, the true bond dissociation energy of the $A_2$ molecule must be

Professor PC Thomas Classes · Answer

(a) : Let us mathematically assign the bond dissociation energies of $A_2$, $B_2$, and $AB$ as $x$, $0.25x$, and $x 	ext{ kJ mol}^{-1}$ respectively, based strictly on the provided ratio. The defining thermochemical equation for the formation of $AB$ is $\frac{1}{2}A_{2(g)} + \frac{1}{2}B_{2(g)} ightarrow AB_{(g)}$, bearing $\Delta H = -150 	ext{ kJ mol}^{-1}$. Applying the bond energy formula: $\Delta H = \sum (	ext{B.E.}_{	ext{reactants}}) - \sum (	ext{B.E.}_{	ext{products}})$. Substituting our assigned variables: $-150 = [\frac{1}{2}(x) + \frac{1}{2}(0.25x)] - [1(x)] = [0.5x + 0.125x] - x = 0.625x - x = -0.375x$. Solving this linear equation for $x$ gives: $x = \frac{-150}{-0.375} = 400 	ext{ kJ mol}^{-1}$.