A2 + B2 2AB. kJ mol–1
AB, A2 and B2 are diatomic molecule. If the bond enthalpies of A2, B2 and AB are in the ratio 1 : 0.5 : 1, then the bond enthalpy of A2 is ____. kJ mol–1 (Nearest integer)
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Solution
$\begin{array}{ll}\text { Sol. } & A_2+B_2 \rightarrow 2 A B \cdot \Delta H_{\mathrm{f}}^{\circ}=-200 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Rightarrow \Delta H_{\mathrm{f}}^{\circ}(A B)=-200 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{array}$
$\therefore \Delta \mathrm{H}_{\mathrm{R}}^{\circ}$ for reaction $\mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB}$ is $=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Given: Bond Enthalpy of $\mathrm{A}_2, \mathrm{~B}_2$ and AB is $1 \sim 0.5 ; 1$ Assuming bond enthalpy of $\mathrm{A}_2$ be $\times \mathrm{kJ} \mathrm{mol}^{-1}$
∴ Bond enthalpy $\mathrm{B}_2=0.5 \times \mathrm{kJ} \mathrm{mol}^{-1}$
∴ Bond enthalpy $\mathrm{AB}=(\mathrm{x}) \mathrm{kJ} \mathrm{mol}^{-1}$
$$
\begin{aligned}
& -400=x+0.5 x-2 x \\
& -400=-0.5 x
\end{aligned}
$$
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