$A B_{1} A_{2}$ and $B_{2}$ are diatomic molecules. If the bond enthalpies of $A_{2}, A B$ and $B_{2}$ are in the ratio $1: 1: 0.5$ and enthalpy of formation of $A B$ from $A_{2}$ and $B_{2}$ is $-100 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$. What is the bond energy of $\mathrm{A}_{2}$ :

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$A B_{1} A_{2}$ and $B_{2}$ are diatomic molecules. If the bond enthalpies of $A_{2}, A B$ and $B_{2}$ are in the ratio $1: 1: 0.5$ and enthalpy of formation of $A B$ from $A_{2}$ and $B_{2}$ is $-100 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$. What is the bond energy of $\mathrm{A}_{2}$ :

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kritika · Answer 1 · 2021-12-24T04:53:46+0000

Correct option: (d) $400 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Let bond energy of $A_{2}$ be $x$ then bond energy of $A B$ is also $x$ and bond energy of $B_{2}$ is $x / 2$.
Enthalpy of formation of $A B$ is $-100 \mathrm{KJ} /$ mole:
$$
\begin{aligned}
&\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}, \\
&\frac{1}{2} \mathrm{~A}_{2}+\frac{1}{2} \mathrm{~B}_{2} \rightarrow \mathrm{AB} ; \\
&\Delta=-100 \mathrm{KJ}
\end{aligned}
$$
$$
\begin{aligned}
&\text { or }-100=\left(\frac{x}{2}+\frac{x}{4}\right)-x \\
&\therefore-100=\frac{2 x+x-4 x}{4} \\
&\therefore x=400 \mathrm{KJ}
\end{aligned}
$$

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