Q57 / 180 · KCET 2021 · Chemistry
Bond enthalpies of $A_{2}, B_{2}$ and $A B$ are in the ratio $2: 1: 2$. If bond enthalpy of formation of $A B$ is $-100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The bond enthalpy of $B_{2}$ is
- A$100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerC$200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$150 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
Given,
$(\mathrm{BE})_{A_{2}}:(\mathrm{BE})_{B_{2}}:(\mathrm{BE})_{A B}=2: 1: 2$
where, $B E=$ Bond enthalpy
$\left(\Delta H_{f}\right)_{A B}=-100 \mathrm{~kJ}$
Let the BE of $A_{2}, B_{2}$ and $A B$ be $2 x, x$ and $2 x$ respectively.
As we know, $\Delta_{f} H^{\circ}=\Sigma \Delta_{\text {diss }} H_{R}^{\circ}-\Sigma \Delta_{\text {diss }} H_{P}^{\circ}$ where, $\quad \Lambda_{\text {diss }} H^{\circ}=$ Enthalpy of dissociation
$\begin{aligned}
\frac{1}{2} A_{2} &+\frac{1}{2} B_{2} \rightarrow A B \\
\therefore \quad-100 &=\left(\frac{1}{2} 2 x+\frac{1}{2} x\right)-x \\
x &=200
\end{aligned}$
$\therefore$ Bond enthalpy of $B_{2}$ is $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Hence, the correct answer is option: 'C'.