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The bond dissociation energies of ${{\text{X}}_{2}}\text{, }{{\text{Y}}_{2}}\text{ and XY}$ are in the ratio of $1:0.5:1$ .$\Delta H$ for the formation of $XY~is\text{ }-200\text{ }kJ/mol$. The bond dissociation energy of ${{\text{X}}_{2}}$ will be :A. $800\text{ }kJ/mol$B. $200\text{ }kJ/mol$C. $400\text{ }kJ/mol$D. $100\text{ }kJ/mol$

Last updated date: 15th Jun 2024
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Hint: Use Hess’s law of constant heat summation. According to Hess’s law of constant heat summation, the enthalpy change for a reaction is the same whether the reaction takes place in one or a series of steps.

The bond dissociation energies of ${{\text{X}}_{2}}\text{, }{{\text{Y}}_{2}}\text{ and XY}$ are in the ratio of $1:0.5:1$.
Let a kJ/mol be the bond dissociation energy of ${{\text{X}}_{2}}$. The bond dissociation energy of ${{\text{Y}}_{2}}$ will also be a kJ/mol. The bond dissociation energy of $\text{XY}$will be $\text{0}\text{.5 }kJ/mol$.
Write balance chemical equations that represent bond dissociation processes.
\begin{align} & \text{XY}\to \text{X+Y }\Delta H\text{ = a kJ/mol }...\text{ }...\text{(1)} \\ & {{\text{X}}_{2}}\to 2\text{X }\Delta H\text{ = a kJ/mol }...\text{ }...\text{(2) } \\ & {{\text{Y}}_{2}}\to 2\text{Y }\Delta H\text{ = 0}\text{.5a kJ/mol }...\text{ }...\text{(3) } \\ \end{align}

Write the reaction for the formation of $\text{XY}$.
$\frac{1}{2}{{\text{X}}_{2}}+\frac{1}{2}{{\text{Y}}_{2}}\to \text{XY}...\text{ }...(4)$

Add equations (2) and (3) and divide the result with 2.
\begin{align} & \frac{{{\text{X}}_{2}}+{{\text{Y}}_{2}}\to 2\text{X+}2\text{Y }\Delta H\text{ = a kJ/mol+0}\text{.5a kJ/mol}}{2}\text{ } \\ & \frac{1}{2}{{\text{X}}_{2}}+\frac{1}{2}{{\text{Y}}_{2}}\to \text{X+Y }\Delta H\text{ = 0}\text{.75a kJ/mol }...\text{ }...\text{(5) } \\ \end{align}
Subtract equation (5) from equation (1) to obtain equation (4)
\begin{align} & \frac{1}{2}{{\text{X}}_{2}}+\frac{1}{2}{{\text{Y}}_{2}}\to \text{X+Y }\Delta H\text{ = 0}\text{.75a kJ/mol }...\text{ }...\text{(5) } \\ & -\left[ \text{XY}\to \text{X+Y }\Delta H\text{ = a kJ/mol }...\text{ }...\text{(1)} \right] \\ & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ & \frac{1}{2}{{\text{X}}_{2}}+\frac{1}{2}{{\text{Y}}_{2}}\to \text{XY}...\text{ }...(4) \\ \end{align}

Calculate the enthalpy change for reaction (4) by subtracting the enthalpy change for reaction (1) from the enthalpy change for reaction (1)
\begin{align} & \Delta H\text{ = 0}\text{.75a kJ/mol}-\text{a kJ/mol} \\ & \Delta H\text{ = }-\text{0}\text{.25a kJ/mol} \\ \end{align}

But $\Delta H$for the formation of $\text{XY}$ is $-200\text{ }kJ/mol$.
Hence,

\begin{align} & -\text{200 kJ/mol = }-\text{0}\text{.25a kJ/mol} \\ & \text{a=}\frac{-200\text{ kJ/mol}}{-0.25} \\ & \text{a=800 kJ/mol} \end{align}

Hence, the option A) $800\text{ }kJ/mol$is the correct answer.

Note:
When two reactions are added, the values of the enthalpy changes are also added. When two reactions are subtracted, the values of the enthalpy changes are also subtracted. When a reaction is divided with a number, the enthalpy change value is also divided with the same number.