Question

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. \[100\text{ }kJ/mol\]

Answer 1

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.

Complete step by step answer:
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.