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

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.

& \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{aligned}$$ Write the reaction for the formation of $$\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{aligned} & \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{aligned}$$ Subtract equation (5) from equation (1) to obtain equation (4) $$\begin{aligned} & \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{aligned}$$ Calculate the enthalpy change for reaction (4) by subtracting the enthalpy change for reaction (1) from the enthalpy change for reaction (1) $$\begin{aligned} & \Delta H\text{ = 0}\text{.75a kJ/mol}-\text{a kJ/mol} \\\ & \Delta H\text{ = }-\text{0}\text{.25a kJ/mol} \\\ \end{aligned}$$ But $$\Delta H$$for the formation of $$\text{XY}$$ is $$-200\text{ }kJ/mol$$. Hence, $$\begin{aligned} & -\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{aligned}$$ **Hence, the option A) $$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.