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Step-by-Step Solution
Step 1: Identify the given information
• Let the bond dissociation energies of X2, Y2, and XY be denoted by
$x$, $0.5x$, and $x$ kJ mol–1 respectively (based on the ratio 1 : 0.5 : 1).
• The enthalpy change for the formation of XY from its elements is given as
$\Delta H = -200\text{ kJ mol}^{-1}$.
Step 2: Write the formation reaction
The reaction for the formation of XY from X2 and Y2 can be written
as:
$$ \frac{1}{2} X_2 + \frac{1}{2} Y_2 \to XY $$
Step 3: Express the enthalpy change in terms of bond energies
According to the relationship for enthalpy change from bond energies,
$$
\Delta H = \sum(\text{Bond energies of reactants}) \;-\; \sum(\text{Bond energies of products}).
$$
Step 4: Substitute the values into the enthalpy equation
For the reactants, we have half of the bond X2 (energy $\frac{1}{2} x$) and half of the bond Y2 (energy $\frac{1}{2} \times 0.5x = 0.25x$).
For the product, we have one XY bond of energy $x$.
Therefore,
$$
-200 = \left[ \frac{1}{2} x + \frac{1}{2} \times 0.5x \right] - [x].
$$
Step 5: Solve for x
Combine the terms in the brackets:
$$
\frac{1}{2} x + 0.25x = 0.5x + 0.25x = 0.75x.
$$
So the equation becomes:
$$
-200 = 0.75x - x = -0.25x.
$$
Rearranging,
$$
0.25x = 200 \quad \Longrightarrow \quad x = \frac{200}{0.25} = 800.
$$
Therefore, the bond dissociation energy of $X_2$ is
$$800\text{ kJ mol}^{-1}.$$
Final Answer
The bond dissociation energy of $X_2$ is
800 kJ mol–1.
