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Step-by-Step Detailed Solution
Step 1: Write the reaction for forming 2 moles of XY from X₂ and Y₂
We are given that the reaction is:
$X_2 + Y_2 \rightarrow 2XY$
The enthalpy change for this reaction ($\Delta H$) can be found by using the enthalpy of formation of XY.
Step 2: Calculate the total enthalpy change for the reaction
The enthalpy of formation of 1 mole of XY is given as $-200\,\text{kJ mol}^{-1}$. For producing 2 moles of XY, the total enthalpy change is:
$\Delta H = 2 \times (-200)\,\text{kJ mol}^{-1} = -400\,\text{kJ mol}^{-1}.$
Step 3: Express the reaction enthalpy in terms of bond dissociation energies
Let $x$ be the bond dissociation energy of $X_2$. According to the ratio provided in the question:
Bond dissociation energy of $X_2 = x$
Bond dissociation energy of $Y_2 = 0.5x$
Bond dissociation energy of $XY = x$ (given the ratio is 1 : 1 : 0.5 for XY : X₂ : Y₂, and we assume each is proportional to $x$; hence XY has the same bond energy as X₂ or we can let XY = x, X₂ = x, Y₂ = 0.5x).
The enthalpy change of the reaction in terms of bond enthalpies is:
$\Delta H = \big[(\text{bond dissociation energy of } X_2) + (\text{bond dissociation energy of } Y_2)\big] - 2 \times (\text{bond dissociation energy of } XY).$
Substituting the values, we get:
$\Delta H = x + 0.5x - 2(x) = (x + 0.5x - 2x).$
So,
$\Delta H = -0.5x.$
Step 4: Solve for x using the known enthalpy change
We have $\Delta H = -400\,\text{kJ mol}^{-1}$ from the overall reaction. Hence,
$-400 = -0.5x.$
Solving for $x$:
$x = \frac{400}{0.5} = 800\,\text{kJ mol}^{-1}.$
Step 5: State the final answer
The bond dissociation energy of $X_2$ is
800 kJ mol-1.
