Question

The bond dissociation energies of X₂ , Y₂ and XY are in the ratio of 1 : 0·5 : 1. ΔH for the formation of XY is – 200 kJ mol^–1 . The bond dissociation energy of X₂ will be

• A. 200 kJ mol^–1
• B. 800 kJ mol^–1
• C. 100 kJ mol^–1
• D. 400 kJ mol^–1

Answer 1

The Correct Option is B

To determine the bond dissociation energy of X₂ given the ratio and the enthalpy change for the reaction forming XY, we can use the concept of Hess's law, which states that the total enthalpy change during a chemical reaction is the same regardless of the number of stages in which the reaction is effected. The reaction in question can be represented as:

1. Break X₂ into 2X:
   \( \mathrm{X_2 \rightarrow 2X} \) with dissociation energy \( E_{X_2} \)
2. Break Y₂ into 2Y:
   \( \mathrm{Y_2 \rightarrow 2Y} \) with dissociation energy \( E_{Y_2} \)
3. Formation of XY:
   \( \mathrm{X + Y \rightarrow XY} \) with formation enthalpy \( -E_{XY} \)

According to Hess's Law, the enthalpy change for the formation of XY can be given by the following equation:

\( \Delta H = \frac{E_{X_2}}{2} + \frac{E_{Y_2}}{2} - E_{XY} \)

We are given:

• The ratio of bond dissociation energies: \( E_{X_2} : E_{Y_2} : E_{XY} = 1 : 0.5 : 1 \)
• \( \Delta H = -200 \, \text{kJ mol}^{-1} \)

If we let \( E_{X_2} = x \), then \( E_{Y_2} = 0.5x \) and \( E_{XY} = x \).

Substitute these into the enthalpy equation:

\( -200 = \frac{x}{2} + \frac{0.5x}{2} - x \)

Simplifying the equation:

\( -200 = \frac{x}{2} + \frac{0.5x}{2} - x = \frac{x + 0.5x}{2} - x = \frac{1.5x}{2} - x \)

\( -200 = \frac{1.5x - 2x}{2} = \frac{-0.5x}{2} = -0.25x \)

Solving for \( x \):

\( -200 = -0.25x \Rightarrow x = \frac{200}{0.25} = 800 \)

Therefore, the bond dissociation energy of X₂ is \( 800 \, \text{kJ mol}^{-1} \), matching the given correct answer.