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

The given problem involves understanding the bond dissociation energies and calculating the energy based on the given reaction enthalpy. We have the dissociation energies of X₂, Y₂, and XY in the ratio 1 : 0.5 : 1. Additionally, the enthalpy change for the formation of XY (\( \Delta H \)) is given as \( -200 \,\text{kJ mol}^{-1} \). We need to find out the bond dissociation energy of X₂.

Let's denote the bond dissociation energies of X₂, Y₂, and XY as D_X-X, D_Y-Y, and D_X-Y respectively. According to the ratio given, we have:

• D_X-X : D_Y-Y : D_X-Y = 1 : 0.5 : 1

Assume D_X-X = \( x \) \( \text{kJ mol}^{-1} \). Then, according to the ratio:

• D_Y-Y = \( 0.5x \)
• D_X-Y = \( x \)

The reaction for the formation of XY can be written as:

\( \mathrm{X_2 + Y_2 \rightarrow 2XY} \)

The enthalpy change (\( \Delta H \)) for this reaction is given as the difference between the energy required to break the bonds and the energy released by the formation of bonds:

\( \Delta H = [D_{X-X} + D_{Y-Y}] - 2D_{X-Y} \)

Substituting the given values and \( \Delta H = -200 \,\text{kJ mol}^{-1} \), we get:

\( x + 0.5x - 2x = -200 \)

This equation simplifies to:

\( 1.5x - 2x = -200 \)

Which further simplifies to:

\( -0.5x = -200 \)

Solving this equation, we find:

\( x = 400 \,\text{kJ mol}^{-1} \)

Thus, the bond dissociation energy of X₂ (\( D_{X-X} \)) is:

\( 400 \,\text{kJ mol}^{-1} \)

Therefore, the correct answer is 400 kJ mol^-1.