Question

The bond dissociation energies of \( X_2, Y_2, \) and \( XY \) are in the ratio of 1:0.5:1. If \( \Delta H \) for the formation of \( XY \) is -200 kJ mol\(^{-1}\), what is the bond dissociation energy of \( X_2 \)?

• A. \( 200 \) kJ mol\(^{-1} \)
• B. \( 100 \) kJ mol\(^{-1} \)
• C. \( 400 \) kJ mol\(^{-1} \)
• D. \( 800 \) kJ mol\(^{-1} \)

Hint:

Use the enthalpy equation: \( \Delta H = \sum {Bond Energy (Reactants)} - \sum {Bond Energy (Products)} \).

Answer 1

The Correct Option is D

Step 1: Bond dissociation energy and formation enthalpy.
The bond dissociation energies of \( X_2 \), \( Y_2 \), and \( XY \) are in the ratio of 1:0.5:1, respectively. Let's denote the bond dissociation energy of \( X_2 \) as \( D(X_2) = x \, {kJ/mol} \).
- The bond dissociation energy of \( Y_2 \), \( D(Y_2) = 0.5x \).
- The bond dissociation energy of \( XY \), \( D(XY) = x \, {kJ/mol} \).
Step 2: Applying the given formation enthalpy.
The enthalpy change for the formation of \( XY \) from \( X_2 \) and \( Y_2 \) is given as: \[ \Delta H = D(X_2) + D(Y_2) - D(XY) \] Substituting the known values: \[ -200 = x + 0.5x - x \] \[ -200 = 0.5x \] Solving for \( x \): \[ x = \frac{-200}{0.5} = -400 \, {kJ/mol} \] Thus, \( D(X_2) = 800 \, {kJ/mol} \).