Question

Bond dissociation energies of \(XY,\;X_2\) and \(Y_2\) (all diatomic molecules) are in the ratio \(1:1:0.5\) and \(\Delta H_f\) of \(XY\) is \(-200\ \text{kJ mol}^{-1}\). The bond dissociation energy of \(X_2\) will be:

• A. \(800\ \text{kJ mol}^{-1}\)
• B. \(200\ \text{kJ mol}^{-1}\)
• C. \(300\ \text{kJ mol}^{-1}\)
• D. \(400\ \text{kJ mol}^{-1}\)

Hint:

For formation of a diatomic molecule \(XY\): \[ \Delta H_f = \frac{1}{2}D_{X_2} + \frac{1}{2}D_{Y_2} - D_{XY} \] Always break reactant bonds and form product bonds when calculating enthalpy changes.

Answer 1

The Correct Option is A

Step 1: Let the bond dissociation energies be expressed using the given ratio. Given ratio: \[ D_{XY} : D_{X_2} : D_{Y_2} = 1 : 1 : 0.5 \] Let: \[ D_{XY} = D,\quad D_{X_2} = D,\quad D_{Y_2} = 0.5D \]
Step 2: Write the expression for enthalpy of formation of \(XY\). Formation reaction: \[ \frac{1}{2}X_2 + \frac{1}{2}Y_2 \rightarrow XY \] \[ \Delta H_f = \text{Bonds broken} - \text{Bonds formed} \] \[ \Delta H_f = \left(\frac{1}{2}D_{X_2} + \frac{1}{2}D_{Y_2}\right) - D_{XY} \]
Step 3: Substitute the bond energies. \[ \Delta H_f = \left(\frac{1}{2}D + \frac{1}{2}(0.5D)\right) - D \] \[ = (0.5D + 0.25D) - D \] \[ = -0.25D \]
Step 4: Use the given value of \(\Delta H_f\). \[ -0.25D = -200 \] \[ D = 800\ \text{kJ mol}^{-1} \]
Step 5: Identify the required quantity. \[ D_{X_2} = D = 800\ \text{kJ mol}^{-1} \]
Hence, the bond dissociation energy of \(X_2\) is \[ \boxed{800\ \text{kJ mol}^{-1}} \]