Question

The bond dissociation enthalpy of \( X_2 \) calculated from the given data is ---- kJ mol\(^{-1}\) (nearest integer). % Given Data: \[ M(s) + X(s) \rightarrow M^+(g) + X^-(g) \quad \Delta H_{\text{lattice}} = 800 \, \text{kJ/mol} \] \[ M(s) \rightarrow M(g) \quad \Delta H_{\text{sub}} = 100 \, \text{kJ/mol} \] \[ M(g) \rightarrow M^+(g) + e^-(g) \quad \Delta H_{\text{i}} = 500 \, \text{kJ/mol} \] \[ X(g) + e^-(g) \rightarrow X^-(g) \quad \Delta H_{\text{eg}} = -300 \, \text{kJ/mol} \] \[ M(s) + X_2(g) \rightarrow M^+X^-(s) \quad \Delta H_{\text{f}} = -400 \, \text{kJ/mol} \]

Hint:

To calculate bond dissociation enthalpy, use the Born-Haber cycle, which involves lattice enthalpy, ionization, electron affinity, and formation enthalpies.

Answer 1

Correct Answer: 200

The bond dissociation enthalpy of \( X_2 \) can be calculated using the Born-Haber cycle and the enthalpies provided: \[ \Delta H_{\text{bond dissociation}} = \Delta H_{\text{sub}} + \Delta H_{\text{i}} + \Delta H_{\text{eg}} - \Delta H_{\text{lattice}} - \Delta H_{\text{f}}. \] Substituting the values: \[ \Delta H_{\text{bond dissociation}} = 100 + 500 - 300 - 800 - (-400) = 200 \, \text{kJ/mol}. \] Thus, the bond dissociation enthalpy is 200 kJ/mol.