The bond dissociation enthalpy of $$X_2$$ (ΔHbond) calculated from the given data is ____________ $$kJmol^{-1}$$. (Nearest integer)

• $$M^+X^-$$(s) → $$M^+$$(g) +$$X^-$$(g)                ΔH°lattice = 800 $$kJmol^{-1}$$
• M(s) → M(g)                                                ΔH°sub = 100 $$kJmol^{-1}$$
• M(g) → $$M^+(g)+e^{_-}(g)$$                       ΔHi = 500 $$kJmol^{-1}$$
• X(g) → $$e^-(g)→X^-(g)$$                           ΔH°eg = -300 $$kJmol^{-1}$$
• M(g) + ½X2(g) → $$M^+X^-$$(s)                   ΔH°f = -400 $$kJmol^{-1}$$

[Given: ($$M^+X^-$$) is a pure ionic compound and X forms a diatomic molecule ($$X_2$$) in gaseous state]

We need to determine the bond dissociation enthalpy of $$X_2$$ from the given thermodynamic data using the Born-Haber cycle.

Given Data:

• Lattice enthalpy ($$\Delta H_{\text{lattice}}^{\circ}$$) for $$M^+X^-(s) \rightarrow M^+(g) + X^-(g) = 800 \text{ kJ mol}^{-1}$$
• Enthalpy of sublimation ($$\Delta H_{\text{sub}}^{\circ}$$) for $$M(s) \rightarrow M(g) = 100 \text{ kJ mol}^{-1}$$
• Ionization enthalpy ($$\Delta H_{i}$$) for $$M(g) \rightarrow M^+(g) + e^-(g) = 500 \text{ kJ mol}^{-1}$$
• Electron gain enthalpy ($$\Delta H_{\text{eg}}^{\circ}$$) for $$X(g) + e^-(g) \rightarrow X^-(g) = -300 \text{ kJ mol}^{-1}$$
• Enthalpy of formation ($$\Delta H_{f}^{\circ}$$) for $$M(s) + \frac{1}{2}X_2(g) \rightarrow M^+X^-(s) = -400 \text{ kJ mol}^{-1}$$

Born-Haber Cycle Step-by-Step Breakdown:

The standard enthalpy of formation of an ionic solid from its elemental states can be expressed as the algebraic sum of the individual energetic steps according to Hess's Law:

1. Sublimation of metal: $$M(s) \rightarrow M(g) \quad \Delta H = \Delta H_{\text{sub}}^{\circ} = +100 \text{ kJ mol}^{-1}$$
2. Dissociation of non-metal gas: $$\frac{1}{2}X_2(g) \rightarrow X(g) \quad \Delta H = \frac{1}{2}\Delta H_{\text{bond}}$$
3. Ionization of metal gas: $$M(g) \rightarrow M^+(g) + e^-(g) \quad \Delta H = \Delta H_{i} = +500 \text{ kJ mol}^{-1}$$
4. Electron gain by non-metal gas: $$X(g) + e^-(g) \rightarrow X^-(g) \quad \Delta H = \Delta H_{\text{eg}}^{\circ} = -300 \text{ kJ mol}^{-1}$$
5. Lattice formation: $$M^+(g) + X^-(g) \rightarrow M^+X^-(s) \quad \Delta H = -\Delta H_{\text{lattice}}^{\circ} = -800 \text{ kJ mol}^{-1}$$

Calculation:

Applying Hess's Law formula:

$$\Delta H_{f}^{\circ} = \Delta H_{\text{sub}}^{\circ} + \frac{1}{2}\Delta H_{\text{bond}} + \Delta H_{i} + \Delta H_{\text{eg}}^{\circ} - \Delta H_{\text{lattice}}^{\circ}$$

Substitute the given numerical values into the equation:

$$-400 = 100 + \frac{1}{2}\Delta H_{\text{bond}} + 500 + (-300) - 800$$

Simplify the right side of the equation:

$$-400 = \frac{1}{2}\Delta H_{\text{bond}} + (100 + 500 - 300 - 800)$$

$$-400 = \frac{1}{2}\Delta H_{\text{bond}} - 500$$

Isolate the bond dissociation term:

$$\frac{1}{2}\Delta H_{\text{bond}} = -400 + 500$$

$$\frac{1}{2}\Delta H_{\text{bond}} = 100$$

$$\Delta H_{\text{bond}} = 100 \times 2 = 200 \text{ kJ mol}^{-1}$$

Therefore, the bond dissociation enthalpy of $$X_2$$ is $$200 \text{ kJ mol}^{-1}$$.

Answer: 200