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.

Answer 2

Bond Dissociation Enthalpy Calculation Using Born-Haber Cycle 

The bond dissociation enthalpy of \( X_2 \) can be calculated using the Born-Haber cycle. The formula is: \[ \Delta H_{\text{bond dissociation}} = \Delta H_{\text{sub}} + \Delta H_{\text{i}} + \Delta H_{\text{eg}} - \Delta H_{\text{lattice}} - \Delta H_{\text{f}} \] Where: - \( \Delta H_{\text{sub}} \) is the enthalpy of sublimation, - \( \Delta H_{\text{i}} \) is the ionization energy, - \( \Delta H_{\text{eg}} \) is the electron gain enthalpy, - \( \Delta H_{\text{lattice}} \) is the lattice energy, - \( \Delta H_{\text{f}} \) is the enthalpy of formation.

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Step 1: Substitute the given values into the formula

We are given the following values: - \( \Delta H_{\text{sub}} = 100 \, \text{kJ/mol} \), - \( \Delta H_{\text{i}} = 500 \, \text{kJ/mol} \), - \( \Delta H_{\text{eg}} = -300 \, \text{kJ/mol} \), - \( \Delta H_{\text{lattice}} = -800 \, \text{kJ/mol} \), - \( \Delta H_{\text{f}} = -400 \, \text{kJ/mol} \). Substitute these values into the Born-Haber cycle equation: \[ \Delta H_{\text{bond dissociation}} = 100 + 500 - 300 - 800 - (-400) \]

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Step 2: Simplify the expression

Simplifying the equation: \[ \Delta H_{\text{bond dissociation}} = 100 + 500 - 300 - 800 + 400 \] \[ \Delta H_{\text{bond dissociation}} = 200 \, \text{kJ/mol} \]

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Final Answer:

\[ \boxed{\Delta H_{\text{bond dissociation}} = 200 \, \text{kJ/mol}} \]

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Quick Concept:

The bond dissociation enthalpy is the amount of energy required to break a bond in one mole of a molecule. It can be calculated using the Born-Haber cycle, which combines the enthalpies of sublimation, ionization, electron gain, lattice energy, and formation.