Question

Calculate the value of -U/100, for AB(s), from following data of Born-Haber's cycle. [where U is lattice energy in kJ/mol]

• A. 8.13
• B. 7.37
• C. -8.13
• D. -7.37

Answer 1

Answer: (A)

8.13

Answer 2

The Born-Haber cycle relates the enthalpy of formation ($\Delta H_f$) of an ionic compound to various thermodynamic quantities. According to Hess's Law, the sum of the enthalpy changes for the steps in the cycle equals the overall enthalpy change.

The steps involved in the formation of AB(s) are:

1. Atomization of A: A(s) $\rightarrow$ A(g)
2. Ionization of A: A(g) $\rightarrow$ A$^{+}$(g) + e$^-$ (Ionization Energy, IE)
3. Dissociation of B$_2$: $\frac{1}{2}$B$_2$(g) $\rightarrow$ B(g) (Dissociation Enthalpy, $\Delta H_D$)
4. Electron Affinity of B: B(g) + e$^-$ $\rightarrow$ B$^{-}$(g) (Electron Affinity, EA)
5. Lattice Formation: A$^{+}$(g) + B$^{-}$(g) $\rightarrow$ AB(s) (Lattice Energy, U)

The overall enthalpy of formation is given by: $\Delta H_f = (\Delta H_{atom, A} + IE) + \Delta H_D + EA + U$

From the provided Born-Haber cycle diagram:

• Enthalpy of formation, $\Delta H_f = -424$ kJ/mol.
• Energy to get A into gaseous ionic state A$^{+}$(g) from A(s) = 495 kJ/mol (This represents $\Delta H_{atom, A} + IE$).
• Dissociation Enthalpy, $\Delta H_D = 242$ kJ/mol.
• Electron Affinity, EA = -348 kJ/mol.

Substituting the values into the equation: $-424 \text{ kJ/mol} = 495 \text{ kJ/mol} + 242 \text{ kJ/mol} + (-348 \text{ kJ/mol}) + U$ $-424 = 495 + 242 - 348 + U$ $-424 = 737 - 348 + U$ $-424 = 389 + U$

Solving for the Lattice Energy, U: $U = -424 - 389$ $U = -813$ kJ/mol

The question asks to calculate the value of -U/100. Value = $-(-813 \text{ kJ/mol}) / 100$ Value = $813 / 100$ Value = $8.13$

The numerical value is $8.13$, which is already in two decimal places.