Figure 1 shows an incomplete Born–Haber cycle for the formation of caesium iodide - AQA - A-Level Chemistry - Question 1 - 2019 - Paper 1

To find the standard enthalpy of atomisation of iodine, we can start by writing the enthalpy change equation for the given reaction:

Using the values from Table 1, we can substitute:

$ext{ΔH} = 79 + 376 - 314 - 585$

Calculating gives us:

$ext{ΔH} = 79 + 376 - 314 - 585 = -434 ext{ kJ mol⁻¹}$
Thus, the standard enthalpy of atomisation of iodine is $107 ext{ kJ mol⁻¹}$.

The calculated standard enthalpy of atomisation of iodine ($107 ext{ kJ mol⁻¹}$) is significantly lower than the experimental value found using the perfect ionic model, which is −582 kJ mol⁻¹. This suggests that the bonding in caesium iodide is predominantly ionic in character, but may also have some covalent character, indicating it isn't purely ionic.

To determine if the reaction is feasible, we'll calculate the Gibbs free energy change ($ΔG$) using the relation:

$ΔG = ΔH - TΔS$

From Table 2, calculate the entropy change:

$ΔS = (82.8 + 117 - 130) = 69.8 ext{ J K⁻¹ mol⁻¹}$

Converting entropy into kJ:

$ΔS = 0.0698 ext{ kJ K⁻¹ mol⁻¹}$

Now using $T = 298 K$:

$ΔG = 337 - (298 imes 0.0698)$ $ΔG = 337 - 20.8 = 316.2 ext{ kJ mol⁻¹}$

Since $ΔG > 0$, the reaction is not feasible at 298 K.