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Figure 1 shows an incomplete Born-Haber cycle for the formation of caesium iodide - AQA - A-Level Chemistry - Question 1 - 2019 - Paper 1
Question 1
Figure 1 shows an incomplete Born-Haber cycle for the formation of caesium iodide. The diagram is not to scale. Cs(g) + rac{1}{2} I_2(g) Csl(s) Table 1 gives val... show full transcript
Worked Solution & Example Answer:Figure 1 shows an incomplete Born-Haber cycle for the formation of caesium iodide - AQA - A-Level Chemistry - Question 1 - 2019 - Paper 1
Step 1
Step 2
Use Figure 1 and the data in Table 1 to calculate the standard enthalpy of atomisation of iodine.
Answer
Using Hess’s law, we can write the following equation:
ΔH° = ΔH(a) + 376 + (-314) + (-585) + 337
Substituting the values from the table:
ΔH° = ΔH(a) + 376 - 314 - 585 + 337 = ΔH(a) + 107 kJ/mol
Given that the standard enthalpy change for the entire reaction when forming caesium iodide is approximately zero, the value for the enthalpy of atomisation of iodine calculates to:
ΔH(a) = -107 kJ/mol.
Step 3
The enthalpy of lattice formation for caesium iodide in Table 1 is a value obtained by experiment. Deduced what these values indicate about the bonding in caesium iodide.
Answer
The calculated lattice formation value of -582 kJ/mol suggests a strong ionic character in the bonding of caesium iodide. On the other hand, the experimental value of -585 kJ/mol indicates the real-world stability and strong interaction between Cs extsuperscript{+} and I extsuperscript{-}, showing that the bonding is almost entirely ionic, rather than having significant covalent character.
Step 4
Use data from Table 2 to show that this reaction is not feasible at 298 K.
Answer
To determine whether the reaction is feasible at 298 K, we can calculate the change in Gibbs free energy (ΔG):
ΔG = ΔH - TΔS
From Table 2:
ext{ΔS} = ((82.8 + 117) - 130) = 69.8 ext{ J K}^{-1} ext{ mol}^{-1}
Converting this to kJ:
ΔS = 0.0698 ext{ kJ K}^{-1} ext{ mol}^{-1}
Calculating ΔG:
ΔG = 337 kJ/mol - (298 K × 0.0698 kJ K⁻¹ mol⁻¹) ΔG = 337 kJ/mol - 20.8 kJ/mol = 316.2 kJ/mol
Since ΔG > 0, the reaction is not feasible at 298 K.