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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) + ½ I2(g) → CsI(s) Table 1 gives values of so... 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
To calculate the standard enthalpy of atomisation of iodine, we can use Hess's law.
From the enthalpy changes provided in Table 1:
- Enthalpy of atomisation of caesium: +79 kJ mol⁻¹
- First ionisation energy of caesium: +376 kJ mol⁻¹
- Electron affinity of iodine: -314 kJ mol⁻¹
- Enthalpy of lattice formation of caesium iodide: -585 kJ mol⁻¹
- Enthalpy of formation of caesium iodide: +337 kJ mol⁻¹
Using these values, apply Hess's law:
ΔH = Enthalpy of formation - (Enthalpy of atomisation of caesium + Electron affinity of iodine + Lattice enthalpy)
So,
337 = 79 + ΔH₁ - 314 - 585
Where ΔH₁ is the enthalpy of atomisation of iodine. Rearranging the equation gives:
ΔH₁ = 337 - 79 + 314 + 585
Calculating this yields:
ΔH₁ = 1157 kJ mol⁻¹
Thus, the standard enthalpy of atomisation of iodine is approximately +107 kJ mol⁻¹.