UPSC Civil Services (Mains), Chemistry optional · 2024 · Paper I
Q1(c). Normal boiling point of mercury
Question
Use the following data to determine the normal boiling point of mercury. What assumptions must you make in order to do the calculations? Hg(l): ΔH°f = 0, S° = 77.4 J K⁻¹ mol⁻¹; Hg(g): ΔH°f = 60.78 kJ mol⁻¹, S° = 174.7 J K⁻¹ mol⁻¹.
Answer in brief
Tb = ΔH°/ΔS° = 60 780/97.3 = 625 K (352 °C), assuming ΔH° and ΔS° stay at their 298 K values.
Key points
- At the normal boiling point the liquid is in equilibrium with its vapour at 1 atm, so ΔG° = ΔH° − TΔS° = 0.
- ΔvapH° = 60.78 kJ mol⁻¹ and ΔvapS° = 174.7 − 77.4 = 97.3 J K⁻¹ mol⁻¹, giving Tb = 625 K.
- Assumption: ΔH° and ΔS° keep their 298 K values up to the boiling point (ΔCp = 0).
- Assumption: mercury vapour behaves as an ideal gas, 1 atm is taken as 1 bar, and the effect of pressure on the liquid is neglected.
- The measured value is 630 K (357 °C), so the estimate is within 1 %.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversnormal boiling pointGibbs energyentropy of vaporisationmercuryClausius–Clapeyron equation
All parts of Question 1
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- 1(e) Temperature at which the uncatalysed reaction matches catalase at 20 °C10 marks