Notifications
Clear all
1
03/10/2021 1:18 pm
Topic starter
What is the normal boiling point of mercury?
Given : ΔH°f (Hg, l) = 0; S°(Hg, l) = 77.4 J/Kmol ΔH°f (Hg, g) = 60.8 kJ/mol; S°(Hg, g) = 174.4 J/K–mol
(a) 624.8 K
(b) 626.8 K
(c) 636.8 K
(d) None of these
1 Answer
1
03/10/2021 1:23 pm
Correct answer: (b) 626.8 K
Explanation:
Hg(l) ⇌ Hg(g),
ΔRS° = 174.4 - 77.4 = 97 J/K-mol
∵ ΔG° = ΔH° - TΔS° = 0
T = \(\frac{\Delta H°}{\Delta S°}\)
= \(\frac{60.8 \times 1000}{97}\)
= 626.8 K