Mercury is often used in synthetic organic and inorganic chemistry in the contex

Question

Mercury is often used in synthetic organic and inorganic chemistry in the context of high vacuum distillation lines, Schlenk lines, and for Toeppler pumps. a. What is the elemental symbol for mercury? b. What phase is mercury at room temperature? c. Why would mercury be used in those applications? d. If you had a 25 ft tall column of mercury, at what temperature would you expect the bottom to freeze? Consider that Delta H_fus = 0.548 kcal/mol for mercury, and its normal freezing point is 234.3 K. Mercury experiences a molar volumetric expansion upon melting of 0.517 mL/mol, and has a density of 13.6 g/mL. e. At what temperature would you expect the top of the column to freeze? f. How high would your column be if the entire column was frozen? State any assumptions. g. In recent years, environmental and health concerns have led to huge increases in cost and regulations of having large quantities of mercury. As a result, scientists have had to identify effective substitutes for mercury in those pieces of equipment. Suggest possible substitutes for mercury in each application and identify advantages and disadvantages for each.

Explanation / Answer

a)

Mercury has Hg, which comse from latin, Hydrargyrum, meaning, liquid silver

b)

phase of mercury at room T is LIQUID, since it is below its boiling point and above its freezing

Tfreezing = -38°C, Tboil = 356°C

c)

Mercury has a very large density compared with respect to air and even water, 13600 kg/m3 approx. This help us apply pressure and avoid large displacements of fluids (for cases such as water, we will require 13x times more length!)

Also, the cohesion of mercury is high, yet adhesion is low, which makes it perfect for flow in tubes.

d)

h = 25 ft Hg

find T for bottom to freeze...

so..

apply Clasius Clapeyron

ln(P2/P1) = dH/R*(1/T1-1/T2)

T1 = 234.3 K, P1 = 760 mm HG

1ft Hg = 304.8 mm Hg

change to 25 ft of HG --> 25*304.8 = 7620 mm Hg of Pressure

Apply vapor equation

dH = 0.548 kcal/mol --> 0.548 *4.184 = 2.29064 kJ/mol

so:

ln(7620 /760) = 2290.64/8.314 * (1/234.3 - 1/T2)

2.31 = 275.5 *  (1/234.3 - 1/T2)

1/T2 = -(2.31 /275.5 - 1/234.3)

T2 = 242.9 K = -30.1K

e)

T for tp to freeze... at the top, the Pressue is 1 atm, s assume normal T --> 234.3K, -38°C

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