Molecules such as H2, N2, O2, CO2, and CH4 in Problem 5 are formed through chemical bonding of atoms — Physical Chemistry — Kinetics Chemistry Question
Discovery of the noble gases
Molecules such as H2, N2, O2, CO2, and CH4 in Problem 5 are formed through chemical bonding of atoms. Even though valency was known in the 19th century, the underlying principle behind chemical bonding had not been understood for a long time. Ironically, the discovery of the noble gases with practically zero chemical reactivity provided a clue as to why elements other than the noble gases combine chemically. In 1882, Rayleigh decided to redetermine accurately gas densities in order to test Prout's hypothesis.
What is Prout's hypothesis? What evidence did he use to support his hypothesis? (Search the Internet or other sources.)
Model Answer
In 1816 Prout published a hypothesis that all matter is composed ultimately of hydrogen. (Later, Harlow Shapley, an eminent astronomer, said that if God did create the world by a word, the word would have been hydrogen.) Prout cited as evidence the fact that the specific gravities of gaseous elements appeared to be whole–number multiples of the value for hydrogen.
To remove oxygen and prepare pure nitrogen, Rayleigh used a method recommended by Ramsay. Air was bubbled through liquid ammonia and was passed through a tube containing copper at red heat where the oxygen of the air was consumed by hydrogen of the ammonia. Excess ammonia was removed with sulfuric acid. Water was also removed. The copper served to increase the surface area and to act as an indicator. As long as the copper remained bright, one could tell that the ammonia had done its work.
Write a balanced equation for the consumption of oxygen in air by hydrogen from ammonia. Assume that air is 78 % of nitrogen, 21 % of oxygen, and 1 % of argon by volume (unknown to Rayleigh) and show nitrogen and argon from the air in your equation.
Model Answer
28 NH3 + 21 O2 + 78 N2 + Ar → 92 N2 + 42 H2O + Ar
Calculate the molecular mass of nitrogen one would get from the density measurement of nitrogen prepared as above. Note that argon in the sample, initially unknown to Rayleigh, did contribute to the measured density.
(Atomic masses: Ar(N) = 14.0067, and Ar(Ar) = 39.948).
Model Answer
((92 × 2 × 14.0067) + 39.948) / 93 = 28.142
Rayleigh also prepared nitrogen by passing air directly over red–hot copper.
Write a balanced equation for the removal of oxygen from air by red–hot copper. Again show nitrogen and argon from the air in your equation.
Model Answer
78 N2 + 21 O2 + Ar + 42 Cu → 78 N2 + 42 CuO + Ar
Calculate the molecular mass of nitrogen one would get from the density measurement of the nitrogen prepared by the second method.
Model Answer
((78 × 2 × 14.0067) + 39.948) / 79 = 28.164
To Rayleigh’s surprise, the densities obtained by the two methods differed by a thousandth part – a difference small but reproducible. Verify the difference from your answers in 6.3 and 6.5.
Model Answer
= 1.0008 (about 0.1%)
To magnify this discrepancy, Rayleigh used pure oxygen instead of air in the ammonia method. How would this change the discrepancy?
Model Answer
4 NH3 + 3 O2 → 2 N2 + 6 H2O
The relative molecular mass of pure nitrogen = 2 × 14.0067 = 28.013
= 1.0054
The discrepancy would increase about 7–fold (0.0054 / 0.0008).
Nitrogen as well as oxygen in the air was removed by the reaction with heated Mg (more reactive than copper). Then a new gas occupying about 1 % of air was isolated. The density of the new gas was about x–times that of air. Calculate x.
Model Answer
40 : 29 = 1.4
A previously unseen line spectrum was observed from this new gas separated from 5 cm3 of air. The most remarkable feature of the gas was the ratio of its specific heats (Cp/Cv), which proved to be 5/3, i. e. the highest possible,. The observation showed that the whole of the molecular motion was (*). Thus, argon is a monatomic gas.
Choose (*): (1) electronic (2) vibrational (3) rotational (4) translational
Model Answer
5 R : 3 R = 1.67 translational
Calculate the mass of argon in a 10 m×10 m×10 m hall at STP.
Model Answer
Volume of air = 1000 m3 = 1.0×10^6 dm3
4.5×10^4 mol of air
Mass of argon = 4.5×10^4 × 0.01 × 40 = 1.8×10^4 g = 18 kg
In 1894, Rayleigh and Ramsay announced the discovery of Ar. Discovery of other noble gases (He, Ne, Kr, Xe) followed and a new group was added to the periodic table. As a result, Rayleigh and Ramsay received the Nobel Prizes in physics and in chemistry, respectively, in 1904.
Element names sometimes have Greek or Latin origin and provide clues as to their properties or means of discovery. Match the element name with its meaning.
Helium • • new
Neon • • stranger
Argon • • lazy
Krypton • • hidden
Xenon • • sun
Model Answer
helium – sun
neon – new
argon – lazy
krypton – hidden
xenon – stranger