In nature, the long-lived radioactive elements, Th and U, give rise to sequences of shorter-lived ra — Physical Chemistry — Kinetics Chemistry Question
First Order Rate Processes and Radioactivity
In nature, the long-lived radioactive elements, Th and U, give rise to sequences of shorter-lived radioactive isotopes. If nuclear decay occurs in closed systems, activities of daughter nuclides become equal to parent activities on a time scale related to the daughter’s half-life. Departures from this rule indicate that other processes in addition to radioactive decay are affecting the daughter’s abundance. Opportunities to identify and study the rates of these processes arise.
In water from a lake, the rate of radioactive decay of dissolved 222Rn (half-life, t½, 3.8 d) is found to be 4.2 atoms·min–1·(100 dm3)–1. All of this 222Rn is produced by decay of dissolved 226Ra (t½ 1600 y), which has an activity of 6.7 atoms min–1 (100 dm3)–1. These activities do not change measurably with time. Because every atom of 226Ra that decays produces an atom of 222Rn, the deficit in 222Rn activity implies that 222Rn is being lost from the lake by an unknown process in addition to radioactive decay.
Calculate the concentration of 222Rn in the lake in units of both atoms (100 dm3)–1 and moles dm–3.
Model Answer
Let λ222Rn be the radioactive decay constant for 222Rn in min–1 and (222Rn) be the radon concentration. As is true for any first order rate constant, λ = ln(2) / t½. Thus
λ222Rn = 0.693 / 3.8 d = 0.18 d–1 which is equal to 1.3 · 10–4 min–1.
Since the rate of decay of 222Rn is 4.2 atoms min–1 (100 dm3) –1, we obtain:
4.2 atoms min–1·(100 dm3) –1 = λ222Rn( 222Rn) = (1.3 · 10–4 min–1)(222Rn)
and (222Rn) = 3.2 · 104 atoms (100 dm3) –1. Dividing by Avagadro’s number and by 100 gives 5.5 · 10–22 moles· dm–3. (Note what remarkably small concentrations can be measured by radioactivity!)
Supposing that the unknown process obeys a first order rate law, calculate the rate constant for this process in units of min–1.
Model Answer
Since the activity of 222Rn is not changing with time, its addition to the lake by 226Ra decay must be exactly balanced by its loss through radioactive decay and the unknown first order process. Mass balance therefore requires that:
6.7 atoms min–1 (100 dm3) –1 = (λ222Rn + k)(222Rn) where k is the rate constant for the unknown process. Solving gives: k = 0.79 · 10–4 min–1
Based on periodic properties of elements, is the unknown process most likely a biological, chemical or physical process?
Model Answer
Since Rn is an inert gas, it could be affected only by physical processes at temperatures and pressures found in lakes; it could be involved in no chemical or biological reactions. (In fact, the “unknown process” is known to be diffusion of Rn through the water/air interface and escape to the atmosphere).
222Rn decays exclusively by alpha emission. Identify its radioactive decay product (including the mass).
Model Answer
Alpha emission reduces atomic number by 2 and atomic mass by 4, so the product is 218Po.