Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces: $$_{15}^{32} \mathrm{P} \longrightarrow_{16}^{32} \mathrm{S}+\mathrm{e}^{-}$$ rate \(=4.85 \times 10^{-2} \mathrm{day}^{-1}\left[^{32} \mathrm{P}\right]\) What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of \(0.0033 \mathrm{M}\) ?

Short Answer

Expert verified
The instantaneous rate of production of electrons is 1.60 x 10^(-4) M/day.

Step by step solution

01

Identify given information

The provided exercise gives us the decay reaction of phosphorus-32 to sulfur-32 emitting an electron and a rate expression for the production of electrons. The rate of the reaction is given as a first-order decay process with a rate constant of 4.85 × 10^(-2) day^(-1) and the initial concentration of phosphorus-32 as 0.0033 M.
02

Understand the rate equation

The rate of production of electrons from the decay of phosphorus-32 is given by the rate equation rate = k[^(32)P]. In this equation, k is the rate constant and [^(32)P] is the concentration of phosphorus-32.
03

Calculate the instantaneous rate of electron production

Using the rate equation, the instantaneous rate of electron production can be calculated by substituting the values for the rate constant (k = 4.85 x 10^(-2) day^(-1)) and the concentration of phosphorus-32 ([^(32)P] = 0.0033 M) into the equation.
04

Perform the calculation

rate = k[^(32)P] = (4.85 x 10^(-2) day^(-1)) x (0.0033 M). Multiply the rate constant by the concentration to find the rate at which electrons are being produced.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Phosphorus
Understanding the role of radioactive phosphorus is essential when delving into biochemical studies. Phosphorus, being a crucial element in various biochemical molecules such as DNA and RNA, offers vital insights when labeled with a radioactive isotope. When phosphorus-32, a radioactive isotope, decays, it emits beta particles (electrons), allowing researchers to trace the pathways and locations of biochemical reactions. This nuclear property provides a unique window into the dynamic processes occurring within living organisms. It’s especially useful because it’s both detectable and measurable, enabling precise assessments of biochemical mechanisms.
Biochemical Reaction Mechanisms
The biochemical reaction mechanisms in cells are complex and highly regulated processes that govern life at the molecular level. To understand these mechanisms, scientists often rely on tracers like radioactive phosphorus. By integrating into biological molecules without altering their function, these tracers can reveal the inner workings of cellular activities. The path of the tracer through metabolic reactions indicates how substances are synthesized, broken down, or transported within cells, thereby elucidating details of the biochemical pathways involved.
First-Order Decay Process
A first-order decay process is one where the rate of decay is directly proportional to the amount of the substance that remains. In mathematical terms, the rate can be described by the equation:
\( rate = k[^\text{32}P] \),
where \( k \) is the rate constant and \( [^\text{32}P] \) is the concentration of the radioactive substance, in this case, phosphorus-32. This kind of decay signifies that the half-life of the substance – the time it takes for half of the material to decay – remains constant irrespective of how much of the substance is present. This predictability makes it a foundational concept in understanding radioactive substances and their behavior over time in both natural and experimental settings.
Instantaneous Rate of Electron Production
To determine the instantaneous rate of electron production, one must capture the rate at a specific moment, which reflects the activity of the radioactive decay at that instance. In our case, with phosphorus-32, this instantaneous rate reflects how quickly electrons are produced as the phosphorus undergoes radioactive decay. By measuring this rate, researchers can infer the amount of phosphorus-32 present at the time of measurement. This data is critical in biomedical research where it’s important to know the immediate rate of change, rather than average rates, to understand the dynamics of the biochemical reactions being studied.

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Most popular questions from this chapter

The annual production of \(\mathrm{HNO}_{3}\) in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel. (a) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)\) The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in \(\mathrm{O}_{2},\) what is the rate of formation of \(\mathrm{NO}_{2}\) when the oxygen concentration is \(0.50 \mathrm{M}\) and the nitric oxide concentration is \(0.75 \mathrm{M}\) ? The rate constant for the reaction is \(5.8 \times\) \(10^{-6} \mathrm{L}^{2} \mathrm{mol}^{-2} \mathrm{s}^{-1}\).

Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are \(6 \mathrm{h}\) and \(73 \mathrm{h}\), respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?

Write the rate law for each of the following elementary reactions: (a) \(\mathrm{O}_{3} \stackrel{\text { sunlight }}{\longrightarrow} \mathrm{O}_{2}+\mathrm{O}\) (b) \(\mathrm{O}_{3}+\mathrm{Cl} \longrightarrow \mathrm{O}_{2}+\mathrm{ClO}\) (c) \(\mathrm{ClO}+\mathrm{O} \longrightarrow \mathrm{Cl}+\mathrm{O}_{2}\) (d) \(\mathrm{O}_{3}+\mathrm{NO} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}\) (e) \(\mathrm{NO}_{2}+\mathrm{O} \longrightarrow \mathrm{NO}+\mathrm{O}_{2}\)

For the reaction \(Q \longrightarrow W+X,\) the following data were obtained at \(30^{\circ} \mathrm{C}:\) $$\begin{array}{|c|c|c|c|} \hline[Q]_{\text {initial }}(M) & 0.170 & 0.212 & 0.357 \\ \hline \text { Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}\right)\) } & 6.68 \times 10^{-3} & 1.04 \times 10^{-2} & 2.94 \times 10^{-2} \\ \hline \end{array}$$ (a) What is the order of the reaction with respect to \([Q]\), and what is the rate law? (b) What is the rate constant?

A study of the rate of the reaction represented as \(2 A \longrightarrow B\) gave the following data: $$\begin{array}{|c|c|c|c|c|c|c|c|} \hline \text { Time (s) } & 0.0 & 5.0 & 10.0 & 15.0 & 20.0 & 25.0 & 35.0 \\ \hline \text { \([A](M)\) } & 1.00 & 0.775 & 0.625 & 0.465 & 0.360 & 0.285 & 0.230 \\ \hline \end{array}$$ (a) Determine the average rate of disappearance of \(A\) between \(0.0 \mathrm{s}\) and \(10.0 \mathrm{s}\), and between \(10.0 \mathrm{s}\) and \(20.0 \mathrm{s}\). (b) Estimate the instantaneous rate of disappearance of \(A\) at 15.0 s from a graph of time versus \([A]\). What are the units of this rate? (c) Use the rates found in parts (a) and (b) to determine the average rate of formation of \(B\) between 0.00 s and 10.0 s, and the instantaneous rate of formation of \(B\) at 15.0 s.

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