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Chapter 18: Problem 37

Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half life of 5770 years. What is the rate constant (in years \(^{-1}\) ) for the decay? What fraction would remain after 11540 years?

Short Answer

Expert verified
The rate constant is approximately \(1.201 \times 10^{-4}\) years\(^{-1}\). After 11540 years, about 25% remains.

Step by step solution

01

Understanding Half-Life and Rate Constant Relationship

Radioactive decay is a first-order reaction. For first-order reactions, the relationship between the half-life \( t_{1/2} \) and the rate constant \( k \) is given by the formula \( t_{1/2} = \frac{0.693}{k} \).
02

Calculate the Rate Constant

We know the half-life \( t_{1/2} \) is 5770 years. We can rearrange the formula to solve for \( k \):\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5770} \approx 1.201 \times 10^{-4} \; \text{years}^{-1} \]
03

Understanding the Amount Remaining Formula

The expression for a first-order process to find the remaining amount of substance \( N \) after time \( t \) is \( N = N_0 \times e^{-kt} \), where \( N_0 \) is the initial amount.
04

Calculate the Fraction Remaining After 11540 Years

Given \( t = 11540 \) years and \( k = 1.201 \times 10^{-4} \; \text{years}^{-1} \), substitute these into the formula \( N = N_0 \times e^{-kt} \) to find the fraction remaining:\[ N/N_0 = e^{-1.201 \times 10^{-4} \times 11540} \approx e^{-1.3858} \approx 0.250 \]

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Key Concepts

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

Half-life calculation
The concept of half-life is central to understanding radioactive decay. The half-life of a radioactive substance is the time it takes for half of the material to decay. In our example, the half-life is 5770 years.
This means that every 5770 years, half of the carbon in the wood sample will have decayed into another element.
Half-life can be used to determine the rate at which a substance decays, a quantity known as the rate constant.
  • Half-life is constant and does not change over time.
  • It provides a straightforward way to compare the stability of different radioactive isotopes.
Understanding the half-life allows us to determine many important characteristics of radioactive substances, such as how long they might remain active or how quickly they convert into new elements.
First-order reactions
Radioactive decay follows the rules of a first-order reaction. This type of reaction has a rate that depends on the concentration of only one reactant.
This means the reaction rate is directly proportional to the remaining quantity of the substance.
In our exercise, the decay of radioactive carbon is considered first-order, making its behavior predictable and mathematically simple to model.
  • The rate law for a first-order reaction is expressed as: \( \frac{d[A]}{dt} = -k[A] \).
  • Here, \([A]\) represents the concentration of the substance, and \(k\) is the rate constant.
First-order reactions are common in various areas of chemistry and physics because they simplify how changes in concentration affect reaction rates, relying solely on the current state and not past concentrations.
Rate constant determination
The rate constant \(k\) is an essential element in understanding decay processes. It offers insight into the speed of the reaction.
For first-order processes, the rate constant can be calculated if we know the half-life of the substance.
  • The relationship is \( t_{1/2} = \frac{0.693}{k} \).
  • In our example, rearranging this formula gives us \( k = \frac{0.693}{5770} \), which equals approximately \(1.201 \times 10^{-4} \text{ years}^{-1} \).
This rate constant is important because it allows us to predict how much of the radioactive material will remain after a given period, as seen in the formula for exponential decay.
Exponential decay formula
The exponential decay formula is critical for calculating how much of a radioactive substance remains after a certain period. This formula links the initial amount of a substance to the remaining amount after time \( t \).
The formula is stated as: \( N = N_0 \times e^{-kt} \). Here, \( N_0 \) is the initial quantity, \( N \) is the remaining quantity after time \( t \), and \( k \) is the rate constant.
  • This formula shows that the quantity decreases exponentially over time.
  • For our exercise, using the formula with \( k = 1.201 \times 10^{-4} \text{ years}^{-1} \) and \( t = 11540 \text{ years} \) gives the remaining fraction of approximately 0.250.
This formula is powerful because it can predict the substance's behavior over any timeframe, making it an essential tool in fields such as archaeology, geology, and environmental science.

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Most popular questions from this chapter

A nuclear explosion has taken place leading to increase in concentration of \(C^{14}\) in nearby areas. \(C^{14}\) concentration is \(C_{1}\) in nearby areas and \(C_{2}\) in areas far away. If the age of the fossil is determined to be \(T_{1}\) and \(T_{2}\) at the respective places then (a) The age of the fossil will increase at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (b) The age of the fossil will decrease at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (c) The age of fossil will be determined to be same (d) \(\frac{T_{1}}{T_{2}}=\frac{C_{1}}{C_{2}}\)

The decomposition reaction\(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After \(Y \times 10^{3} \mathrm{~s}\), the pressure inside the cylinder is found to be \(1.45 \mathrm{~atm}\). If the rate constant of the reaction is \(5 \times 10^{-4} \mathrm{~s}^{-1}\), assuming ideal gas behaviour, the value of \(Y\) is [Adv. 2019]

The rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\), is \(3.0 \times 10^{-5}\) \(\mathrm{sec}^{-1}\). If the rate is \(2.40 \times 10^{-5}\) mol litre \(^{-1} \mathrm{sec}^{-1}\), then the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (in mol litre \(^{-1}\) ) is [2000S] (a) \(1.4\) (b) \(1.2\) (c) \(0.04\) (d) \(0.8\)

Two reactions (i) \(A \rightarrow\) products, (ii) \(B \rightarrow\) products, follows first order kinetics. The rate of the reaction: \((i)\) is doubled when the temperature is raised from \(300 \mathrm{~K}\) to \(310 \mathrm{~K}\). The half life for this reaction at \(310 \mathrm{~K}\) is 30 minutes. At the same temperature \(B\) decomposes twice as fast as \(A\). If the energy of activation for the reaction, (ii) is half that of reaction (i), calculate the rate constant of the reaction (ii) at \(300 \mathrm{~K}\).

The gas phase decomposition of dimethyl ether follows first order kinetics. $$ \mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) $$ The reaction is carried out in a constant volume container at \(500^{\circ} \mathrm{C}\) and has a half life of \(14.5\) minutes. Initially, only dimethyl ether is present at a pressure of \(0.40\) atmosphere. What is the total pressure of the system after 12 minutes? Assume ideal gas behaviour. [1993 - 4 Marks]
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