All radioactive decay processes follow first-order kinetics. The half-life of the radioactive isotope tritium \(\left({ }^{3} \mathrm{H}\right.\), or \(\left.\mathrm{T}\right)\) is \(12.3\) years. How much of a \(25.0\)-mg sample of tritium would remain after \(10.9\) years?

Short Answer

Expert verified
Approximately 13.4 mg of tritium would remain after 10.9 years.

Step by step solution

01

Understanding First-Order Kinetics and Half-Life

First-order kinetics implies that the rate of the decay process depends on the concentration of the substance that is decaying. The half-life is the time taken for half of the radioactive material to decay. Knowing that the half-life of tritium is 12.3 years, we can use this information along with the first-order decay formula to calculate the remaining amount of tritium.
02

Calculating the Decay Constant

For first-order kinetics, the decay constant, denoted as \(k\), is related to the half-life (\(t_{1/2}\)) by the formula \(k = \frac{\ln(2)}{t_{1/2}}\). We will use this formula to find \(k\) using the given half-life of tritium (12.3 years).
03

Use the First-Order Decay Formula

The remaining amount of a radioactive isotope is given by \(N = N_0 \cdot e^{-kt}\), where \(N_0\) is the initial quantity, \(t\) is the time elapsed, and \(k\) is the decay constant. We'll substitute the values of \(k\), \(N_0 = 25.0 \text{ mg}\), and \(t = 10.9 \text{ years}\) to find \(N\).
04

Calculate the Remaining Quantity of Tritium

Substituting the known values into the decay equation, we solve for \(N\) to find the amount of tritium remaining after 10.9 years.

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Key Concepts

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

Half-Life Calculation
Understanding the concept of half-life is essential in the study of radioactive materials. The half-life is the time required for exactly half of a sample of a radioactive isotope to decay. It is a constant value that is characteristic to each isotope. This means that regardless of the amount of the substance you start with, after one half-life, you will have 50% of your original substance left.

In our example with tritium (\( {}^{3}\text{H} \) or T), with a half-life of 12.3 years, it shows that after 12.3 years, any quantity of tritium would be reduced to half of its original amount. Let's say we started with 25.0 mg of tritium; after 12.3 years, we'd expect to have 12.5 mg remaining. If we needed to calculate how much tritium would be left after a time period that is not a multiple of the half-life, then we'd need the decay constant and the first-order decay formula, which will be discussed next.

It's also worth mentioning that half-life calculation doesn't account for chemical reactions or physical processes that the material may undergo, only its inherent radioactivity.
Decay Constant
The decay constant, represented as \( k \), is a probability that characterizes the rate of decay of a radioactive isotope. It is directly related to the half-life and is used to describe how quickly an isotope will decay. For first-order kinetics, the relationship between the decay constant and the half-life is given by the formula \( k = \frac{\ln(2)}{t_{1/2}} \). The natural logarithm of 2 (\( \ln(2) \)) is used because decay processes are described by the exponential function.

In our tritium example, we can calculate the decay constant using the provided half-life of 12.3 years. By plugging this into our decay constant formula, we find that \( k \) is a small fraction per year, which indicates that tritium decays slowly. The decay constant gives us a quantitative measure that can be applied across any amount of material and allows us to calculate the remaining quantity of the isotope at any point in time using the first-order decay formula, which brings us to our final concept.
First-Order Decay Formula
To determine the exact amount of a radioactive element remaining after any given time, not just at its half-life, we employ the first-order decay formula, which is expressed as \( N = N_0 \cdot e^{-kt} \). \( N \) represents the remaining amount of the isotope, \( N_0 \) is the initial amount, \( t \) is the time that has passed, and \( k \) is the decay constant.

So for our tritium with an initial mass of 25.0 mg and wanting to know how much remains after 10.9 years, we substitute these values into our formula to solve for \( N \). Because the exponential term \( e^{-kt} \) shows how much of the substance has decayed over time, it will always be a fraction less than 1 for positive values of \( kt \), reflecting the remaining proportion of the undecayed material.

This formula demonstrates that the amount of substance does not decrease linearly over time but rather at a rate proportional to its current amount—hence 'first-order'. The continuous decay over time creates an exponential decay graph, which is key in understanding radioactivity and the predictable yet non-linear nature of radioactive decay.

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Most popular questions from this chapter

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes by first- order kinetics with a rate constant of \(3.7 \times 10^{-5} \mathrm{~s}^{-1}\) at \(298 \mathrm{~K}\). (a) What is the half-life (in hours) for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(298 \mathrm{~K}\) ? (b) If \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}=0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), what will be the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) after \(3.5 \mathrm{~h}\) ? (c) How much time (in minutes) will elapse before the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration decreases from \(0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.0135 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) ?

Determine whether each of the following statements is true or false. If a statement is false, explain why. (a) For a reaction with a very large equilibrium constant, the rate constant of the forward reaction is much larger than the rate constant of the reverse reaction. (b) At equilibrium, the rate constants of the forward and reverse reactions are equal. (c) Increasing the concentration of a reactant increases the rate of a reaction by increasing the rate constant in the forward direction.

The second-order rate constant for the decomposition of \(\mathrm{NO}_{2}\) (to \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) ) at \(573 \mathrm{~K}\) is \(0.54 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}\). Calculate the time for an initial \(\mathrm{NO}_{2}\) concentration of \(0.20 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to decrease to (a) one-half; (b) one-sixteenth; (c) one-ninth of its initial concentration.

The data below were collected for the reaction \(2 \mathrm{HI}(\mathrm{g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\) at \(580 \mathrm{~K}\). $$ \begin{array}{lccccc} \text { Time }(\mathrm{s}) & 0 & 1000 . & 2000 . & 3000 . & 4000 . \\ {[\mathrm{HI}]\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right)} & 1.0 & 0.11 & 0.061 & 0.041 & 0.031 \end{array} $$ (a) Using a graphing calculator or graphing software, such as that on the Web site for this book, plot the data in an appropriate fashion to determine the order of the reaction. (b) From the graph, determine the rate constant for (i) the rate law for the loss of HI and (ii) the unique rate law.

Determine whether each of the following statements is true or false. If a statement is false, explain why. (a) The equilibrium constant for a reaction equals the rate constant for the forward reaction divided by the rate constant for the reverse reaction. (b) In a reaction that is a series of equilibrium steps, the overall equilibrium constant is equal to the product of all the forward rate constants divided by the product of all the reverse rate constants. (c) Increasing the concentration of a product increases the rate of the reverse reaction, and so the rate of the forward reaction must then increase, too.

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