Half-Life: Videos & Practice Problems

Half-life, the time required for half of a substance to decay or decompose, varies with reaction order. In zero-order reactions, half-life is directly proportional to the initial reactant concentration and inversely proportional to the rate constant (k), shortening as concentration decreases. For first-order reactions, common in radioactive decay, half-life is constant, independent of initial concentration, and calculated using:

$T_{\frac{1}{2}}=\frac{\ln \left(2\right)}{k}$

Second-order reactions have a half-life that increases as the initial concentration decreases, as it is dependent on both the rate constant and the initial concentration. Understanding these relationships is crucial for predicting reaction behavior over time and is represented graphically with half-life on the y-axis and time on the x-axis.

Zero-Order Half-life

Half-life is defined as the time required for a reactant to decrease to half of its initial concentration during a decay or decomposition process. The half-life function varies depending on the order of the reaction, which can be zero, first, or second order. For zero-order reactions, the half-life can be calculated using the equation:

t_1/2 = &frac{[A]₀}{2k}

In this equation, [A]₀ represents the initial concentration of the reactant, and k is the rate constant, measured in units of molarity per time (M·s^-1). It is important to note that for zero-order reactions, the half-life is directly dependent on the initial concentration of the reactant. As the concentration decreases, the half-life also decreases. This means that the time taken to lose half of the initial amount becomes shorter as the concentration diminishes.

Graphically, this relationship can be represented with half-life plotted on the y-axis and time on the x-axis. As time progresses, the half-life decreases, resulting in a negative slope on the graph. This illustrates that the time required to reach half of the initial concentration shortens over time, reflecting the nature of zero-order kinetics.

Zero-Order Half-life Example

The reverse Haber reaction is characterized as a zero-order reaction, which is indicated by its rate constant units of molarity per second (M/s). To determine the half-life of this reaction, we can use the formula for the half-life of a zero-order reaction, which is given by:

\( t_{1/2} = \frac{[A]_0}{2k} \)

In this equation, \( t_{1/2} \) represents the half-life, \([A]_0\) is the initial concentration of the reactant (in this case, ammonia), and \( k \) is the rate constant. Given that the initial concentration of ammonia is \( 2.47 \times 10^{-2} \) moles per liter and the rate constant \( k \) is \( 1.45 \times 10^{-6} \) M/s, we can substitute these values into the equation.

Substituting the values, we have:

\( t_{1/2} = \frac{2.47 \times 10^{-2}}{2 \times 1.45 \times 10^{-6}} \)

Calculating this gives:

\( t_{1/2} = \frac{2.47 \times 10^{-2}}{2.90 \times 10^{-6}} \approx 8.52 \times 10^{3} \text{ seconds} \)

Thus, the half-life for the reverse Haber reaction at the specified conditions is approximately \( 8.52 \times 10^{3} \) seconds. This calculation illustrates the relationship between the initial concentration of a reactant and the rate constant in determining the time required for half of the reactant to be consumed in a zero-order reaction.

Decomposition of a certain substance Y at 45°C was found to be zero order. What is the half-life of substance Y if it took 15.5 minutes to decompose 67% of this substance? [Y]₀ = 0.25 M.

• A. 0.058 min
• B. 0.029 min
• C. 12 min
• D. 23.5 min

Answer: (C)

First-Order Half-Life

In the study of radioactive processes, it is essential to understand that they adhere to a first-order rate law. The half-life of a first-order reaction can be calculated using the equation:

t_1/2 = \frac{\ln{2}}{k}

Here, t_1/2 represents the half-life, k is the rate constant, and ln 2 is a constant value approximately equal to 0.693. This means that when you see 0.693 on a formula sheet, it is derived from the natural logarithm of 2. The units of the rate constant k are time^-1, indicating that it is inversely related to time.

One of the key insights regarding the half-life equation is that it does not include the initial concentration of the reactant. This indicates that the half-life remains constant throughout the reaction, independent of the concentration of the reactant. As a result, the half-life will not change over time, making it a reliable measure for first-order reactions.

When graphing this relationship, the half-life can be plotted against time, where the y-axis represents the half-life and the x-axis represents time. This visualization reinforces the concept that the half-life remains consistent, as both ln 2 and k are constants in this context. Keeping these principles in mind will aid in understanding the behavior of first-order reactions and their half-lives.

First-Order Half-Life Example

In chemical kinetics, the half-life of a reaction is a crucial concept, particularly for first-order reactions. The half-life is defined as the time required for the concentration of a reactant to decrease to half of its initial value. For first-order reactions, the half-life can be calculated using the formula:

\( t_{1/2} = \frac{\ln(2)}{k} \)

In this case, the rate constant \( k \) is given as \( 2.3 \times 10^{-3} \, \text{s}^{-1} \) at a temperature of 35 degrees Celsius. To find the half-life, we substitute the value of \( k \) into the formula:

\( t_{1/2} = \frac{\ln(2)}{2.3 \times 10^{-3}} \)

Calculating this gives:

\( t_{1/2} \approx 301 \, \text{s} \)

It is important to note that while the initial concentration of nitrogen dioxide is provided as \( 1.4 \times 10^{-1} \, \text{M} \), it does not affect the calculation of the half-life for a first-order reaction. The half-life is solely dependent on the rate constant \( k \). This illustrates that in some problems, not all provided information is necessary for solving the question at hand.

Radioactive plutonium-239 (t_1/2 = 2.41 × 10⁵ yr) is used in nuclear reactors and atomic bombs. If there are 5.70 × 10² g of plutonium isotope in a small atomic bomb, how long will it take for the substance to decay to 3.00 × 10² g?

• A. 2.23×10⁵ yr
• B. 2.60×10⁵ yr
• C. 517 yr
• D. 600 yr

Answer: (A)

Which of the following statements is False?

• A. The average rate of a reaction decreases during a reaction.
• B. The rate of zero order reactions are not dependent on concentrations.
• C. The rate of a first order reaction is dependent on concentrations.
• D. The half-life of a first order reaction is dependent on the initial concentration of reactant.

Answer: (D)

Second-Order Half-Life

In second order reactions, the relationship between the rate of reaction and the concentration of reactants is defined by a specific rate law. The half-life (\(t_{1/2}\)) for these reactions can be expressed with the formula:

\[ t_{1/2} = \frac{1}{k \cdot [A]_0} \]

Here, \(k\) represents the rate constant, and \([A]_0\) is the initial concentration of the reactant. The units for the rate constant \(k\) in second order reactions are typically expressed as molarity^-1 time^-1.

One key aspect of second order reactions is that the half-life is inversely related to the initial concentration. As the initial concentration decreases, the half-life increases. This occurs because the formula indicates that as \([A]_0\) becomes smaller, the value of \(t_{1/2}\) becomes larger. Essentially, the half-life extends over time, meaning it takes progressively longer to reduce the concentration of the reactant by half.

This characteristic of second order reactions contrasts with first order reactions, where the half-life remains constant regardless of the concentration. This consistency in first order processes is why they are often preferred in radioactive decay scenarios. Understanding how the half-life varies with concentration in second order reactions is crucial for predicting the behavior of chemical systems over time.

Second-Order Half-Life Example

In a second-order reaction, the relationship between half-life and initial concentration is defined by the formula:

\( t_{1/2} = \frac{1}{k \cdot [A]_0} \)

Where \( t_{1/2} \) is the half-life, \( k \) is the rate constant, and \( [A]_0 \) is the initial concentration of the reactant. To find the initial concentration, we can rearrange this equation:

\( [A]_0 = \frac{1}{k \cdot t_{1/2}} \)

Given that the half-life \( t_{1/2} \) is 0.45 seconds, we need to determine the rate constant \( k \). In this case, \( k \) is provided as \( 3.5 \times 10^{-3} \, \text{s}^{-1} \). Plugging in the values, we can calculate the initial concentration:

\( [A]_0 = \frac{1}{(3.5 \times 10^{-3}) \cdot (0.45)} \)

Calculating this gives:

\( [A]_0 = \frac{1}{0.001575} \approx 634.92 \, \text{M} \)

Considering significant figures, the initial concentration of the reactant is approximately:

\( [A]_0 \approx 630 \, \text{M} \)

This calculation illustrates the relationship between the half-life of a second-order reaction and the initial concentration, emphasizing the importance of the rate constant in determining the concentration of reactants at the start of the reaction.

Use the data below to determine the half-life of decomposition of NOCl reaction which follows 2nd order kinetics.

• A. 1.00×10⁴ s
• B. 1.80×10⁵ s
• C. 2.02×10⁴ s
• D. 1.022×10³ s

Answer: (A)

Here’s what students ask on this topic:

Which expression relates h, the number of half-lives, and y, the number of years?

The expression that relates h (the number of half-lives) and y (the number of years) is derived from the definition of a half-life, which is the time required for half of a sample of a radioactive substance to decay. The expression can be written as:

$h=\frac{y}{t_{1/2}}$

Here, t1/2 represents the half-life of the radioactive substance in years. This equation shows that the number of half-lives (h) is equal to the total time elapsed (y) divided by the half-life of the substance. If you know the half-life of a substance and the amount of time that has passed, you can use this expression to calculate how many half-lives have occurred during that period. This is particularly useful in fields like geology, archaeology, and physics, where understanding the decay of radioactive isotopes is important for dating materials and studying processes over time.

How can I find the number of half-lives?

To find the number of half-lives that have elapsed, you need to know two key pieces of information: the total time that has passed and the half-life duration of the substance in question. The half-life is the time it takes for half of the original amount of a radioactive substance to decay.

Here's a step-by-step method to calculate the number of half-lives:

1. Determine the half-life of the substance. This information is usually provided in the problem or can be found in scientific literature.
2. Measure or find out the total time that has passed since the substance began to decay.
3. Divide the total time by the half-life of the substance to find the number of half-lives.

For example, if the half-life of a substance is 5 years and 20 years have passed, you would calculate the number of half-lives as follows:

$\text{Number of half-lives}=\frac{\text{Total time passed}}{\text{Half-life duration}}$ $\text{Number of half-lives}=\frac{20\text{years}}{5\text{years}}$ $\text{Number of half-lives}=4$

So, four half-lives have

How do you calculate the number of half-lives?

To calculate the number of half-lives that have passed for a sample of a radioactive substance, you'll use the formula:

$\text{Number of half-lives}=\frac{\text{Total time elapsed}}{\text{Half-life of the substance}}$

Here's how you do it step by step:

1. Determine the total time that has elapsed since the substance began to decay.
2. Find out the half-life of the substance, which is the time it takes for half of the substance to decay. This information is typically provided in textbooks or reference materials.
3. Divide the total time elapsed by the half-life of the substance.

For example, if you have a substance with a half-life of 4 years and you want to know how many half-lives have passed in 16 years, you would calculate it as follows:

$\text{Number of half-lives}=\frac{\text{16 years}}{\text{4 years/half-life}}=\text{4 half-lives}$

This means that in 16 years, the substance has gone through 4 half-lives.