1. Understanding First-Order Reactions

First-order reactions are chemical reactions where the rate is directly proportional to the concentration of a single reactant. Mathematically, this relationship is expressed as:

$\text{Rate} = k[A]$

Here, $k$ represents the rate constant, and $[A]$ denotes the concentration of the reactant A. The defining feature of first-order reactions is that their rate remains dependent solely on the concentration of one reactant, making them relatively straightforward to analyze compared to higher-order reactions.

2. The Concept of Half-life ($t_{1/2}$)

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial value. For first-order reactions, the half-life is unique because it remains constant regardless of the initial concentration. This characteristic distinguishes first-order reactions from reactions of other orders, where half-life can vary with concentration.

The half-life for a first-order reaction can be derived from the integrated rate law:

$$\ln\left(\frac{[A]_0}{[A]}\right) = kt$$

Setting $[A] = \frac{[A]_0}{2}$ and solving for $t$ gives:

$$t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}$$

3. Integrated Rate Law for First-Order Reactions

The integrated rate law allows us to relate the concentration of a reactant at any time $t$ to its initial concentration. For first-order reactions, it is expressed as:

$$\ln [A] = -kt + \ln [A]_0$$

Or alternatively:

$$[A] = [A]_0 e^{-kt}$$

This equation is pivotal in calculating the concentration of reactants or determining the rate constant when half-life and initial concentrations are known.

4. Determining the Rate Constant ($k$)

The rate constant is a crucial parameter that quantifies the speed of a reaction. For first-order reactions, it can be calculated using the half-life formula:

$$k = \frac{\ln 2}{t_{1/2}}$$

Alternatively, by rearranging the integrated rate law, $k$ can be determined if the concentrations at two different times are known:

$$k = \frac{\ln [A]_0 - \ln [A]}{t}$$

5. Graphical Representation

First-order reactions exhibit a linear plot when plotting $\ln [A]$ versus time ($t$). The slope of this line equals $-k$, allowing for the straightforward determination of the rate constant. This linear relationship is a diagnostic tool to confirm the order of a reaction experimentally.

$$\ln [A] = -kt + \ln [A]_0$$

Graph Image Placeholder: A straight line with $\ln [A]$ on the y-axis and time on the x-axis.

6. Relationship Between Half-life and Rate Constant

The half-life and rate constant in first-order reactions are inversely related. As the rate constant increases, the half-life decreases, indicating a faster reaction. This relationship is fundamental in fields like pharmacokinetics, where the half-life of drugs determines dosing intervals.

$$t_{1/2} = \frac{\ln 2}{k}$$

Thus, knowing either the half-life or the rate constant allows the calculation of the other, facilitating the analysis of reaction kinetics.

7. Applications of Half-life in First-Order Reactions

Understanding half-life in first-order reactions has practical applications across various disciplines:

• Pharmacology: Determines how long a drug remains effective in the bloodstream.
• Environmental Science: Assesses the degradation rate of pollutants.
• Chemical Engineering: Aids in the design of reactors and the scaling of industrial processes.
• Radioactive Decay: Although a separate domain, the mathematics of half-life is similar, facilitating interdisciplinary understanding.

8. Calculation Examples

**Example 1:** Given a first-order reaction with a rate constant of $0.03 \, \text{min}^{-1}$, calculate its half-life.

Using the formula:

$$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{0.03} \approx 23.1 \, \text{minutes}$$

**Example 2:** If the half-life of a first-order reaction is 10 hours, determine the rate constant.

$$k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{10} = 0.0693 \, \text{hr}^{-1}$$

1. Derivation of the Half-life Formula for First-Order Reactions

The derivation begins with the integrated rate law for first-order reactions:

$$\ln [A] = -kt + \ln [A]_0$$

To find the half-life, set $[A] = \frac{[A]_0}{2}$:

$$\ln \left(\frac{[A]_0}{2}\right) = -kt_{1/2} + \ln [A]_0$$

Subtract $\ln [A]_0$ from both sides:

$$\ln \left(\frac{1}{2}\right) = -kt_{1/2}$$

Since $\ln \left(\frac{1}{2}\right) = -\ln 2$, we have:

$$\ln 2 = kt_{1/2}$$

Therefore:

$$t_{1/2} = \frac{\ln 2}{k}$$

2. Temperature Dependence of the Rate Constant

The rate constant $k$ is temperature-dependent, typically increasing with temperature. This relationship is quantitatively described by the Arrhenius equation:

$$k = A e^{-Ea/(RT)}$$

Where:

• $A$ = Frequency factor
• $Ea$ = Activation energy
• $R$ = Gas constant
• $T$ = Temperature in Kelvin

As temperature increases, the exponential term increases, leading to a larger $k$ and, consequently, a shorter half-life:

$$t_{1/2} = \frac{\ln 2}{k}$$

This interplay is critical in designing chemical processes that require precise control over reaction rates.

3. Complex Problem-Solving: Multi-step Calculation

**Problem:** A first-order reaction has a half-life of 5 hours. If the initial concentration of reactant A is 0.8 M, what will be its concentration after 15 hours?

**Solution:** First, determine the rate constant $k$:

$$k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{5} = 0.1386 \, \text{hr}^{-1}$$

Using the integrated rate law:

$$[A] = [A]_0 e^{-kt} = 0.8 \times e^{-0.1386 \times 15}$$

Calculate the exponent:

$$-0.1386 \times 15 = -2.079$$

Thus:

$$[A] = 0.8 \times e^{-2.079} \approx 0.8 \times 0.125 = 0.10 \, \text{M}$$

After 15 hours, the concentration of A will be approximately 0.10 M.

4. Interdisciplinary Connections: Pharmacokinetics

In pharmacokinetics, the half-life of a drug determines how long it remains active in the body. First-order kinetics is often applicable, where the rate of drug elimination is proportional to its concentration. Understanding half-life aids in dosing regimen design, ensuring therapeutic efficacy while minimizing side effects.

For instance, a drug with a short half-life may require multiple doses per day, whereas one with a longer half-life can be administered less frequently. Additionally, variations in half-life among individuals necessitate personalized medicine approaches.

5. Impact of Catalysts on Half-life

Catalysts accelerate the rate of chemical reactions without being consumed in the process. For first-order reactions, the introduction of a catalyst increases the rate constant $k$, thereby decreasing the half-life:

$$t_{1/2} = \frac{\ln 2}{k}$$

This reduction in half-life can be crucial in industrial processes where time efficiency translates to cost savings and increased productivity. The catalytic effect is central to heterogeneous catalysis in automotive converters and enzymatic reactions in biological systems.

6. Radioactive Decay as a First-Order Process

Radioactive decay is a natural first-order process where unstable nuclei lose energy by emitting radiation. Similar to chemical first-order reactions, the decay rate is proportional to the number of undecayed nuclei:

$$N = N_0 e^{-kt}$$

Where $N$ is the number of nuclei at time $t$, and $N_0$ is the initial number. The half-life concept is directly applicable, allowing the prediction of the time required for half of the radioactive atoms to decay, which is essential in fields like geology for radiometric dating and medicine for diagnostic imaging.

• Half-life in first-order reactions remains constant regardless of initial concentration.
• The rate constant ($k$) is inversely related to half-life.
• Integrated rate laws facilitate the calculation of reactant concentrations over time.
• Graphical analysis confirms reaction order and aids in determining $k$.
• Applications span diverse fields, including pharmacology, environmental science, and radioactive decay.