Problem 101 - Chemistry: The Central Science

Question

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{~nm}\). The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{~nm}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is \(0.605\) at the beginning of the reaction. (b) The absorbance falls to \(0.250\) at \(30.0 \mathrm{~min}\). Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100\) ?

Step-by-Step Solution

Answer

The initial concentration of the colored reactant is \(1.08 \times 10^{-4}\) M. The rate constant for this first-order reaction is \(6.48 \times 10^{-4}\) s^{-1}. The half-life of the reaction is \(1.07 \times 10^{3}\) s, and it takes \(4.63 \times 10^{3}\) s for the absorbance to fall to \(0.100\).

Step 1: (a) Calculate the initial concentration of the colored reactant

To find the initial concentration of the colored reactant, we can use the Beer-Lambert law, which relates the absorbance (A) to the concentration (c) of the colored species as follows: \[ A = \varepsilon c l \] Here, \( \varepsilon \) is the extinction coefficient of the colored species (with units M^{-1} cm^{-1}), and \( l \) is the path length through which the light passes (with units in cm). Rearranging the equation for the concentration, we get: \[ c = \frac{A}{\varepsilon l} \] Since all values are given in the problem, simply substitute them into the equation: \[ c = \frac{0.605}{(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1})(1.00 \mathrm{~cm})} \]

Step 2: (a) Initial concentration calculation

Now, perform the calculation: \[ c = \frac{0.605}{5.60 \times 10^{3} \mathrm{~M}^{-1} \mathrm{~cm}^{-1}} = 1.08 \times 10^{-4} \mathrm{M} \] Thus, the initial concentration of the colored reactant is \(1.08 \times 10^{-4}\) M.

Step 3: (b) Calculate the rate constant in units of s^{-1}

To find the rate constant, we can use the first-order reaction equation for concentration and absorbance: \[ A = A_0 e^{-kt} \] Here, \(A_0\) is the initial absorbance, \(A\) is the absorbance at time \(t\), and \(k\) is the rate constant (with units s^{-1} since the time given is in seconds). The exercise gives us the initial absorbance (\(0.605\)) and tells us that the absorbance falls to \(0.250\) at \(30.0\ \min\). Rearranging the equation for the rate constant: \[ k = \frac{-\ln{\frac{A}{A_0}}}{t} \] Substituting the given values into the equation: \[ k = \frac{-\ln{\frac{0.250}{0.605}}}{30.0 \times 60} \]

Step 4: (b) Rate constant calculation

Now, perform the calculation: \[ k = \frac{-\ln{\frac{0.250}{0.605}}}{30.0 \times 60} = 6.48 \times 10^{-4} \mathrm{s}^{-1} \] Thus, the rate constant of this first-order reaction is \(6.48 \times 10^{-4 }\) s^{-1}.

Step 5: (c) Calculate the half-life of the reaction

To find the half-life (t_{1/2}) of the reaction, we can use the following equation for first-order reactions: \[ t_{1/2} = \frac{\ln{2}}{k} \] Substitute the rate constant calculated in part (b) into the equation: \[ t_{1/2} = \frac{\ln{2}}{6.48 \times 10^{-4} \mathrm{s^{-1}}} \]

Step 6: (c) Half-life calculation

Now, perform the calculation: \[ t_{1/2} = \frac{\ln{2}}{6.48 \times 10^{-4} \mathrm{s^{-1}}} = 1.07 \times 10^{3} \mathrm{s} \] Thus, the half-life of the reaction is \(1.07 \times 10^{3}\) s.

Step 7: (d) Calculate the time it takes for the absorbance to fall to 0.100

To find the time it takes for the absorbance to fall to \(0.100\), we can use the first-order reaction equation for concentration and absorbance: \[ A = A_0 e^{-kt} \] We already have the initial absorbance and rate constant, so we can rearrange the equation to find the time: \[ t = \frac{-\ln{\frac{A}{A_0}}}{k} \] Substitute the given values for the absorbance and the calculated rate constant: \[ t = \frac{-\ln{\frac{0.100}{0.605}}}{6.48 \times 10^{-4} \mathrm{s}^{-1}} \]

Step 8: (d) Time calculation

Now, perform the calculation: \[ t = \frac{-\ln{\frac{0.100}{0.605}}}{6.48 \times 10^{-4} \mathrm{s}^{-1}} = 4.63 \times 10^{3} \mathrm{s} \] Thus, it takes \(4.63 \times 10^{3}\) s for the absorbance to fall to \(0.100\).

Key Concepts

Beer-Lambert Law

The Beer-Lambert Law is a fundamental principle in the science of spectrophotometry, which is a key tool used in spectroscopy. As per this law, there's a direct, linear relationship between the concentration of colored species in a solution and the absorbance of light passing through that solution. This relationship is given by the equation:

\[ A = \varepsilon c l \]
where \( A \) represents the absorbance, \( \text { \textepsilon } \) is the molar absorptivity or extinction coefficient, \( c \) is the concentration, and \( l \) is the path length of the cuvette holding the sample.
It's crucial to understand that absorbance is unitless, and therefore, the units of the extinction coefficient must cancel out with those of concentration and path length, resulting in a versatile and broadly applicable equation. This law forms the basis for analyzing concentrations in solutions, making it instrumental for students and professionals in biochemistry, pharmacology, and other fields that involve chemical analysis.

• Different substances absorb light differently at specific wavelengths.
• Experimentally, a known path length and extinction coefficient allow us to determine an unknown concentration.
• In practical applications, a spectrophotometer is used to measure absorbance.

Spectroscopy

Spectroscopy is a branch of science focused on the study of how light interacts with matter. By measuring the intensity of light as a function of wavelength, researchers can gain insights into the properties of different substances. As an analytical method, spectroscopy can identify compounds, quantify concentrations, and investigate the electronic or molecular structure of various materials.

Spectroscopy encompasses a range of techniques, each suited to different types of analysis. For instance, UV-Vis spectroscopy, which involves ultraviolet and visible light, is the technique mentioned in the exercise that monitors the absorbance of a colored reactant at a specific wavelength. This method is particularly useful for studying substances that absorb light in the UV or visible regions of the electromagnetic spectrum. Students can find spectroscopy applications in physical chemistry, environmental science, astrophysics, and even forensic analysis.

• Through absorbance measurements, spectroscopy can provide data on reaction rates, as demonstrated in the exercise.
• It is vital for investigating the kinetics of chemical reactions — fundamental for developing drugs, food preservatives, and other chemical products.

Half-Life

Half-life, denoted by \( t_{1/2} \), is a term most commonly linked with radioactive decay, but it is just as essential in the context of chemical reactions, especially those that are first-order. In this scenario, the half-life represents the time taken for the concentration of a reactant to decrease to half its initial value.

For a first-order reaction, the half-life is remarkably consistent and unaffected by initial concentrations. It can be determined using the equation:

\[ t_{1/2} = \frac{\ln{2}}{k} \]
Here, \( k \) is the rate constant of the reaction. Understanding half-life is crucial, as it is an intrinsic property of a reaction, providing insights for comparison between different reactions, and is essential in disciplines like pharmacology (where it helps determine dosing schedules) and environmental science (to predict the persistence of pollutants).

• Half-life provides a straightforward way to conceptualize the decay or de-concentration of a substance over time.
• It is used to predict the duration needed for a reactant to reach a specific concentration.

Rate Constant

The rate constant, represented by \( k \), is a key parameter in chemical kinetics that provides the speed at which a reaction proceeds. It is unique to each chemical reaction and depends on factors like temperature and the presence of a catalyst.

For first-order reactions, as seen in the example problem, the rate constant can be found from the known concentrations and their corresponding time points. The mathematical expression connecting the rate constant to the absorbance is:

\[ k = \frac{-\ln{\frac{A}{A_0}}}{t} \]
where \( A \) is the absorbance at time \( t \) and \( A_0 \) is the initial absorbance. The significance of the rate constant extends beyond calculations; it's instrumental in comparing the efficacy of chemical processes, designing reactors, and in pharmaceuticals, determining how quickly a drug will act. It's a quantifiable way to express how conditions or catalysts modify reaction speeds, serving as a guideline for laboratory experiments and industrial production.

• A higher rate constant means a faster reaction.
• Temperature changes can greatly influence the value of \( k \).
• Calculating the rate constant is essential for optimizing chemical processes and ensuring product quality.