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 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 \(0.100 ?\)

Short answer

The initial concentration of the colored reactant is \(1.08 \times 10^{-4} M\). The rate constant is approximately \(1.135 \times 10^{-4} s^{-1}\). The half-life of the reaction is approximately 102 minutes. Finally, it takes about 91.17 minutes for the absorbance to fall to 0.100.

Step-by-step solution

Calculate Initial Concentration

To find the initial concentration of the colored reactant, we can use the formula A = εcl. We have A = 0.605, ε = 5.60 × 10³ M⁻¹ cm⁻¹, and l = 1 cm. Rearrange the formula to solve for the initial concentration (c₀): c₀ = A / (ε × l) = 0.605 / (5.60 × 10³ M⁻¹ cm⁻¹ × 1 cm) Now, substitute the given values and solve for c₀: c₀ = 0.605 / (5.60 × 10³) = 1.08 × 10⁻⁴ M.

Calculate Rate Constant

First, we need to find the concentration after 30 minutes, when the absorbance has fallen to 0.250. We can use the same formula from the previous step: c₁ = A₁ / (ε × l) = 0.250 / (5.60 × 10³ M⁻¹ cm⁻¹ × 1 cm) c₁ = 0.250 / (5.60 × 10³) = 4.46 × 10⁻⁵ M. Now, we can use the first-order reaction equation to find the rate constant k: ln(c₀ / c₁) = kt where t is the time in seconds. In this case, t = 30 min × 60 s/min = 1800 s. Solve for k: k = ln(c₀ / c₁) / t = ln(1.08 × 10⁻⁴ / 4.46 × 10⁻⁵) / 1800 k ≈ 1.135 × 10⁻⁴ s⁻¹.

Calculate Half-Life

Now that we have the rate constant k, we can calculate the half-life t_(1/2) using the formula: t_(1/2) = ln(2) / k = ln(2) / 1.135 × 10⁻⁴ t_(1/2) ≈ 6,108 s ≈ 102 min.

Time for Absorbance to Fall to 0.100

To find the time it takes for the absorbance to fall to 0.100, first find the final concentration c₂: c₂ = A₂ / (ε × l) = 0.100 / (5.60 × 10³ M⁻¹ cm⁻¹ × 1 cm) c₂ = 0.100 / (5.60 × 10³) = 1.786 × 10⁻⁵ M. Use the first-order reaction equation again and solve for time t: t = ln(c₀ / c₂) / k = ln(1.08 × 10⁻⁴ / 1.786 × 10⁻⁵) / 1.135 × 10⁻⁴ t ≈ 5470 s ≈ 91.17 min.

Key concepts

Spectroscopy

Spectroscopy is a fascinating technique used to study the interaction of matter with electromagnetic radiation. It helps chemists and researchers to deduce important molecular information about substances. In our case, we're leveraging a specific type - UV/Vis spectroscopy, which is used to observe the absorbance of light by a colored reactant.

By measuring absorbance, which is the amount of light absorbed at a specific wavelength, we can understand the concentration of a compound in a solution. This is crucial because absorbance is directly related to concentration, as explained by Beer's Law. The information gathered through spectroscopy helps in calculating important reaction parameters, such as the initial concentration and rate constant for reactions.

In our exercise, we monitor the change in absorbance at a wavelength of 520 nm, which enables us to track the reaction progress.

Absorbance

Absorbance is a key parameter that reflects how much light a sample absorbs at a certain wavelength. The higher the absorbance, the more light is being absorbed by the sample, and this often indicates a higher concentration of the absorbing species.

We use the Beer-Lambert Law to relate absorbance \(A\) to concentration \(c\) through the equation: \(A = \varepsilon c l\). Here, \(\varepsilon\) is the molar extinction coefficient, which indicates how strongly a substance absorbs light at a given wavelength, and \(l\) is the path length of the sample - typically the width of the cuvette.

This relationship means that by measuring the absorbance, we can calculate the concentration of the reactant, provided we know the extinction coefficient and path length. For our reaction, monitoring changes in absorbance allowed us to determine the initial and intermediate concentrations, which were then used to calculate the rate constant.

Rate Constant

The rate constant, often denoted as \(k\), is a crucial part of the rate law for reactions. It gives insight into the speed of the reaction at a given temperature. In a first-order reaction, the rate constant has units of \(\text{s}^{-1}\), implying how the concentration of a reactant changes with time.

The rate constant provides valuable insights because different reactions have different rate constants, reflecting how quickly or slowly they proceed under the same conditions.

For our exercise, we calculated the rate constant using the formula: \(k = \frac{\ln(c_0 / c_1)}{t}\), where \(c_0\) is the initial concentration, \(c_1\) is the concentration at a time \(t\), and the natural log \(\ln\) ensures the relationship holds for exponential decay typical of first-order reactions.

The rate constant helps us not only understand the current reaction but also predict how the reaction will proceed over time.

Half-Life

Half-life is the time it takes for half of the reactant to be consumed in a reaction. This is a very useful metric, especially for reactions that follow first-order kinetics, like in our example.

For first-order reactions, the half-life \(t_{1/2}\) is constant and doesn't depend on the initial concentration, unlike zero- or second-order reactions. The formula used to calculate the half-life in these instances is: \(t_{1/2} = \frac{\ln(2)}{k}\), where \(\ln(2)\) is the natural log of 2, approximately 0.693.

Knowing the half-life of a reaction can help in making predictions about how long a reaction will take to reach a certain stage or to be completed, which is very useful in practical laboratory settings and industrial applications.