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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-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 \(0.100 ?\)

Short Answer

Expert verified
\( (a) \, c = 1.08 \times 10^{-4}\,\mathrm{M} \newline (b) \, k = 1.14 \times 10^{-3}\,\mathrm{s^{-1}} \newline (c) \, t_{1/2} = 607\,\mathrm{s} \newline (d) \, t = 4373\,\mathrm{s} \)

Step by step solution

01

Part (a): Initial concentration

First, apply the Beer-Lambert Law to find the initial concentration: \(A = \epsilon lc\), where: A - the absorbance \( \epsilon \) - the extinction coefficient (\(5.60 \times 10^{3}\, \mathrm{M^{-1}cm^{-1}}\)) l - path length (1.00 cm) c - concentration of the solution (in Molarity) Rearrange the equation to find the concentration: \(c = \frac{A}{\epsilon l}\) Plug the values from the exercise: \(c = \frac{0.605}{(5.60 \times 10^{3}\,\mathrm{M^{-1}cm^{-1}})(1.00\, \mathrm{cm})}\) Calculate the initial concentration: \(c = 1.08 \times 10^{-4}\,\mathrm{M}\)
02

Part (b): Rate constant

Next, apply the first-order reaction kinetics to find the rate constant: \(A_t = A_0e^{-kt}\), where: \(A_t\) - the absorbance at a given time (0.250) \(A_0\) - initial absorbance (0.605) k - rate constant (in s^-1) t - time elapsed (30.0 min = 1800 s) Rearrange the equation to solve for k: \(k = \frac{-\ln{\frac{A_t}{A_0}}}{t}\) Plug values from the exercise and the initial concentration: \(k = \frac{-\ln{\frac{0.250}{0.605}}}{1800\,\mathrm{s}}\) Calculate the rate constant: \(k = 1.14 \times 10^{-3}\,\mathrm{s^{-1}}\)
03

Part (c): Half-life

Now calculate the half-life (t_1/2) using the first-order reaction formula: \(t_{1/2} = \frac{\ln{2}}{k}\) Plug the value of k from the previous part: \(t_{1/2} = \frac{\ln{2}}{1.14 \times 10^{-3}\,\mathrm{s^{-1}}}\) Calculate the half-life: \(t_{1/2} = 607\,\mathrm{s}\)
04

Part (d): Time for the absorbance to fall to 0.100

Finally, calculate the time it takes for the absorbance to fall to 0.100 using the first-order reaction kinetics equation: Rearrange the equation to solve for time: \(t = \frac{-\ln{\frac{A_t}{A_0}}}{k}\) Plug the values from the exercise and the rate constant: \(t = \frac{-\ln{\frac{0.100}{0.605}}}{1.14 \times 10^{-3}\,\mathrm{s^{-1}}}\) Calculate the time: \(t = 4373\,\mathrm{s}\)

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Key Concepts

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

Beer-Lambert Law
The Beer-Lambert Law is a fundamental principle in spectroscopy that relates the absorption of light to the properties of the material through which the light is traveling. It helps in understanding how the concentration of a solute can affect the measurement of light passing through a sample. The law is expressed in the formula:
  • \(A = \epsilon lc\)
Here:
  • \(A\) is the absorbance, indicating the amount of light absorbed by the sample.
  • \(\epsilon\) is the extinction coefficient, a constant that denotes how strongly a substance can absorb light at a given wavelength.
  • \(l\) is the path length of the sample, usually measured in centimeters.
  • \(c\) is the concentration of the sample in molarity (M).
This simple yet powerful relation allows us to determine unknown concentrations of colored solutions by measuring their absorbance at specific wavelengths.
Absorbance
Absorbance is a measure of the amount of light absorbed by a solution at a particular wavelength. It is a dimensionless quantity that indicates how much of the incident light is not transmitted through the sample. When light encounters a solution, certain wavelengths are absorbed by the solute, leading to changes in intensity. Absorbance is given by the equation:
  • Absorbance (\(A\)) = \(\log_{10}(\frac{I_0}{I})\)
Where:
  • \(I_0\) is the initial intensity of the light entering the solution.
  • \(I\) is the intensity of the light leaving the solution.
Absorbance correlates directly to concentration thanks to the Beer-Lambert Law, meaning as concentration increases, absorbance typically increases. This makes absorbance a crucial variable in determining the concentration of solutions in chemical reactions.
Extinction Coefficient
The extinction coefficient, often symbolized by \(\epsilon\), is a constant that defines how strongly a chemical species absorbs light at a particular wavelength. It is expressed in units of \(\mathrm{M^{-1}cm^{-1}}\) to indicate its effect on concentration and path length. A higher extinction coefficient means the substance is more effective at absorbing light, which results in higher absorbance values for any given concentration. This coefficient is intrinsic to the substance and the specific wavelength used. In the context of the Beer-Lambert Law, the extinction coefficient helps us link absorbance to concentration, allowing calculations such as determining unknown concentrations from measured absorbances.
Rate Constant
In kinetics, the rate constant (\(k\)) is a crucial factor that helps determine the speed of a chemical reaction. For first-order reactions, where the rate depends linearly on the concentration of one reactant, the relationship between absorbance and time can be described through the equation:
  • \(A_t = A_0e^{-kt}\)
Where:
  • \(A_t\) is the absorbance at time \(t\).
  • \(A_0\) is the initial absorbance.
  • \(e\) is the base of the natural logarithm.
  • \(t\) is time.
  • \(k\) is the rate constant.
Rearranging this equation allows us to solve for \(k\), indicating how fast the reaction proceeds. A large rate constant corresponds to a rapid reaction rate, whereas a small rate constant indicates a slower reaction.
Half-life
Half-life (\(t_{1/2}\)) is the time required for the concentration of a reactant to decrease to half its initial value. In first-order reactions, half-life is independent of the starting concentration, making it a useful characteristic to describe the reaction's progress. The formula for half-life in first-order kinetics is:
  • \(t_{1/2} = \frac{\ln{2}}{k}\)
This indicates that half-life is solely dependent on the rate constant \(k\). The concept of half-life is widely applied in various fields, from pharmacokinetics to radioactive decay, providing a straightforward measure for reaction durations.

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Most popular questions from this chapter

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{array}{l} \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ \mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{array} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

The reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) was performed and the following data were obtained under conditions of constant \(\left[\mathrm{Cl}_{2}\right]:\) (a) Is the following mechanism consistent with the data? $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) & \rightleftharpoons \mathrm{NOCl}_{2}(g) \quad(\text { fast }) \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) & \longrightarrow 2 \mathrm{NOCl}(g) &(\text { slow }) \end{aligned} $$ (b) Does the linear plot guarantee that the overall rate law is second order?

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{array}{l} \mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{array} $$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3} .\) (b) Do we consider \(\mathrm{NO}_{2}\) a catalyst or an intermediate in this reaction? (c) Would you classify NO as a catalyst or as an intermediate? (d) Is this an example of homogeneous catalysis or heterogeneous catalysis?

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \rightarrow \mathrm{B}(a q)\). A flask is charged with \(0.065 \mathrm{~mol}\) of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero and that A cleanly converts to B with no intermediates. (b) Calculate the average rate of disappearance of A for each 10 -min interval in units of \(M /\) s. (c) Between \(t=0 \mathrm{~min}\) and \(t=30 \mathrm{~min},\) what is the average rate of appearance of \(\mathrm{B}\) in units of \(\mathrm{M} / \mathrm{s}\) ? Assume that the volume of the solution is constant.

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. \((\mathbf{b})\) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathbf{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M},\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M} ?\) (d) What is the reaction rate at \(1000 \mathrm{~K}\) if \([\mathrm{NO}]\) is decreased to \(0.010 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to \(0.030 \mathrm{M} ?\)

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