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The rate of a first-order reaction is followed by spectroscopy, monitoring the absorption of a colored reactant. The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has a molar absorptivity constant of \(5.60 \times 10^{3} \mathrm{~cm}^{-1} \mathrm{M}^{-1}\). (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 \(0.250\) within \(30.0\) 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

Expert verified
(a) The initial concentration of the colored reactant is \(c = \frac{0.605}{(5.60 \times 10^3 \mathrm{cm}^{-1}\mathrm{M}^{-1})(1.00 \thinspace cm)}\). (b) The rate constant is \(k = \frac{\ln(\frac{0.605}{(5.60 \times 10^3 \mathrm{cm}^{-1}\mathrm{M}^{-1})(1.00 \thinspace cm)})}{(30.0 \times 60)}\). (c) The half-life of the reaction is \(t_{1/2} = \frac{\ln(2)}{k}\). (d) The time it takes for the absorbance to fall to \(0.100\) is \(t = \frac{\ln(\frac{c_{initial}}{c_{target}})}{k}\).

Step by step solution

01

Determine the initial concentration

Using the Beer-Lambert Law, we can find the initial concentration of the colored reactant by: \(Absorbance = \epsilon \times b \times c\) where \(\epsilon\) = molar absorptivity constant (\(5.60 \times 10^3 \mathrm{cm}^{-1}\mathrm{M}^{-1}\)) \(b\) = path length (\(1.00 \thinspace cm\)) \(c\) = initial concentration Given that the initial absorbance is \(0.605\), we can solve for the initial concentration: \(0.605 = (5.60 \times 10^3 \mathrm{cm}^{-1}\mathrm{M}^{-1})(1.00 \thinspace cm)c\) Solving for \(c\), we get: \(c = \frac{0.605}{(5.60 \times 10^3 \mathrm{cm}^{-1}\mathrm{M}^{-1})(1.00 \thinspace cm)}\)
02

Calculate the rate constant

We are given that the absorbance falls to \(0.250\) within \(30.0\) minutes. In order to determine the rate constant, we first need to find the concentrations corresponding to these absorbances. As described in Step 1: \(c_{initial} = \frac{0.605}{(5.60 \times 10^3 \mathrm{cm}^{-1}\mathrm{M}^{-1})(1.00 \thinspace cm)}\) \(c_{final} = \frac{0.250}{(5.60 \times 10^3 \mathrm{cm}^{-1}\mathrm{M}^{-1})(1.00 \thinspace cm)}\) Now, we can plug those values into the formula for a first-order reaction: \(\ln(\frac{c_{initial}}{c_{final}}) = k \times t\) where \(k\) = rate constant \(t\) = time in seconds (\(30.0 \times 60\)) Solving for \(k\), we get: \(k = \frac{\ln(\frac{0.605}{(5.60 \times 10^3 \mathrm{cm}^{-1}\mathrm{M}^{-1})(1.00 \thinspace cm)})}{(30.0 \times 60)}\)
03

Calculate the half-life of the reaction

With the rate constant from Step 2, we can find the half-life of the reaction using the first-order half-life formula: \(t_{1/2} = \frac{\ln(2)}{k}\)
04

Calculate the time for the absorbance to fall to a specific value

Now, we will determine the time it takes for the absorbance to fall to \(0.100\). First, we will calculate the concentration corresponding to this absorbance: \(c_{target} = \frac{0.100}{(5.60 \times 10^3 \mathrm{cm}^{-1}\mathrm{M}^{-1})(1.00 \thinspace cm)}\) Now, we can use the first-order reaction formula to find the time: \(\ln(\frac{c_{initial}}{c_{target}}) = k \times t\) Solving for \(t\), we get: \(t = \frac{\ln(\frac{c_{initial}}{c_{target}})}{k}\)

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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, a fundamental principle in spectroscopy, connects the absorption of light to the properties of the material it passes through. This law is crucial for determining concentrations of solutions in chemistry. It states that absorbance (A) is directly proportional to the concentration (c) of the absorbing species in the sample and the path length (b) of the light through the sample: \[A = \epsilon \times b \times c\]where \(\epsilon\) is the molar absorptivity or extinction coefficient, which signifies how strongly a substance absorbs light at a given wavelength. The initial exercise utilized this law to determine the concentration of a colored reactant, allowing students to calculate effectively using given absorbance and path length values.
Molar absorptivity
Molar absorptivity, represented by \(\epsilon\), is a measure of how well a chemical species absorbs light at a particular wavelength. It plays a crucial role in the Beer-Lambert Law for calculating concentrations. In the context of the exercise, with a molar absorptivity of \(5.60 \times 10^3\, \text{cm}^{-1}\text{M}^{-1}\), it helped determine the initial concentration of the colored reactant from the absorbance. Here are a few important points about molar absorptivity:
  • Units are \(\text{cm}^{-1}\text{M}^{-1}\), ensuring consistency across measurements.
  • Higher \(\epsilon\) values indicate stronger absorption, useful in identifying and quantifying substances.
Conveying the substance's efficiency in absorbing light, this concept is pivotal for quantitative spectroscopic analysis.
Half-life of reaction
The half-life of a reaction is the time required for half of the reactant to be consumed in a chemical reaction. For first-order reactions, the half-life is independent of the initial concentration and can be calculated using the rate constant \(k\). The formula for the half-life of a first-order reaction is:\[t_{1/2} = \frac{\ln(2)}{k}\]In the exercise, once the rate constant was found, this formula was used to ascertain the time it takes for the absorbance to fall to half its initial value. This concept helps chemists understand reaction dynamics and predict how a reaction proceeds over time, making it especially useful when observing concentration changes.
Rate constant
The rate constant \(k\) is an essential factor in the rate laws of chemical reactions, defining the speed of a reaction. For first-order reactions, the rate constant can be determined by tracking how the concentration of a reactive substance decreases over time. In the exercise, the relation:\[\ln\left(\frac{c_{\text{initial}}}{c_{\text{final}}}\right) = k \times t\]was used to calculate \(k\) given the decrease in absorbance over a specific period. Understanding this constant is crucial in predicting reaction rates, which are fundamental for processes in industry and research.
  • Units for first-order reactions are \(\text{s}^{-1}\), ensuring time is appropriately considered.
  • Its value gives insight into how quickly a reaction approaches completion.
Spectroscopy in chemistry
Spectroscopy is a powerful analytical tool in chemistry, utilized for observing and quantifying substances. It involves measuring how a sample interacts with electromagnetic radiation, often leading to insights into a molecule’s structure or concentration. In the original problem, spectroscopy followed the absorption of a colored reactant to study a first-order reaction. Spectroscopy enhanced understanding by:
  • Providing information on concentration changes over time.
  • Allowing calculation of essential kinetic parameters like the rate constant and half-life.
Due to its versatility and non-destructive nature, spectroscopy is widely used in both qualitative and quantitative analysis, significantly progressing research and application in various chemical fields.

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Most popular questions from this chapter

The enzyme invertase catalyzes the conversion of sucrose, a disaccharide, to invert sugar, a mixture of glucose and fructose. When the concentration of invertase is \(4.2 \times 10^{-7} \mathrm{M}\) and the concentration of sucrose is \(0.0077 \mathrm{M}\) invert sugar is formed at the rate of \(1.5 \times 10^{-4} \mathrm{M} / \mathrm{s}\). When the sucrose concentration is doubled, the rate of formation of invert sugar is doubled also. (a) Assuming that the enzyme-substrate model is operative, is the fraction of enzyme tied up as a complex large or small? Explain. (b) Addition of inositol, another sugar, decreases the rate of formation of invert sugar. Suggest a mechanism by which this occurs.

The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{NO}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ \mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism. (c) Identify any intermediates in the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\). If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-} .\) This rapid reaction gives the following rate data: \begin{tabular}{lll} \hline \(\left.\mathrm{OCl}^{-}\right](\mathrm{M})\) & {\(\left[\mathrm{I}^{-}\right](M)\)} & Rate \((M / \mathrm{s})\) \\ \hline \(1.5 \times 10^{-3}\) & \(1.5 \times 10^{-3}\) & \(1.36 \times 10^{-4}\) \\ \(3.0 \times 10^{-3}\) & \(1.5 \times 10^{-3}\) & \(2.72 \times 10^{-4}\) \\ \(1.5 \times 10^{-3}\) & \(3.0 \times 10^{-3}\) & \(2.72 \times 10^{-4}\) \\ \hline \end{tabular} (a) Write the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=\) \(2.0 \times 10^{-3} M\) and \(\left[1^{-}\right]=5.0 \times 10^{-4} M\)

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{HCl}(g)\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{CH}_{3}(g)+\mathrm{HCl}(g)\) $$ \text { Reaction 4: } \mathrm{CH}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g) $$ The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{k} \mathrm{J} / \mathrm{mol}) & \mathrm{E}_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \\ \hline \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. \(\ln\) which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(g)\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that it is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.

NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{array}{r} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{array} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Why is NO considered a catalyst and not an intermediate? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism? If you think not, suggest what might be going on.

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