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A reaction of the form $$aA \longrightarrow Products$$gives a plot of \(\ln [\mathrm{A}]\) versus time (in seconds), which is a straight line with a slope of \(-7.35 \times 10^{-3} .\) Assuming \([\mathrm{A}]_{0}=\) \(0.0100 M,\) calculate the time (in seconds) required for the reaction to reach \(22.9 \%\) completion.

Short Answer

Expert verified
The time required for the reaction to reach 22.9% completion is approximately 392.85 seconds.

Step by step solution

01

Determine the order of the reaction

Given that a plot of \(\ln[\mathrm{A}]\) versus time yields a straight line, this tells us that the reaction is a first-order reaction.
02

Identify the relevant integrated rate law for first-order reactions

For a first-order reaction of the form \(A \longrightarrow Products\), the integrated rate law is \(\ln[\mathrm{A}] = -kt + \ln[\mathrm{A}]_0\), where \(k\) is the rate constant, \([\mathrm{A}]_0\) is the initial concentration of A, and \([\mathrm{A}]\) is the concentration of A at time \(t\).
03

Determine the rate constant

The problem provides the slope of the \(\ln[\mathrm{A}]\) versus time plot, which is equal to the rate constant with a negative sign, i.e. -k = \(-7.35 \times 10^{-3}\). Calculate the value of the rate constant, \(k\): \(k = 7.35 \times 10^{-3}\ \mathrm{s^{-1}}\)
04

Calculate the concentration at 22.9% completion

To find the concentration of A at 22.9% completion (\([\mathrm{A}]_{22.9}\)), use the following formula: \([\mathrm{A}]_{22.9} = [\mathrm{A}]_{0}(1 - 0.229)\) Substitute the given value for \([\mathrm{A}]_{0}\): \([\mathrm{A}]_{22.9} = (0.0100 M)(1 - 0.229) \approx 0.00771 M\)
05

Use the integrated rate law to solve for the time

Now that we have \([\mathrm{A}]_{0}\), \([\mathrm{A}]_{22.9}\), and \(k\), we can use the integrated rate law to solve for the time \(t\). \(\ln[\mathrm{A}]_{22.9} = -kt + \ln[\mathrm{A}]_{0}\) Plug in the values for \([\mathrm{A}]_{22.9}\), \([\mathrm{A}]_{0}\), and \(k\): \(\ln(0.00771) = -(7.35 \times 10^{-3}\ \mathrm{s^{-1}})(t) + \ln(0.0100)\) Now solve for \(t\): \(t \approx 392.85\ \mathrm{s}\) The time required for the reaction to reach 22.9% completion is approximately 392.85 seconds.

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Key Concepts

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

Understanding the Rate Law
In chemistry, the rate law provides a mathematical relationship between the concentration of reactants and the rate at which a chemical reaction proceeds. It's like a recipe that tells us how the amount of ingredients we have affects the speed of cooking a dish. The rate law for a first-order reaction, for example, can be expressed as
\[ \text{Rate} = k[\text{A}] \]
Here, \(k\) represents the rate constant, a unique value that determines the speed of a given reaction under specific conditions. The concentration of the reactant A, denoted as \([\text{A}]\), affects the reaction rate linearly, meaning if you double the concentration of A, the rate of the reaction doubles as well.
For students tackling problems involving rate laws, it is critical to recognize the order of the reaction, as it directly influences how concentration changes over time. Identifying the order correctly leads to proper use of the integrated rate law, which in turn allows for calculations of reaction progress at various time points.
Delving into Reaction Kinetics
Reaction kinetics is all about the speed of chemical reactions. Think of it as the study of the 'race' between reactants turning into products. Factors such as concentration, temperature, and the presence of a catalyst influence this 'race.' It's like understanding the conditions that would make a car go faster or slower on a track.
Kinetics also delves into the mechanisms of reactions, or the 'steps' that take place during the transformation from reactants to products. The slope of a \( \ln[\text{A}]\) versus time plot, for instance, directly relates to the first-order rate constant \(k\). In the exercise we discussed, a negative slope indicates that the concentration of A decreases over time. By analyzing such plots, chemists can infer valuable information about the speed and nature of reactions, crucial for industries relying on chemical processes, such as pharmaceuticals and materials manufacturing.
Chemical Reaction Completion
The completion percentage of a chemical reaction tells us how far along the reaction has progressed towards forming products from reactants. In our exercise example, reaching 22.9% completion means almost a quarter of the initial reactant has been transformed into products.
To visualize this, imagine filling a glass of water to a specific mark – when the water reaches this mark, we know exactly how much of the glass is full, just like knowing what fraction of the reactant has been used up at a certain time.
Calculating the time required to reach a certain completion percentage demands an understanding of the relevant rate law and kinetics principles. By knowing the initial concentration and rate constant, one can find out how long it takes for a given fraction of the reactant to react, which is invaluable in industries for optimizing production timelines and ensuring quality control in reactions.

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Most popular questions from this chapter

Sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) decomposes to sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) and chlorine \(\left(\mathrm{Cl}_{2}\right)\) by reaction in the gas phase. The following pressure data were obtained when a sample containing \(5.00 \times 10^{-2}\) mol sulfury 1 chloride was heated to \(600 . \mathrm{K}\) in a \(5.00 \times 10^{-1}-\mathrm{L}\) container. Defining the rate as $$-\frac{\Delta\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]}{\Delta t}$$ a. determine the value of the rate constant for the decomposition of sulfuryl chloride at \(600 .\) K. b. what is the half-life of the reaction? c. what fraction of the sulfuryl chloride remains after \(20.0 \mathrm{h} ?\)

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \space\mathrm{kJ} / \mathrm{mol} .\) In the presence of a catalyst at \(37^{\circ} \mathrm{C}\) the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of the rate constant. c. The half-life for a reaction, a \(A \longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0}\) d. The half-life for a second-order reaction increases as the reaction proceeds.

The reaction $$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ was studied by performing two experiments. In the first experiment the rate of disappearance of NO was followed in the presence of a large excess of \(\mathrm{O}_{3}\). The results were as follows \(\left(\left[\mathrm{O}_{3}\right]\right.\) remains effectively constant at \(1.0 \times 10^{14}\) molecules/cm \(^{3}\) ): In the second experiment [NO] was held constant at \(2.0 \times 10^{14}\) molecules/cm \(^{3}\). The data for the disappearance of \(\mathbf{O}_{3}\) are as follows: a. What is the order with respect to each reactant? b. What is the overall rate law? c. What is the value of the rate constant from each set of experiments? $$\text { Rate }=k^{\prime}[\mathrm{NO}]^{x} \quad \text { Rate }=k^{\prime \prime}\left[\mathrm{O}_{3}\right]^{y}$$ d. What is the value of the rate constant for the overall rate law? $$\text { Rate }=k[\mathrm{NO}]^{\mathrm{x}}\left[\mathrm{O}_{3}\right]^y$$

A certain first-order reaction is \(45.0 \%\) complete in 65 s. What are the values of the rate constant and the half-life for this process?

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