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

A proposed mechanism for a reaction is $$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{Br}^{-} \quad \text { Slow }$$ $$\mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+} \quad \text { Fast }$$ $$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{H}_{3} \mathrm{O}^{+}\quad \text { Fast }$$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction? What are the intermediates in the proposed mechanism?

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The decomposition of iodoethane in the gas phase proceeds according to the following equation: $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HI}(g)$$ At \(660 . \mathrm{K}, k=7.2 \times 10^{-4} \mathrm{s}^{-1} ;\) at \(720 . \mathrm{K}, k=1.7 \times 10^{-2} \mathrm{s}^{-1}\) What is the value of the rate constant for this first-order decomposition at \(325^{\circ} \mathrm{C} ?\) If the initial pressure of iodoethane is 894 torr at \(245^{\circ} \mathrm{C},\) what is the pressure of iodoethane after three half-lives?

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