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

Short Answer

Expert verified
The rate constant (k) for this first-order reaction is approximately \(0.0101 \thinspace s^{-1}\), and the half-life (t鈧/鈧) is approximately 68.6 seconds.

Step by step solution

01

Determine the fraction remaining after 45.0% completion

To determine the fraction remaining after the reaction is 45.0% complete, we need to subtract the completion percentage from 100%. This will give us the remaining percentage of the reactants present after the reaction. Fraction remaining = 1 - 0.450 = 0.550
02

Apply the integrated rate law for first-order reactions

The integrated rate law for a first-order reaction is: \[\ln{\frac{[A]_0}{[A]}} = kt\] Now, we know that the fraction remaining is 0.550, so: \[\ln{\frac{[A]_0}{0.550[A]_0}} = kt\] Since we need to find the rate constant, and the initial concentration will cancel out anyway, we can rewrite the equation as: \[\ln{\frac{1}{0.550}} = k \times 65\]
03

Calculate the rate constant (k)

Now we can solve for k: \[k = \frac{\ln{\frac{1}{0.550}}}{65} \approx 0.0101 \thinspace s^{-1}\] The rate constant k for this first-order reaction is approximately 0.0101 s鈦宦.
04

Calculate the half-life (t鈧/鈧)

Now that we have the rate constant, we can calculate the half-life using the equation: \[t_{1/2} = \frac{0.693}{k}\] Plug in the value of k: \[t_{1/2} = \frac{0.693}{0.0101} \approx 68.6 \thinspace s\] The half-life for this first-order reaction is approximately 68.6 seconds. In conclusion, the rate constant for this process is approximately 0.0101 s鈦宦, and the half-life is approximately 68.6 seconds.

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Key Concepts

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

Rate Constant
In a first-order reaction, the rate constant, often denoted as \( k \), is a crucial factor. It determines the speed at which a reaction progresses. The rate constant is unique to each reaction and is influenced by factors such as temperature and catalysts. For a first-order reaction, the rate constant can be calculated using the formula derived from the integrated rate law:- \[k = \frac{\ln(1/x)}{t}\]- where \( x \) is the fraction of reactants remaining after a given time \( t \).
In our exercise, we calculated the rate constant as \( k \approx 0.0101 \, s^{-1} \). This value indicates how rapidly the reaction occurs, with higher values signifying faster reactions.
Half-Life
The half-life of a reaction, symbolized as \( t_{1/2} \), is the time required for half of the reactants to be converted into products. It is a helpful measure for understanding reaction kinetics, especially in biological and environmental contexts. For a first-order reaction, the half-life is related to the rate constant through the formula:- \[t_{1/2} = \frac{0.693}{k}\]- This equation shows that, regardless of the initial concentration, the half-life only depends on the rate constant.
In our example, with a rate constant of \( k = 0.0101 \, s^{-1} \), the half-life is calculated to be approximately \( 68.6 \, ext{seconds} \). This means that every \( 68.6 \, ext{seconds} \), the amount of reactant is halved.
Integrated Rate Law
The integrated rate law for first-order reactions is a key equation that links concentrations of reactants to time and rate constant. In first-order reactions, the integrated rate law is written as:- \[\ln{\left( \frac{[A]_0}{[A]} \right)} = kt\]- Here, \( [A]_0 \) is the initial concentration and \( [A] \) is the concentration at time \( t \). This formula allows us to calculate the amount of product formed at any given time point.
In our problem, we used the integrated rate law to find the fraction of reactants remaining and to calculate the rate constant \( k \). By simplifying and rearranging the equation using known values, we solve for \( k \) intuitively.
Fraction Remaining
The fraction remaining in a first-order reaction is a simple yet crucial concept. It depicts what portion of the initial reactant remains after a certain period of time. As reactions proceed, reactants convert into products, thus decreasing their quantity. In our exercise, the fraction remaining was calculated as:- \[1 - ext{percent complete} = ext{fraction remaining}\]- For example, when the reaction is \( 45 ext{%} \) complete, the fraction remaining is \( 0.55 \). This metric is vital for further calculations, as it directly impacts the rate constant and helps assess reaction progress efficiently. Understanding this allows us to track how quickly a reaction is progressing and plan accordingly, especially in laboratory settings or chemical manufacturing.

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Most popular questions from this chapter

Write the rate laws for the following elementary reactions. a. \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) b. \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)\) c. \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) d. \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

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 sulfuryl 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 \(600 . \mathrm{K}\). b. what is the half-life of the reaction? c. what fraction of the sulfuryl chloride remains after \(20.0 \mathrm{~h}\) ?

The type of rate law for a reaction, either the differential rate law or the integrated rate law, is usually determined by which data is easiest to collect. Explain.

You and a coworker have developed a molecule that has shown potential as cobra antivenin (AV). This antivenin works by binding to the venom (V), thereby rendering it nontoxic. This reaction can be described by the rate law $$ \text { Rate }=k[\mathrm{AV}]^{1}[\mathrm{~V}]^{1} $$ You have been given the following data from your coworker: $$ \begin{aligned} [\mathrm{V}]_{0} &=0.20 \mathrm{M} \\ [\mathrm{AV}]_{0} &=1.0 \times 10^{-4} \mathrm{M} \end{aligned} $$ A plot of \(\ln [\mathrm{AV}]\) versus \(t\) (s) gives a straight line with a slope of \(-0.32 \mathrm{~s}^{-1}\). What is the value of the rate constant \((k)\) for this reaction?

The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant? If a reactant has a \(-1\) order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?

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