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(a) For the generic reaction \(\mathrm{A} \rightarrow \mathrm{B}\) what quantity, when graphed versus time, will yield a straight line for a first- order reaction? (b) How can you calculate the rate constant for a first-order reaction from the graph you made in part (a)?

Short Answer

Expert verified
(a) For a first-order reaction, plotting \(ln[\mathrm{A}]\) against time (\(t\)) will yield a straight line with a slope of \(-k\). (b) To calculate the rate constant (\(k\)), find the slope of the line \(ln[\mathrm{A}]\) versus \(t\) by selecting two points on the line and calculating the change in y divided by the change in x. The rate constant \(k\) is equal to the opposite of the slope: \[k = -\frac{ln[\mathrm{A}]_2 - ln[\mathrm{A}]_1}{t_2 - t_1} \]

Step by step solution

01

Understanding First-order Reactions

A first-order reaction is a type of chemical reaction where the rate of the reaction is directly proportional to the concentration of the reactant. Mathematically, it is represented as: \[rate = k[\mathrm{A}] \] Where \(rate\) is the reaction rate, \(k\) is the rate constant, and \([\mathrm{A}]\) represents the concentration of the reactant A.
02

Determine the Quantity to be Plotted

To obtain a straight line, we need to plot a graph for which the slope is constant. Since the reaction rate relates to the concentration, we should obtain the integrated rate law for the first-order reaction. This equation shows how the concentration of the reactant A changes with time. By integrating the first-order reaction rate equation, it can be shown that: \[ln[\mathrm{A}] = ln[\mathrm{A}]_0 - kt \] Where \([\mathrm{A}]\) represents the concentration of A at a particular time \(t\), \([\mathrm{A}]_0\) is the initial concentration of A, and \(k\) is the rate constant. From this equation, we observe that plotting \(ln[\mathrm{A}]\) against \(t\) will produce a straight line with a slope of \(-k\).
03

Calculate the Rate Constant from the Graph

To calculate the rate constant (\(k\)), we can use the straight line graph created in step 2. Since the slope of the line equals \(-k\), we can find the slope of the line \(ln[\mathrm{A}]\) versus \(t\) by selecting two points on the line and calculating the change in y divided by the change in x. Mathematically, the slope (\(-k\)) can be calculated as: \[-k = \frac{ln[\mathrm{A}]_2 - ln[\mathrm{A}]_1}{t_2 - t_1} \] After calculating the slope, we determine the rate constant \(k\) by switching the sign of the slope (multiplying by -1): \[k = -\frac{ln[\mathrm{A}]_2 - ln[\mathrm{A}]_1}{t_2 - t_1} \] By using the graph to find the slope, we can calculate the rate constant for the first-order reaction.

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Key Concepts

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

Chemical Kinetics
Understanding the speed at which chemical reactions occur is essential for scientists and engineers who need to control and optimize various processes. Chemical kinetics, the branch of physical chemistry that deals with understanding the rates of chemical reactions, is crucial for designing reactors, pharmaceuticals, and even environmental systems.

At the heart of chemical kinetics is the idea that the rate of a chemical reaction is influenced by various factors, including the presence of catalysts, the concentration of reactants, and the temperature. These factors change the way molecules collide and interact, ultimately affecting how quickly products are formed from reactants.

By studying the kinetics of a reaction, one can predict how concentration changes over time and how different conditions affect the speed of the reaction, enabling us to harness chemistry for practical applications.
Reaction Rate
The reaction rate tells us how fast a reactant is being consumed or how fast a product is being created in a chemical reaction. It's measured by the change in concentration of a reactant or product per unit time. For instance, in a first-order reaction like \(\mathrm{A} \rightarrow \mathrm{B}\), the rate is directly proportional to the concentration of the reactant A. That's why the equation for the rate of a first-order reaction can be written as:
\[rate = k[\mathrm{A}] \]
The constant \(k\) in this equation is the rate constant, which is a measure of the intrinsic speed of the reaction and remains constant at a given temperature.

Understanding the reaction rate is not only about knowing how quickly a reactant disappears or a product forms but also about how those changes can be controlled and manipulated for desired outcomes, such as increasing yield in industrial processes or ensuring controlled drug release in pharmaceutical applications.
Integrated Rate Laws
Integrated rate laws are powerful tools in chemical kinetics that relate the concentrations of reactants or products to time. They are obtained by integrating the rate equations, which are based on the order of the reaction. For first-order reactions, where the rate is dependent directly on the concentration of one reactant, the integrated rate law is particularly simple and informative.

As illustrated in the exercise, the integrated rate law for a first-order reaction can be expressed as:
\[ln[\mathrm{A}] = ln[\mathrm{A}]_0 - kt \]
Here, \(ln[\mathrm{A}]_0\) is the natural logarithm of the initial concentration of A, and \(ln[\mathrm{A}]\) is the natural logarithm of the concentration of A at time \(t\). The beauty of this equation is that it gives us a straight-line relationship between \(ln[\mathrm{A}]\) and time, where the slope is equal to the negative of the rate constant \(k\). With a simple experiment and graphing \(ln[\mathrm{A}]\) against \(t\), scientists can quickly determine \(k\), providing valuable insights into the reaction's behavior over time.

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Most popular questions from this chapter

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c),\) has an activation energy of 86.8 \(\mathrm{kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{s}^{-1}\) (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C} .\) (b) A g \(\mathrm{KOH}\) in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reac-solution of \(\mathrm{KOH}\) in ethanol is made up by dissolving 0.335 g KOH in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(35^{\circ} \mathrm{C} ?(\mathbf{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(50^{\circ} \mathrm{C}\) .

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products?

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are 154 \(\mathrm{kJ} / \mathrm{mol}\) and 136 \(\mathrm{kJ} / \mathrm{mol}\) , respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\) . (b) What is the activation energy for the reverse reaction?

Consider the reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} .\) Is each of the following statements true or false? (a) The rate law for the reaction must be Rate \(=k[\mathrm{A}][\mathrm{B}] .\) (b) If the reaction is an elementary reaction, the rate law is second order. (c) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be greater than that for the forward reaction.

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