/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 106 The following scheme in which \(... [FREE SOLUTION] | 91Ó°ÊÓ

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The following scheme in which \(\mathrm{A}\) is converted to \(\mathrm{B}\), which is then converted to \(\mathrm{C}\), is known as a consecutive reaction: \(\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C}\) Assuming that both steps are first order, sketch on the same graph the variations of \([\mathrm{A}],[\mathrm{B}],\) and \([\mathrm{C}]\) with time.

Short Answer

Expert verified
Concentrations of A decrease, B rise and fall, and C increase over time.

Step by step solution

01

Understanding the Reaction

We have a consecutive reaction where substance A is converted to B, and B is then converted to C. Both steps are first-order reactions.
02

Write the Rate Equations

For the reactions, we have the rate equations: 1. For A: \( \frac{d[A]}{dt} = -k_1[A] \)2. For B: \( \frac{d[B]}{dt} = k_1[A] - k_2[B] \)3. For C: \( \frac{d[C]}{dt} = k_2[B] \)
03

Solve the Rate Equations

The solution involves solving these ordinary differential equations:1. \([A] = [A]_0 e^{-k_1 t}\)2. For [B], solve: \( \frac{d[B]}{dt} = k_1[A]_0 e^{-k_1 t} - k_2[B] \)3. For [C], \( [C] = [A]_0 - [A] - [B] \)
04

Sketch the Graph

On a graph, plot the concentrations over time:1. \([A]\) starts at \([A]_0\) and decreases exponentially.2. \([B]\) starts at 0, rises to a peak, and then decreases.3. \([C]\) starts at 0 and increases, approaching \([A]_0\).
05

Interpret the Graph

The concentration of A decreases continuously. B increases initially, peaks, and then decreases as it is converted to C. C increases steadily as B converts to C.

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Key Concepts

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

First-Order Reactions
In chemistry, a first-order reaction is a type of reaction where the rate is directly proportional to the concentration of one reactant. This means that the speed at which the reaction occurs depends only on the concentration of one substance. In first-order reactions, the rate of reaction can be described mathematically by the expression: \[ \text{Rate} = k \cdot [A] \] where \( k \) is the rate constant, and \([A]\) is the concentration of the reactant. ### Characteristics of First-Order Reactions - **Exponential Decay:** The concentration of the reactant decreases exponentially over time. - **Constant Half-Life:** The time taken for the concentration to decrease by half is constant, regardless of the initial concentration. First-order reactions are common and important in understanding the kinetics of many chemical processes. In the case of consecutive reactions, each step can follow first-order kinetics, making them predictable and mathematically describable.
Rate Equations
Rate equations are mathematical representations that describe how the concentration of reactants impacts the speed of a chemical reaction. For complex reactions like consecutive reactions, rate equations can help us understand the interaction between different steps. In our scenario, the two consecutive first-order reactions can be described by the following rate equations: - For A converting to B: \( \frac{d[A]}{dt} = -k_1[A] \) - For B converting to C: \( \frac{d[B]}{dt} = k_1[A] - k_2[B] \) - For the formation of C: \( \frac{d[C]}{dt} = k_2[B] \) These equations highlight how the concentration of each substance changes over time. The equations are crucial for determining how long it takes for certain concentrations to be reached, and they can help scientists predict the outcomes of reactions under different conditions.
Differential Equations
Differential equations are essential tools in modeling how systems change over time, especially in chemical kinetics. For consecutive reactions, the rate equations are expressed as a set of differential equations: - **Decay of A:** \( \frac{d[A]}{dt} = -k_1[A] \) - **Formation and decay of B:** \( \frac{d[B]}{dt} = k_1[A] - k_2[B] \) - **Formation of C:** \( \frac{d[C]}{dt} = k_2[B] \) ### Solving Differential Equations To predict how concentrations change over time, these differential equations need to be solved. For first-order reactions, this typically involves integrating the equations. The solutions provide expressions for the concentrations of A, B, and C as functions of time: - \([A] = [A]_0 e^{-k_1 t}\) - Solving for \([B]\) and \([C]\) involves more complex integration steps. This process helps us model how quickly substances are converted in consecutive reactions.
Concentration Graphs
Concentration graphs visually represent how the concentrations of reactants and products change over time during a reaction. For our consecutive reaction: - **Initial Concentrations:** The graph starts with \([A] = [A]_0\), \([B] = 0\), and \([C] = 0\). - **Changes Over Time:** - \([A]\) decreases exponentially, depicted as a downward slope. - \([B]\) initially increases to a peak as A converts to B, and then decreases as B is transformed to C. - \([C]\) steadily increases as B converts to C, eventually plateauing when conversion completes. ### Interpreting Graphs - **Key Features:** Each curve represents the dynamic change in concentration of A, B, and C. - **Peaks and Plateaus:** The peak of \([B]\) indicates the maximum concentration, while the plateau of \([C]\) shows the completion of the reaction. These graphs provide a visual understanding of the reaction kinetics and make it easier to comprehend the transformation process in consecutive reactions.

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Most popular questions from this chapter

When methyl phosphate is heated in acid solution, it reacts with water: $$ \mathrm{CH}_{3} \mathrm{OPO}_{3} \mathrm{H}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}_{3} \mathrm{PO}_{4} $$ If the reaction is carried out in water enriched with \({ }^{18} \mathrm{O},\) the oxygen- 18 isotope is found in the phosphoric acid product but not in the methanol. What does this tell us about the mechanism of the reaction?

To carry out metabolism, oxygen is taken up by hemoglobin \((\mathrm{Hb})\) to form oxyhemoglobin \(\left(\mathrm{Hb} \mathrm{O}_{2}\right)\) according to the simplified equation: $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times 10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} \mathrm{M}\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} M / \mathrm{s}\) during exercise to meet the demand of the increased metabolism rate. Assuming the \(\mathrm{Hb}\) concentration to remain the same, what must the oxygen concentration be to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

What is the rate-determining step of a reaction? Give an everyday analogy to illustrate the meaning of rate determining.

Determine the overall orders of the reactions to which the following rate laws apply: (a) rate \(=k\left[\mathrm{NO}_{2}\right]^{2},(\mathrm{~b})\) rate \(=k\), (c) rate \(=k\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{Br}_{2}\right]^{1 / 2}\) (d) rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\)

Many reactions involving heterogeneous catalysts are zeroth order; that is, rate \(=k\). An example is the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) over tungsten \((\mathrm{W})\) $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ It is found that the reaction is independent of \(\left[\mathrm{PH}_{3}\right]\) as long as phosphine's pressure is sufficiently high \((\geq 1\) atm \()\). Explain.

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