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Chapter 1: Problem 4

Consider a chemical reaction $$ A+B \longrightarrow A B $$ in which a chemical, \(A,\) combines with a chemical, \(B,\) to form the compound, \(A B\). Assume that the amount of \(B\) greatly exceeds the amount of \(A,\) and that in any second, the amount of \(A B\) that is formed is proportional to the amount of \(A\) present at the beginning of the second. Write a dynamic equation for this reaction, and write a solution equation to the dynamic equation.

Short Answer

Expert verified
The dynamic equation is \( \frac{d[A]}{dt} = -k[A] \); solution is \( [A] = [A]_0 e^{-kt} \).

Step by step solution

01

Understand the Reaction Dynamics

The reaction involves chemicals A and B reacting to form compound AB. Since the amount of B is significantly higher than A, we assume B remains constant. Thus, the rate at which AB is formed is dependent primarily on the amount of A.
02

Define the Rate of Reaction

The rate of formation of AB is proportional to the amount of A present. Mathematically, this rate can be expressed as \( \frac{d[AB]}{dt} = k[A] \), where \( k \) is the rate constant, \( [A] \) is the concentration of A, and \( [AB] \) is the concentration of AB.
03

Set up the Differential Equation for A

Assuming no back reaction or other reactions, the rate of change of A is \( \frac{d[A]}{dt} = -k[A] \) because A is being consumed.
04

Solve the Differential Equation

The differential equation \( \frac{d[A]}{dt} = -k[A] \) is a separable equation. Separating variables: \( \frac{d[A]}{[A]} = -k \, dt \). Integrating both sides gives \( \ln [A] = -kt + C \). Exponentiating both sides, we get \( [A] = [A]_0 e^{-kt} \), where \( [A]_0 \) is the initial concentration of A.
05

Write the Solution for AB

Since \( [AB] = [A]_0 - [A] \), the concentration of AB over time will be \( [AB] = [A]_0 (1 - e^{-kt}) \).

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the area of chemistry that concerns itself with the speed or rate at which chemical reactions occur. It is pivotal to understand how different conditions such as temperature, concentration of reactants, and catalysts can affect the rate of reaction. In the realm of chemical kinetics, there are some key principles to grasp:
  • Rate of Reaction: It refers to how fast or slow a reaction takes place. For instance, this can be described quantitatively by the change in concentration of reactants/products over time.
  • Rate Equations: These equations mathematically describe how the speed of a reaction is dependent on the concentration of reactants. In many reactions, this is represented by the rate constant, denoted as \( k \).
  • Factors Affecting Reaction Rates: Concentration, temperature, surface area, and catalysts are crucial factors that influence how rapidly a reaction proceeds.
Understanding chemical kinetics helps in designing experiments and optimizing reactions to achieve desirable rates, which is particularly important in industrial chemical processes.
Reaction Rate
The reaction rate is a fundamental concept in chemical kinetics established to measure how quickly a chemical reaction progresses. It is linked to the change in concentration of reactants or products over a given period:
  • Instantaneous Rate: This is the rate at a specific moment in the reaction, usually obtained from the slope of a concentration vs. time graph.
  • Average Rate: This is calculated over a time interval by the change in concentration of a reactant or product divided by the change in time.
  • Unit Observations: The units of reaction rate often depend on the specifics of the reaction, commonly measured in moles per liter per second (mol/L/s).
In our exercise, the formation of compound \( AB \) is proportional to the amount of \( A \) present. This means that the reaction rate can be expressed through the equation \( \frac{d[AB]}{dt} = k[A] \). This equation shows that as \( [A] \) decreases over time, due to its conversion into \( AB \), the rate of the reaction will also change accordingly.
Separable Equations
A separable differential equation is a type of differential equation that can be broken down into two distinct parts, each involving a single variable. This makes them relatively easier to solve compared to other equations. The exercise we went through involves a separable differential equation, which models the rate of consumption of chemical \( A \):
  • Form: A separable equation can be expressed as \( \frac{dy}{dx} = g(y)h(x) \).
  • Method: To solve, one rearranges it to \( \frac{dy}{g(y)} = h(x)dx \). This separation simplifies integration.
For example, in the chemical reaction exercise, \( \frac{d[A]}{dt} = -k[A] \) is a classic separable equation. Solving it entails separating and integrating both sides:
  • First, rewrite: \( \frac{d[A]}{[A]} = -k \cdot dt \).
  • Then, integrate: \( \int \frac{d[A]}{[A]} = -\int k \, dt \).
  • This leads to: \( \ln [A] = -kt + C \), solving for \( [A] \), we exponentiate to get \( [A] = [A]_0 e^{-kt} \).
Such exercises teach invaluable skills in handling separable differential equations, which find applications across various scientific fields.

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Most popular questions from this chapter

There is a suggestion that the world human population is growing exponentially. Shown below are the human population numbers in billions of people for the decades \(1940-2010\). \(\begin{array}{lrrrrrrrr}\text { Year } & 1940 & 1950 & 1960 & 1970 & 1980 & 1990 & 2000 & 2010 \\ \text { Index, } t & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text { Human Population } \times 10^{6} & 2.30 & 2.52 & 3.02 & 3.70 & 4.45 & 5.30 & 6.06 & 6.80\end{array}\) 1\. Test the equation $$ P_{t}=2.21 .19^{t} $$ against the data where \(t\) is the time index in decades after 1940 and \(P_{t}\) is the human population in billions. 2\. What percentage increase in human population each decade does the model for the equation assume? 3\. What world human population does the equation predict for the year \(2050 ?\)

The polymerase chain reaction is a means of making multiple copies of a DNA segment from only a minute amounts of original DNA. The procedure consists of a sequence of, say, 30 cycles in which each segment present at the beginning of a cycle is duplicated once; at the end of the cycle that segment and one copy is present. Introduce notation and write a difference equation with initial condition from which the amount of DNA present at the end of each cycle can be computed. Suppose you begin with 1 picogram \(=0.000000000001 \mathrm{~g}\) of DNA. How many grams of DNA would be present after 30 cycles.

Show that the doubling time of \(y=A B^{t}\) is \(1 /\left(\log _{2} B\right)\)

Show that \(y=A B^{t}\) with \(B<1\) has a half-life of $$ T_{\text {Half }}=\frac{\log \frac{1}{2}}{\log B}=\frac{-\log 2}{\log B} $$

The equation, \(B_{t}-B_{t-1}=r B_{t-1},\) carries the same information as \(B_{t+1}-B_{t}=r B_{t}\) a. Write the first four instances of \(B_{t}-B_{t-1}=r B_{t-1}\) using \(t=1, t=2, t=3,\) and \(t=4\). b. Cascade these four equations to get an expression for \(B_{4}\) in terms of \(r\) and \(B_{0}\). c. Write solutions to and compute \(B_{40}\) for (a.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=0.2 B_{t-1}\) (b.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=0.1 B_{t-1}\) (c.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=0.05 B_{t-1}\) (d.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=-0.1 B_{t-1}\)
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