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Magazine Chapter 12: Problem 270
For a first order chemical reaction the law of mass action for a single reacting chemical is given by \(\mathrm{y}^{\prime}=-\mathrm{ky}\) where \(\mathrm{k}\) is the rate constant. The law of mass action for a second order chemical reaction is given by \(y^{\prime}=-k y^{2}\) where \(y(0)=a_{0}\). Solve the differential equations and evaluate the expression for the half life of the two chemical reactions.
Short Answer
Expert verified
The half-life expressions for the two chemical reactions are:
For the first-order reaction: \(t_{1/2} = \frac{\ln(2)}{k}\)
For the second-order reaction: \(t_{1/2} = \frac{1}{k a_0}\)
Step by step solution
01
Solve the first-order differential equation
We will solve the given first-order differential equation: \(y' = -ky\).
We can rewrite this differential equation as: \(\frac{dy}{dt} = -ky\).
To solve this equation, we can use separation of variables. We can move the terms involving y to one side of the equation and the terms involving t to the other side:
\(\frac{dy}{y} = -kdt\)
Now, integrate both sides of the equation:
\(\int_{y_0}^{y} \frac{d y}{y} = -k\int_{0}^{t} dt\)
\(\ln\left(\frac{y}{y_0}\right) = -kt + C_1\)
Solve for y:
\(y(t) = y_0 e^{-kt}\)
This is the general solution for the first-order reaction (note that we could also absorb the constant of integration inside \(y_0\)).
02
Solve the second-order differential equation
We will solve the given second-order differential equation: \(y'= -ky^2\), with the initial condition \(y(0) = a_0\).
We can rewrite this differential equation as: \(\frac{dy}{dt} = -ky^2\).
Use separation of variables:
\(\frac{dy}{y^2} = -kdt\)
Now, integrate both sides of the equation:
\(\int_{a_0}^{y} \frac{d y}{y^2} = -k\int_{0}^{t} dt\)
\(\left[-\frac{1}{y}\right]_{a_0}^{y} = -kt\)
\(\frac{1}{a_0} - \frac{1}{y} = kt\)
Solve for y:
\(y(t) = \frac{a_0}{1 + kt a_0}\)
This is the general solution for the second-order reaction.
03
Find the half-life expressions for each reaction
The half-life of a reaction is the time it takes for the concentration of the chemical to decrease to half of its initial value. Therefore, we need to find the time t when \(y(t) = \frac{1}{2}y_0\) for the first-order reaction and \(y(t) = \frac{1}{2}a_0\) for the second-order reaction.
First-order reaction half-life:
\(\frac{1}{2}y_0 = y_0 e^{-kt_{1/2}}\)
Divide both sides by \(y_0\) and take the natural logarithm of both sides:
\(\ln(1/2) = -kt_{1/2}\)
Solve for \(t_{1/2}\):
\(t_{1/2} = \frac{\ln(2)}{k}\)
This is the half-life expression for the first-order reaction.
Second-order reaction half-life:
\(\frac{1}{2}a_0 = \frac{a_0}{1 + kt_{1/2} a_0}\)
Multiply both sides by \(1 + kt_{1/2} a_0\) and divide by \(\frac{1}{2}a_0\):
\(1 + kt_{1/2} a_0 = 2\)
Solve for \(t_{1/2}\):
\(t_{1/2} = \frac{1}{k a_0}\)
This is the half-life expression for the second-order reaction.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Chemical Reaction
In the realm of chemistry, a first-order chemical reaction refers to a process where the rate at which the reaction occurs is directly proportional to the concentration of a single reactant. The phenomenon is described by the differential equation \(y' = -ky\), which embodies the law of mass action. This law postulates that for a given single reacting chemical, the rate of reaction is contingent on its concentration at any given time.
To solve this equation, the method of separation of variables is employed. After separating the variables and integrating, we are left with the expression \(y(t) = y_0 e^{-kt}\), which represents the concentration of the reactant at any time \(t\). This form is immensely significant as it not only relates the concentration to time but also introduces the rate constant \(k\), a pivotal factor in determining the speed of the reaction.
To solve this equation, the method of separation of variables is employed. After separating the variables and integrating, we are left with the expression \(y(t) = y_0 e^{-kt}\), which represents the concentration of the reactant at any time \(t\). This form is immensely significant as it not only relates the concentration to time but also introduces the rate constant \(k\), a pivotal factor in determining the speed of the reaction.
Second-Order Chemical Reaction
Differing from first-order reactions, a second-order chemical reaction has a rate that is proportional to the square of the concentration of the reactant. Expressed as \(y'=-ky^2\), a more complex relationship emerges. The rate is linked to the concentration in a quadratic manner, indicating a more rapid change as the concentration increases.
When the variables are isolated and integrated, the solution brought forth is \(y(t) = \frac{a_0}{1 + kta_0}\), where \(a_0\) designates the initial concentration. This formula gives us a clearer understanding of how the concentration diminishes over time in a second-order reaction, and it accentuates the impact of both the initial concentration and the rate constant on the reaction's progress.
When the variables are isolated and integrated, the solution brought forth is \(y(t) = \frac{a_0}{1 + kta_0}\), where \(a_0\) designates the initial concentration. This formula gives us a clearer understanding of how the concentration diminishes over time in a second-order reaction, and it accentuates the impact of both the initial concentration and the rate constant on the reaction's progress.
Rate Constant
The rate constant, symbolized as \(k\), is a fundamental aspect in the mathematical expressions for both first-order and second-order reactions. It appraises the reaction's velocity and isn't influenced by the concentration of reactants or the amount of material present. The magnitude of \(k\) defines the reaction speed: a higher \(k\) translates to a swifter reaction.
This constant is unique to each reaction and can be influenced by factors such as temperature, presence of catalysts, or the medium in which the reaction takes place. It's crucial for predicting how the reaction proceeds over time and is essential in determining the half-life of the reaction, which is a measure of how quickly the reactant concentration declines to half of its initial value.
This constant is unique to each reaction and can be influenced by factors such as temperature, presence of catalysts, or the medium in which the reaction takes place. It's crucial for predicting how the reaction proceeds over time and is essential in determining the half-life of the reaction, which is a measure of how quickly the reactant concentration declines to half of its initial value.
Half-Life Expression
The concept of half-life is integral to the understanding of kinetic processes in chemistry. It denotes the time frame required for the concentration of a reactant in a chemical reaction to be reduced by half. This period, symbolized as \(t_{1/2}\), has distinct expressions for first-order and second-order reactions.
For a first-order reaction, the half-life is given by \(t_{1/2} = \frac{\ln(2)}{k}\), showing that it is not dependent on the initial concentration but solely on the constant \(k\). Conversely, the half-life of a second-order reaction is expressed as \(t_{1/2} = \frac{1}{ka_0}\), indicating an inverse relationship with both the rate constant and the starting concentration. These formulas provide critical insights into the duration and progression of chemical reactions.
For a first-order reaction, the half-life is given by \(t_{1/2} = \frac{\ln(2)}{k}\), showing that it is not dependent on the initial concentration but solely on the constant \(k\). Conversely, the half-life of a second-order reaction is expressed as \(t_{1/2} = \frac{1}{ka_0}\), indicating an inverse relationship with both the rate constant and the starting concentration. These formulas provide critical insights into the duration and progression of chemical reactions.
Law of Mass Action
The law of mass action underpins the mathematical relationships for reaction rates. It asserts that the rate of a chemical reaction at equilibrium is proportional to the product of the concentrations of the reactants raised to the power of their respective stoichiometric coefficients in the balanced equation. This law is exemplified in the first-order reaction \(y' = -ky\), signifying rate dependence on a single reactant's concentration.
Furthermore, in the context of a second-order reaction \(y'=-ky^2\), it proposes a deeper complexity with the rate being proportional to the square of the concentration. Its elegance lies in its capacity to condense the complex interactions of reacting species into a straightforward mathematical model.
Furthermore, in the context of a second-order reaction \(y'=-ky^2\), it proposes a deeper complexity with the rate being proportional to the square of the concentration. Its elegance lies in its capacity to condense the complex interactions of reacting species into a straightforward mathematical model.
Separation of Variables
The technique of separation of variables is vital for solving differential equations that arise in chemical kinetics, among many other fields. It involves manipulating the equation to isolate the variables (for instance, \(y\) and \(t\)) on opposite sides, followed by integrating them separately to find a relationship between the variables.
For instance, in resolving the differential equation for a first-order reaction, this method allows us to transition from \(\frac{dy}{dt} = -ky\) to the integrated form鈥攁nd eventually to the explicit function \(y(t) = y_0 e^{-kt}\), detailing the concentration over time. It is a pivotal tool in understanding how concentrations of reactants change during a reaction, unlocking the ability to predict and control chemical processes.
For instance, in resolving the differential equation for a first-order reaction, this method allows us to transition from \(\frac{dy}{dt} = -ky\) to the integrated form鈥攁nd eventually to the explicit function \(y(t) = y_0 e^{-kt}\), detailing the concentration over time. It is a pivotal tool in understanding how concentrations of reactants change during a reaction, unlocking the ability to predict and control chemical processes.
