Question

The reaction of certain chemical compounds can be modeled using a differential equation of the form \(y^{\prime}(t)=-k y^{n}(t),\) where \(y(t)\) is the concentration of the compound for \(t \geq 0, k>0\) is a constant that determines the speed of the reaction, and \(n\) is a positive integer called the order of the reaction. Assume that the initial concentration of the compound is \(y(0)=y_{0}>0\). a. Consider a first-order reaction \((n=1)\) and show that the solution of the initial value problem is \(y(t)=y_{0} e^{-k t}\). b. Consider a second-order reaction \((n=2)\) and show that the solution of the initial value problem is \(y(t)=\frac{y_{0}}{y_{0} k t+1}\). c. Let \(y_{0}=1\) and \(k=0.1 .\) Graph the first-order and secondorder solutions found in parts (a) and (b). Compare the two reactions.

Short answer

Short Answer: In a first-order reaction, the concentration of the compound decreases at a slower rate compared to a second-order reaction for the same initial concentration and reaction rate values. This is demonstrated in their mathematical forms: \(y_1(t) = e^{-0.1t}\) for a first-order reaction and \(y_2(t) = \frac{1}{0.1t + 1}\) for a second-order reaction (given \(y_0 = 1\) and \(k = 0.1\)).

Step-by-step solution

Write down the given differential equation for first-order reaction.

For a first-order reaction, \(n=1\), so the given differential equation becomes: $$y'(t) = -ky(t)$$

Solve the differential equation using the separation of variables method.

With the differential equation in the form \(y'(t) = -ky(t)\), we will separate variables y and t on each side of the equation: $$\frac{dy}{y(t)}=-k dt$$ Now, integrate both sides: $$\int \frac{dy}{y(t)} = -k \int dt$$ This gives us: $$\ln |y(t)| = -kt + C_1$$ Where \(C_1\) is the constant of integration.

Solve for y(t).

Raise both sides as the power of \(e\) in order to remove the natural logarithm: $$y(t) = e^{-kt+C_1}$$ Which can be written as: $$y(t) = e^{-kt}e^{C_1}$$ Now, let \(C_2=e^{C_1}\), so we have: $$y(t) = C_2 e^{-kt}$$

Apply the initial condition to find the particular solution.

Given that \(y(0)=y_0\), substitute \(t=0\) and \(y(t)=y_0\) into our solution: $$y_0 = C_2 e^{-k(0)}$$ Which simplifies to: $$y_0 = C_2 e^0$$ And since \(e^0 = 1\), we have: $$y_0 = C_2$$ Thus, the solution to the initial value problem is: $$y(t) = y_0 e^{-kt}$$ b. Second-order reaction (n=2)

Write down the given differential equation for second-order reaction.

For a second-order reaction, \(n=2\), so the given differential equation becomes: $$y'(t) = -ky^2(t)$$

Solve the differential equation using the separation of variables method.

With the differential equation in the form \(y'(t) = -ky^2(t)\), we will separate variables y and t on each side of the equation: $$\frac{dy}{y^2(t)}=-k dt$$ Now, integrate both sides: $$-\int \frac{dy}{y^2(t)} = k \int dt$$ This gives us: $$\frac{1}{y(t)} = kt + C_3$$ Where \(C_3\) is another constant of integration.

Solve for y(t).

To remove the reciprocal, we can write: $$y(t)=\frac{1}{kt+C_3}$$

Apply the initial condition to find the particular solution.

Given that \(y(0)=y_0\), substitute \(t=0\) and \(y(t)=y_0\) into our solution: $$y_0 = \frac{1}{k(0)+C_3}$$ Which simplifies to: $$y_0 = \frac{1}{C_3}$$ Thus, \(C_3 = \frac{1}{y_0}\), and the solution to the initial value problem is: $$y(t) = \frac{y_0}{y_0 kt+1}$$ c. Comparison of first-order and second-order reactions We will graph the functions we found for the first order \(y_1(t) = e^{-0.1t}\) and second order \(y_2(t) = \frac{1}{0.1t + 1}\) reactions with \(y_0 = 1\) and \(k = 0.1\). Comparing these two reactions, we can see that the first-order reaction decreases from the initial concentration at a slower rate compared to the second-order reaction. This means that in a second-order reaction, the concentration of the compound decreases faster than in a first-order reaction for the same initial concentration and reaction rate values.

Key concepts

Separation of Variables

Separation of Variables is a powerful technique for solving certain types of differential equations. It allows you to solve an equation by "separating" variables, meaning you isolate all terms involving one variable on one side of the equation and all terms involving the other variable on the other side. This makes integration possible on both sides of the equation.

To illustrate this, consider the differential equation for a first-order reaction, which is given by \(y'(t) = -ky(t)\). Using the separation of variables technique, we rearrange this as \(\frac{dy}{y(t)} = -k \, dt\). Now, both sides can be integrated separately, leading to the solution \(\ln |y(t)| = -kt + C\), where \(C\) is the constant of integration. By raising both sides as powers of \(e\), we resolve the natural logarithm to find \(y(t) = Ce^{-kt}\).

This method simplifies solving differential equations, especially when the equation can be rearranged into a format that supports easy integration.

First-Order Reaction

A first-order reaction refers to a chemical reaction that depends linearly on the concentration of only one reactant. Mathematically, the rate at which the concentration changes is proportional to its current concentration. This is expressed by the differential equation: \(y'(t) = -ky(t)\), where \(k\) is the rate constant and \(y(t)\) is the concentration at time \(t\).

The solution to this equation, as derived using separation of variables, is \(y(t) = y_0 e^{-kt}\), where \(y_0\) is the initial concentration. This exponential decay formula illustrates how the concentration decreases over time.

In a first-order reaction:

• The rate at which the concentration changes does not depend on how much time has passed, but solely on the current concentration.
• This results in a logarithmic time relationship, where a plot of \(\ln(y)\) versus \(t\) generally yields a straight line.

Understanding first-order reactions is crucial for predicting the dynamics of many chemical processes, including radioactive decay and some types of drug metabolism in biology.

Second-Order Reaction

Second-order reactions depend on the concentration of one reactant squared, or on the concentrations of two reactants each raised to the first power. Here, we focus on second-order reactions involving a single reactant. The differential equation for such a reaction is \(y'(t) = -ky^2(t)\).

Using separation of variables, we write this as \(-\int \frac{dy}{y^2(t)} = k \int dt\). Integrating both sides gives \(\frac{1}{y(t)} = kt + C_3\), where \(C_3\) is a constant of integration. After applying the initial condition, we find \(y(t) = \frac{y_0}{y_0 kt + 1}\), where \(y_0\) is the initial concentration.

In a second-order reaction:

• The reaction rate is more sensitive to changes in concentration compared to a first-order reaction.
• The concentration decreases more rapidly, especially at higher concentrations.

This higher sensitivity is critical for analyzing reactions where the concentration significantly affects the reaction rate, making it vital for fields like environmental chemistry and enzymatic kinetics.