Question

Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1)\), the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\) c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\)

Short answer

Question: Show the difference in concentration decay for first-order and second-order reactions given the differential equation \(\frac{d y}{d t} = -ky^n\), where \(k = 0.1\) and \(y_0 = 1\). Answer: For a first-order reaction, the concentration decay follows an exponential decay law, given by \(y(t) = e^{-0.1t}\). For a second-order reaction, the concentration decay is given by \(y(t) = \frac{1}{1+0.1t}\). When graphing these functions for different time values, you will notice a difference in decay rates between first-order and second-order reactions.

Step-by-step solution

Insert \(n=1\) into the given differential equation

We are given \(\frac{d y}{d t} = -ky^n\). Set n=1: \(\frac{d y}{d t} = -ky\).

Solve the differential equation

Since this is a first-order linear differential equation, we can use separation of variables to solve it: \(\frac{1}{y} dy = -k dt\). Now we integrate both sides: \(\int \frac{1}{y} dy = \int -k dt\) \(\ln|y| = -kt + C\), where \(C\) is the integration constant.

Find the general solution of the exponential decay law

To find the general solution in terms of exponential decay, we can take the exponent of both sides: \(y(t) = e^{-kt+C} = e^{-kt} e^C\). We can define a constant \(A = e^C\), which gives us the exponential decay law \(y(t) = A e^{-kt}\). b. Second-order reaction

Insert \(n=2\) into the given differential equation

We are given \(\frac{d y}{d t} = -ky^n\). Set n=2: \(\frac{d y}{d t} = -ky^2\).

Solve the initial value problem

To solve the initial value problem, we can use separation of variables again: \(\frac{1}{y^2} dy = -k dt\). Now we integrate both sides given the initial condition \(y(0)=y_0\): \(\int_{y_0}^{y} \frac{1}{\xi^2} d\xi = -k \int_{0}^{t} d\tau\) \(-\frac{1}{y} + \frac{1}{y_0} = -k t\).

Find the solution for the second-order reaction

Now we solve for \(y(t)\): \(y(t) = \frac{1}{\frac{1}{y_0} + kt}\). c. Graph the concentration

Substitute the given values into the solutions

For both first-order and second-order reactions, substitute the values \(k=0.1\) and \(y_0=1\): First-order reaction: \(y(t) = A e^{-0.1t}\) (We use \(A=1\) since \(y(0)=1\)) Second-order reaction: \(y(t) = \frac{1}{1+0.1t}\).

Graph the concentration functions

Using a graphing tool, plot the functions \(y(t) = e^{-0.1t}\) and \(y(t) = \frac{1}{1+0.1t}\) for different time values of \(t\). You will notice the difference in decay for first-order and second-order reactions.

Key concepts

First-order Reaction

In a first-order reaction, the rate at which a reactant is consumed is directly proportional to its concentration. This is reflected in the differential equation \( \frac{dy}{dt} = -ky \), where \( y(t) \) is the concentration of the reactant at time \( t \), and \( k \) is the reaction rate constant. To understand why this leads to exponential decay, consider how over time, the rate of change decreases as the concentration decreases.

### Exponential Decay PatternBy solving this differential equation using separation of variables, we integrate both sides to find that the concentration behaves according to the equation \( y(t) = A e^{-kt} \). Here, \( A \) is the initial concentration when \( t = 0 \). This formula shows how concentration decreases exponentially over time. The value \( k \) influences how fast or slow this decay occurs. Larger \( k \) values indicate a faster reaction and more rapid decay.

### Real-Life ApplicationIn many natural processes and chemical reactions, such as radioactive decay or the decomposition of pollutants, we see first-order kinetics. This is because they decompose or transform at a rate proportional to how much substance is present at any given moment.

Second-order Reaction

Second-order reactions present a more complex behavior as they typically involve two molecules interacting directly, or the concentration of a single molecule squared affecting the rate. The differential equation modeling a second-order reaction is \( \frac{dy}{dt} = -ky^2 \). When separated and integrated with initial conditions, this doesn't simplify to exponential decay but rather a diminished response over time.

### Solving the Second-Order EquationTo solve \( \frac{dy}{dt} = -ky^2 \), we employ separation of variables: \( \int_{y_0}^{y} \frac{1}{\xi^2} d\xi = -k \int_{0}^{t} d\tau \). Upon integrating, we get \( -\frac{1}{y} + \frac{1}{y_0} = -kt \). Rearranging gives \( y(t) = \frac{1}{\frac{1}{y_0} + kt} \). This demonstrates how the reaction results in a hyperbolic decay, unlike the exponential trend in first-order reactions.

### Characteristic CurveThe characteristic curve of second-order reactions reveals how concentration doesn't just decline smoothly over time but instead follows a curve where as \( t \) increases, the concentration approaches zero more gradually than in an exponential model. This has notable implications in reactions where the reactant concentration plays a critical role, ranging from enzyme-substrate interactions in biochemistry to certain mixed chemical systems.

Exponential Decay

Exponential decay happens when the rate of change of a quantity is proportional to the initial amount of that quantity, manifesting in a steady decline over time. Mathematically represented as \( y(t) = A e^{-kt} \), it clearly outlines how crucial the rate constant \( k \) and the initial concentration \( A \) are in determining the behavior of the decay.

### Recognizing Exponential DecayIn the case of chemical reactions, exponential decay is seen in processes such as radioactive decay, where the decay of nuclei occurs at a rate proportional to the present number of undecayed nuclei. This is analogous to first-order reactions in chemistry, making it a crucial concept in both fields.

### Impact of the Rate ConstantThe rate constant \( k \) plays a decisive role: higher values of \( k \) result in faster decay, meaning the substance or entity diminishes quickly. It has implications across various fields, from predicting the half-life of substances to understanding pharmacokinetics in biological systems, where drugs follow first-order decay to decrease concentration in the body's systems.

Understanding exponential decay's pattern aids learners and professionals in predicting how different parameters will influence the lifetimes of substances and their behavior over time.