Question

Chemical rate equations 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 \(\frac{d y}{d t}=-k y^{n},\) for \(t \geq 0,\) 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

Short Answer: For a first-order reaction, the concentration of a substance follows an exponential decay law, given by the equation \(y(t) = Ce^{-kt}\). For a second-order reaction, the equation is \(y(t) = \frac{y_0}{1 + kt y_0}\). When graphing these equations with given parameters, the first-order reaction shows an exponential decay, while the second-order reaction shows a more gradual decay.

Step-by-step solution

Understanding the chemical rate equation

The chemical rate equation given is \(\frac{dy}{dt} = -ky^n\), where \(y(t)\) represents the concentration of a substance, \(t\) represents time, \(k\) is the rate constant, and \(n\) is the order of the reaction. We will look at the cases when \(n=1\) and \(n=2\), which represent first-order and second-order reactions, respectively.

Exponential decay law for the first-order reaction

For a first-order reaction, \(n=1\). Thus, our equation becomes $$\frac{dy}{dt} = -ky.$$ To show that the concentration obeys an exponential decay law, we will solve this differential equation. Separate the variables and integrate both sides: $$\int \frac{1}{y} dy = -k \int dt.$$ After integrating, we get $$\ln y = -kt + C,$$ where \(C\) is the constant of integration. To remove the natural logarithm, we can take the exponent of both sides, so we get $$y(t) = e^{-kt+C} = e^C e^{-kt}.$$ Since \(e^C\) is just another constant, we can write $$y(t) = Ce^{-kt},$$ which represents an exponential decay law. Thus, for a first-order reaction, the concentration obeys an exponential decay law.

Solve the initial value problem for the second-order reaction

For a second-order reaction, \(n=2\). Thus, our equation becomes $$\frac{dy}{dt} = -ky^2.$$ Given that the initial concentration is \(y(0)=y_0\), we will solve the initial value problem. Separate the variables and integrate both sides: $$\int \frac{1}{y^2} dy = -k \int dt.$$ After integrating, we get $$-\frac{1}{y} = -kt + C.''' Now, let's find the value of the constant \(C\) using the initial condition \(y(0) = y_0\). Plugging in the initial condition, we get $$-\frac{1}{y_0} = C$$. Thus, we can rewrite the equation as $$-\frac{1}{y} = -kt - \frac{1}{y_0}.$$ Solve for \(y(t)\): $$y(t) = \frac{y_0}{1 + kt y_0}.$$

Graph the concentration

We will now graph the concentration for first-order and second-order reactions with \(k=0.1\) and \(y_0 = 1\). For the first-order reaction, the equation is: $$y(t) = e^{-0.1t}.$$ For the second-order reaction, the equation is: $$y(t) = \frac{1}{1+0.1t}.$$ To graph these, we can use any graphing calculator or software such as Desmos. Plot both equations on the same graph and observe how they decay over time. For the first-order reaction, the concentration decays exponentially, while for the second-order reaction, it decays in a more gradual manner.

Key concepts

Understanding Differential Equations

Differential equations are mathematical tools that describe relationships involving rates of change. In our context, they are crucial for modeling how the concentration of a substance in a chemical reaction changes over time. Essentially, a differential equation will relate the rate of change of a quantity to the quantity itself, often involving other variables. For example, the equation \(\frac{d y}{d t}=-k y^{n}\) is a differential equation where \(\frac{d y}{d t}\) represents the rate of change of the concentration \(y(t)\) with respect to time (\(t\)). The constants \(k\) and \(n\) determine the specific characteristics of the reaction, with \(k\) being the rate constant and \(n\) indicating the order of the reaction. Solving differential equations often requires integrating and applying initial conditions to find a specific solution that fits the situation at hand.

Understanding how to manipulate and solve these equations is critical in many fields, including physics, engineering, and, as we see here, chemistry. When we solve a differential equation, we convert a statement about rates of change into a formula that gives us the quantity of interest at any time, providing powerful insights into our system's behavior.

First-Order Reactions

First-order reactions are a category where the rate at which a reactant is consumed is directly proportional to the concentration of the reactant itself. This is reflected mathematically when \(n=1\) in our rate equation, giving us \(\frac{d y}{d t} = -ky\). When we integrate and solve this equation, as shown in the step-by-step solution, we find that the concentration over time \(y(t)\) follows an exponential decay law. This law implies that the concentration at any given time can be predicted by a formula featuring \(e^{-kt}\), where the exponent reflects a constant rate of decay multiplied by time. Students should note that, in first-order reactions, the rate at which the reactant concentration decreases is always proportional to its current concentration, meaning that the concentration graph will be a smooth, continuously declining curve without any sudden changes. This specific behavior makes first-order reactions predictable and relatively straightforward to model.

Second-Order Reactions

In contrast to first-order reactions, second-order reactions involve rates that are proportional to the square of the reactant's concentration. We express this by setting \(n=2\) in our rate equation, yielding \(\frac{d y}{d t} = -ky^2\). The step-by-step solution demonstrates how to solve this using integration and applying the initial condition \(y(0)=y_{0}\) to find the constant of integration. The resulting formula \(y(t) = \frac{y_0}{1 + kt y_0}\) describes how the concentration changes over time. Unlike the exponential decay in first-order reactions, the decay curve in second-order reactions is more gradual. It's crucial for students to recognize that, as the concentration decreases, the rate at which it continues to decrease also slows down more significantly than in first-order reactions. This is due to the squared concentration term in the rate equation, resulting in a less steep decline as the reaction progresses. Understanding this nuance allows students to better predict and differentiate between the behaviors of first-order and second-order reactions.

Exponential Decay Law

The exponential decay law is a fundamental concept in understanding how concentrations change in first-order reactions. It states that the amount of a substance decreases at a rate that is constantly proportional to the amount present. Mathematically, we capture this law with the equation \(y(t) = Ce^{-kt}\), where \(C\) is the initial concentration of the substance, \(k\) is the rate constant, and \(t\) is time. This formula represents an exponential function, differentiating it by a constant factor at regular intervals of time.

The graph of an exponential decay function is characteristically a curve that drops sharply at first and then gradually levels off as it approaches zero. This pattern is familiar to those who study radioactive decay, population decline, or any other process where the rate of change is proportional to the current state. In terms of our exercise, for a chemical reaction following first-order kinetics, the exponential decay model allows us to predict how the concentration of a reactant will diminish over time accurately.

Initial Value Problem

An initial value problem in differential equations is a problem in which you're given a differential equation and the value of the function at a specific point, termed the initial condition. It is common practice in science and engineering to compute a solution to a differential equation that satisfies specific initial conditions, as it allows you to tailor the general solution to a particular scenario.

In the chemical rate equation context, the initial condition usually represents the initial concentration of the substance involved in the reaction. By applying this condition, as we did in the exercise, we can find the specific constant \(C\) that makes the solution unique to the problem we're studying. For example, with the second-order reaction, knowing that \(y(0)=y_{0}\) allows us to solve for \(C\) and get the particular solution that reflects how the concentration of the substance will behave starting from \(y_{0}\). Students must understand how to apply initial conditions to such problems, as it's a fundamental step in moving from abstract general solutions to practical, specific applications.