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Chapter 7: Problem 39

Chemical Reactions In a certain chemical reaction a substance is converted into another substance at a rate proportional to the square of the amount of the first substance present at any time \(t\). Initially \((t=0), 50 \mathrm{~g}\) of the first substance was present; 1 hr later, only \(10 \mathrm{~g}\) of it remained. Find an expression that gives the amount of the first substance present at any time \(t\). What is the amount present after \(2 \mathrm{hr}\) ?

Short Answer

Expert verified
The amount of the first substance present at any time $t$ is given by the expression \(A(t) = \frac{1}{\frac{1}{5}t + \frac{1}{50}}\). After 2 hours, the amount present is approximately \(\frac{50}{41} \mathrm{g}\).

Step by step solution

01

Define the variables and constants

Let: - A(t) be the amount of the first substance present at any time t. - k be the constant of proportionality. We also know that initially (t=0), A(0) = 50 g and after 1 hour (t=1), A(1) = 10 g.
02

Write down the rate equation

Since the rate of the reaction is proportional to the square of the amount of the first substance present, we can write the rate equation as: Rate = -k [A(t)]^2 The negative sign indicates that the amount of the first substance is decreasing as the reaction progresses.
03

Convert the rate equation into a differential equation

As the rate is the change of amount with respect to time, we can rewrite the rate equation as a differential equation: dA/dt = -k [A(t)]^2
04

Solve the differential equation using the initial conditions

To solve the differential equation, we need to separate the variables and integrate both sides: (1/[A(t)]^2) dA = -k dt Integrating both sides with respect to their respective variables, we get: \(\int\) (1/[A(t)]^2) dA = \(\int\) -k dt -1/[A(t)] = -kt + C, where C is the integration constant. Now we use the initial conditions to find the values of k and C: 1. At t = 0, A(0) = 50 g: -1/50 = -k(0) + C C = -1/50 2. At t = 1 hour, A(1) = 10 g: -1/10 = -k(1) + (-1/50) k = 1/5 Now that we have the values of k and C, we can find the expression for A(t).
05

Find the amount of substance present at any time t

Using the values of k and C in the equation, we get: -1/[A(t)] = -1/5*t - 1/50 Now, solve for A(t): 1/A(t) = 1/5*t + 1/50 A(t) = 1 / (1/5*t + 1/50)
06

Find the amount of substance present after 2 hours

Now, we will find the amount of substance present after 2 hours by substituting t = 2 in the equation: A(2) = 1 / (1/5*2 + 1/50) A(2) = 1 / (2/5 + 1/50) A(2) = 1 / (41/50) A(2) 鈮� 50/41 g Therefore, the amount of the first substance present after 2 hours is approximately 50/41 g.

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Key Concepts

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

Chemical Reactions
In the world of chemistry, reactions involve the transformation of substances from one form to another. A chemical reaction often follows laws that describe the concentration of reactants and products over time.
The example provided showcases a reaction where one substance is transformed into another, and this transformation is dependent on how much of the reactant is present.
  • The specific chemical reaction discussed has a rate proportional to the square of the substance's amount. This implies a rapid decrease in reactant as it gets converted.
  • Understanding the behavior of this reaction requires interpreting the rate of change and using differential equations to model it.
Recognizing these principles is crucial for predicting how and when substances change during reactions.
Rate of Change
The rate of change in a chemical reaction describes how quickly the concentration of a substance decreases or increases over time.
In the given exercise, the rate of change is modeled by the differential equation \( \frac{dA}{dt} = -k [A(t)]^2 \).
  • The negative sign indicates a decrease, as the reactant is being consumed in the reaction.
  • The square of \( A(t) \) in the equation shows that the rate isn't constant but accelerates as more of the substance is present initially.
This non-linear rate tells us that abundant substances deplete swiftly, which aligns with many natural chemical processes.
Initial Conditions
Initial conditions are crucial for solving differential equations as they allow us to find specific solutions that describe particular scenarios.
For the chemical reaction problem, the initial conditions at \( t = 0 \) and \( t = 1 \) helped us determine the constants \( k \) and \( C \).
  • At \( t = 0 \), \( A(0) = 50 \) g provides a start point, forming an equation to find the integration constant \( C \).
  • At \( t = 1 \), \( A(1) = 10 \) g gives data to solve for the rate constant \( k \).
These specific conditions are essential for predicting the system's behavior at any given time in the future.
Integration Methods
Integration methods are mathematical techniques used to solve differential equations. They allow us to find a general solution that satisfies the given rate equation.
For our chemical reaction, integration involved separating variables and then integrating each side with respect to its own variable.
  • The equation \( \int \frac{1}{[A(t)]^2} \, dA = \int -k \, dt \) allowed us to integrate and express the relation in terms of \( A(t) \).
  • Post integration, seeding in initial conditions helps tailor this general solution to match specific scenarios of the reaction.
Understanding integration methods is critical for finding the time-dependent behavior of chemical concentrations.

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Most popular questions from this chapter

Lambert's Law of Absorption According to Lambert's Law of Absorption, the percentage of incident light \(L\), absorbed in passing through a thin layer of material \(x\), is proportional to the thickness of the material. For a certain material, if \(\frac{1}{2}\) in. of the material reduces the light to half of its intensity, how much additional material is needed to reduce the intensity to one fourth of its initial value?

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Refer to Exercise \(20 .\) A population of 20 goldfish was introduced into a pond that has an estimated carrying capacity of 200 fish. After 1 month, the population of goldfish had grown to \(80 .\) If the pattern of growth of the population followed the Gompertz curve, how many goldfish were in the pond after 3 months?

Doomsday Equation Suppose that the population \(P\) satisfies the differential equation \(d P / d t=k P^{1.0 i}\), where \(k\) is a positive constant and \(P(0)=1\). a. Solve the initial-value problem. b. Suppose that \(k=0.1\). Plot the graph of \(P(t)\). c. Why is \(d P / d t=k P^{1.01}\) called the "doomsday equation"?
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