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Chapter 4: Problem 25

In a chemical reaction, substance \(A\) decomposes at a rate proportional to the amount of \(A\) present. a) Write an equation relating \(A\) to the amount left of an initial amount \(A_{0}\) after time \(t\). b) It is found that \(8 \mathrm{~g}\) of \(A\) will reduce to \(4 \mathrm{~g}\) in \(3 \mathrm{hr}\). After how long will there be only \(1 \mathrm{~g}\) left?

Short Answer

Expert verified
The equation is: \( A = A_0 e^{-kt} \). It will take 9 hours for the amount to decrease to 1 g.

Step by step solution

01

- Understanding the Decomposition Rate

The decomposition rate of substance A is proportional to the amount of A present. This relationship can be expressed using a differential equation: \( \frac{dA}{dt} = -kA \)where \(k\) is a positive constant of proportionality.
02

- Solving the Differential Equation

To solve the differential equation, separate the variables: \( \frac{dA}{A} = -kdt \)Now, integrate both sides: \( \int \frac{1}{A} dA = -k \int dt \)This yields the natural logarithm: \( \text{ln}(A) = -kt + C \)where \(C\) is the integration constant.
03

- Exponentiate Both Sides

Exponentiate both sides to solve for \(A\): \( e^{\text{ln}(A)} = e^{-kt + C} \)Since \(e^{\text{ln}(A)} = A \) and \(e^C\) is a constant, we can write: \( A = A_0 e^{-kt} \)where \(A_0\) represents the initial amount of substance A.
04

- Using Given Data to Find the Rate Constant

Use the given data to find the proportionality constant \(k\). We know that when \(t = 3 \) hours, \(A\) reduces from 8 g to 4 g: \( 4 = 8 e^{-3k} \)Solve for \(k\): \( \frac{4}{8} = e^{-3k} \) or \( \frac{1}{2} = e^{-3k} \)Take the natural logarithm of both sides: \( \text{ln}(\frac{1}{2}) = -3k \)Thus, \( k = -\frac{\text{ln}(\frac{1}{2})}{3} = \frac{\text{ln}(2)}{3} \)
05

- Finding the Time for Concentration to Decrease to 1g

Now, determine the time \(t\) for the concentration to decrease to \(1\) g. Let \(A = 1 \) g: \( 1 = 8 e^{-kt} \)Substitute the value of \(k\) found earlier:\( 1 = 8 e^{ -\left( \frac{\text{ln}(2)}{3} \right) t} \) Solve for \(t\):\( \frac{1}{8} = e^{ -\left( \frac{\text{ln}(2)}{3} \right) t} \)Take the natural logarithm of both sides: \( \text{ln}(\frac{1}{8}) = -\left( \frac{\text{ln}(2)}{3} \right) t \)Since \( \text{ln}(\frac{1}{8}) = -\text{ln}(8) = -3\text{ln}(2) \), we have: \( -3\text{ln}(2) = -\frac{t}{3} \text{ln}(2) \)Thus, \( t = 9 \text{ hours} \)

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Key Concepts

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

Decomposition Rate
The rate at which a substance decomposes in a chemical reaction is a key concept in kinetics. In this exercise, we are dealing with the decomposition rate of substance A. The rate is said to be proportional to the amount of substance A present. Mathematically, this can be written as:
  • \( \frac{dA}{dt} = -kA \)
Here, \( \frac{dA}{dt} \) is the rate of change of the amount of substance A over time, and \( k \) is the proportionality constant. The negative sign indicates that the amount of A is decreasing over time. Understanding the decomposition rate is essential for solving the differential equation that models this chemical reaction.
Proportionality Constant
The proportionality constant \( k \) is a crucial factor in the differential equation representing the decomposition. This constant \( k \) determines how quickly the substance decomposes. To find the value of \( k \), we use given data from the problem, specifically that 8 grams of A reduces to 4 grams over 3 hours. We set up the equation and solve for \( k \):
  • \( 4 = 8 e^{-3k} \)
By taking the natural logarithm of both sides, we can isolate and solve for \( k \):
  • \( k = \frac{\text{ln}(2)}{3} \)
This process shows how the proportionality constant is derived from experimental data.
Natural Logarithm
The natural logarithm (\( \text{ln} \)) is commonly used in solving differential equations involving exponential functions. In this exercise, after integrating both sides of the separated differential equation and getting:
  • \( \text{ln}(A) = -kt + C \)
The natural logarithm helps simplify the relationship between the variables. When we need to find the constant \( k \) from empirical data, we take the natural logarithm of both sides of the equation to solve for unknowns:
Integration
Integration is the mathematical process used to solve differential equations like \( \frac{dA}{dt} = -kA \). By separating the variables and integrating, we obtain:
  • \( \int \frac{1}{A} dA = -k \int dt \)
This integral can be solved to give us the natural logarithm:
  • \( \text{ln}(A) = -kt + C \)
Integration helps us find a general solution to the differential equation, which then can be further manipulated to find specific solutions based on initial conditions.
Exponential Function
The exponential function \( e^{x} \) frequently appears in solutions to differential equations as it effectively handles growth and decay processes. After solving the integrated equation, we exponentiate both sides to eliminate the natural logarithm:
  • \( e^{\text{ln}(A)} = e^{-kt + C} \)
Since \( e^{\text{ln}(A)} \) simplifies to \( A \) and \( e^C \) is a constant, we get:
  • \( A = A_0 e^{-kt} \)
This representation shows how the amount of substance A changes exponentially over time, driven by the decay constant \( k \). Using the exponentiation step is vital for finding how much of the substance remains after a certain period.

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