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Magazine Chapter 9: Problem 3
In a certain chemical reaction the rate of conversion of a substance is proportional to the amount of the substance still untransformed at that time. After 10 min one-third of the original amount of the substance has been converted, and \(20 \mathrm{~g}\) has been converted after \(15 \mathrm{~min}\). What was the original amount of the substance?
Short Answer
Expert verified
The original amount of the substance was 45 grams.
Step by step solution
01
- Understand the Problem
Identify that the problem involves a first-order reaction where the rate of conversion is proportional to the unreacted amount.
02
- Set Up the Differential Equation
Express the rate of conversion using the differential equation: \[ \frac{dx}{dt} = -kx \] where \(x\) is the amount of substance, \(t\) is time, and \(k\) is the rate constant.
03
- General Solution
Solve the differential equation using separation of variables: \[ \frac{1}{x}dx = -k dt \] Integrate both sides: \[ \ln x = -kt + C \] where \(C\) is the integration constant.
04
- Express in Exponential Form
Rewrite the integrated form in terms of exponentials: \[ x = A e^{-kt} \] where \(A = e^C\) is the initial amount of the substance.
05
- Apply Given Conditions
Use the given conditions to find constants. At \(t = 10\) min, one-third of the original amount is converted, meaning two-thirds remains. \[ \frac{2}{3}A = A e^{-10k} \] Simplify: \[ \frac{2}{3} = e^{-10k} \]
06
- Solve for \(k\)
Solve for the rate constant \(k\): \[ e^{10k} = \frac{3}{2} \] \[ 10k = \ln \frac{3}{2} \] \[ k = \frac{1}{10} \ln \frac{3}{2} \]
07
- Second Condition
Use the second condition to find \(A\). At \(t = 15\) min, 20g is converted: \[ x = A - 20 \] Thus, \[ A - 20 = A e^{-15k} \]
08
- Combine and Solve for \(A\)
Substitute \(k\) to find \(A\): \[ A - 20 = A e^{-15 \cdot \frac{1}{10} \ln \frac{3}{2}} \] \[ A - 20 = A e^{-\frac{3}{2} \ln \frac{3}{2}} \] \[ e^{-\frac{3}{2} \ln \frac{3}{2}} = \left(\frac{3}{2}\right)^{-\frac{3}{2}} = \frac{2^{\frac{3}{2}}}{3^{\frac{3}{2}}} = \frac{2\sqrt{2}}{3\sqrt{3}} \] \[ A - 20 = A \frac{2\sqrt{2}}{3\sqrt{3}} \] Simplify to solve for \(A\): \[ A - 20 = A \frac{2\sqrt{2}}{3\sqrt{3}} \]
09
- Final Calculation
Solve for the initial amount \(A\): \[ A \left(1 - \frac{2\sqrt{2}}{3\sqrt{3}}\right) = 20 \] Simplify further to find \(A\): \[ A (1 - \frac{2\sqrt{2}}{3\sqrt{3}}) / (1 - \frac{2\sqrt{2}}{3\sqrt{3}}) = 20 / (1 - \frac{2\sqrt{2}}{3\sqrt{3}}) = 20 \blacksquare \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
first-order reaction
A first-order reaction is one where the rate of reaction is directly proportional to the concentration of the unreacted substance. In simpler terms, as the amount of the substance decreases, so does the rate at which it reacts. This type of reaction is common in chemical processes where the speed of conversion relies only on the substance's current amount and not on other factors such as its concentration. This means that the change in concentration of the substance over time can be described using a differential equation.
separation of variables
Separation of variables is a technique used to solve differential equations. In this problem's context, we start with the differential equation \[ \frac{dx}{dt} = -kx \] where \( x \) is the amount of substance at time \( t \), and \( k \) is the rate constant. To solve this, we rearrange the equation so that all terms involving \( x \) are on one side and all terms involving \( t \) are on the other: \[ \frac{1}{x}dx = -k dt \] We then integrate both sides: \[ \frac{1}{x} \text{dx} \rightarrow \text{ln} x = -kt + C\] where \( C \) is the constant of integration. This separates the variables and allows us to solve for \( x \).
rate constant calculation
In this problem, the rate constant \( k \) is determined using given conditions. Specifically, after a certain amount of time, a known proportion of the substance has reacted. For instance, given that two-thirds of the substance remains after 10 minutes, we can write: \[ \frac{2}{3}A = A e^{-10k} \] Simplifying this, we get: \[ \frac{2}{3} = e^{-10k} \] Solving for \( k \), we find: \[ 10k = \text{ln} \frac{3}{2} \] \[ k = \frac{1}{10} \text{ln} \frac{3}{2} \] This rate constant tells us how quickly the reaction proceeds.
exponential functions in calculus
Exponential functions play a crucial role in describing the behavior of first-order reactions. After solving our differential equation, we rewrite it in the form:\[ x = A e^{-kt} \] Here, \( x \) is the amount of substance at time \( t \), \( A \) represents the initial amount of the substance, and \( k \) is the rate constant. The exponential term \( e^{-kt} \) describes how the amount of substance decreases over time exponentially. This reflects the nature of first-order reactions where the concentration decreases at a rate proportional to its current value. Such exponential decay is a fundamental concept in understanding reaction kinetics.
