The rate constant for this first-order reaction is 0.0700 s- at 400 °C. A products > After how many seconds will 16.2% of the reactant remain? t =
The rate constant for this first-order reaction is 0.0700 s- at 400 °C. A products > After how many seconds will 16.2% of the reactant remain? t =
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**First-Order Reaction Kinetics**
In this exercise, we are given a first-order chemical reaction where reactant A is converted into products. The rate constant (k) for this first-order reaction is 0.0700 s⁻¹ at 400 °C.
\[ A \longrightarrow \text{products} \]
We are asked to determine the time (t) after which only 16.2% of the reactant remains.
To find the time t, we can use the integrated rate law for a first-order reaction, which is given by:
\[ [A]_t = [A]_0 e^{-kt} \]
Where:
- \([A]_t\) is the concentration of reactant A at time t
- \([A]_0\) is the initial concentration of reactant A
- \(k\) is the first-order rate constant
- \(t\) is the time
Given that we know what percentage of the reactant remains, we can rearrange the equation as:
\[ \frac{[A]_t}{[A]_0} = e^{-kt} \]
By substituting \(\frac{[A]_t}{[A]_0} = 0.162\):
\[ 0.162 = e^{-kt} \]
Taking the natural logarithm (ln) of both sides:
\[ \ln(0.162) = -kt \]
We then solve for t:
\[ t = \frac{\ln(0.162)}{-k} \]
Finally, by plugging in the given rate constant \( k = 0.0700 \, \text{s}^{-1} \):
\[ t = \frac{\ln(0.162)}{-0.0700\, \text{s}^{-1}} \]
\[
t = \boxed{\ \ \ \ \ \ \ }
\]
This calculated value represents the time in seconds at which 16.2% of the original amount of reactant A remains.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff00f8b75-2bac-4335-a200-62a4fa07a89d%2F27b5ace0-526e-4cf4-99ae-9390627e78a8%2F0uo32_processed.png&w=3840&q=75)
Transcribed Image Text:**First-Order Reaction Kinetics**
In this exercise, we are given a first-order chemical reaction where reactant A is converted into products. The rate constant (k) for this first-order reaction is 0.0700 s⁻¹ at 400 °C.
\[ A \longrightarrow \text{products} \]
We are asked to determine the time (t) after which only 16.2% of the reactant remains.
To find the time t, we can use the integrated rate law for a first-order reaction, which is given by:
\[ [A]_t = [A]_0 e^{-kt} \]
Where:
- \([A]_t\) is the concentration of reactant A at time t
- \([A]_0\) is the initial concentration of reactant A
- \(k\) is the first-order rate constant
- \(t\) is the time
Given that we know what percentage of the reactant remains, we can rearrange the equation as:
\[ \frac{[A]_t}{[A]_0} = e^{-kt} \]
By substituting \(\frac{[A]_t}{[A]_0} = 0.162\):
\[ 0.162 = e^{-kt} \]
Taking the natural logarithm (ln) of both sides:
\[ \ln(0.162) = -kt \]
We then solve for t:
\[ t = \frac{\ln(0.162)}{-k} \]
Finally, by plugging in the given rate constant \( k = 0.0700 \, \text{s}^{-1} \):
\[ t = \frac{\ln(0.162)}{-0.0700\, \text{s}^{-1}} \]
\[
t = \boxed{\ \ \ \ \ \ \ }
\]
This calculated value represents the time in seconds at which 16.2% of the original amount of reactant A remains.
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