We have an initial amount of H3N2C14 of 72 milligrams, and a half-life of 30 days. Find the decay constant, k, and determine how long it takes for 95% of the material to decay.
We have an initial amount of H3N2C14 of 72 milligrams, and a half-life of 30 days. Find the decay constant, k, and determine how long it takes for 95% of the material to decay.
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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![### Problem Statement:
We have an initial amount of \( H_3N_2Cl_4 \) of 72 milligrams, and a half-life of 30 days. Find the decay constant, \(k\), and determine how long it takes for 95% of the material to decay.
### Solution:
#### 1. Finding the Decay Constant, \(k\):
The relationship between the half-life \(T_{1/2}\) and the decay constant \(k\) is given by the formula:
\[
T_{1/2} = \frac{\ln(2)}{k}
\]
Given:
\[
T_{1/2} = 30 \ \text{days}
\]
Rearrange the formula to solve for \(k\):
\[
k = \frac{\ln(2)}{T_{1/2}}
\]
Substitute the given half-life:
\[
k = \frac{\ln(2)}{30} \approx \frac{0.693}{30} \approx 0.0231 \ \text{days}^{-1}
\]
#### 2. Determining the Time for 95% Decay:
We need to find the time \(t\) when 95% of the material has decayed, which means 5% of the original material remains.
The decay formula is:
\[
N(t) = N_0 e^{-kt}
\]
Where:
- \(N(t)\) is the amount of substance remaining at time \(t\),
- \(N_0\) is the initial amount of the substance,
- \(k\) is the decay constant,
- \(t\) is the time.
Since 5% of the original amount remains:
\[
0.05 N_0 = N_0 e^{-kt}
\]
Simplifying the equation by dividing both sides by \(N_0\):
\[
0.05 = e^{-kt}
\]
Take the natural logarithm of both sides:
\[
\ln(0.05) = -kt
\]
Solve for \(t\):
\[
t = \frac{\ln(0.05)}{-k}
\]
Substitute the value of \(k\):
\[
t = \frac{\ln(0.05)}{-0.0231}
\]
Calculate the value of \(\ln(0.05)\):
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F969d96d2-01e9-4872-bf96-ee1f42578560%2F53a29a9f-44eb-4fdf-a442-92e0d150bf9b%2Ftq970w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
We have an initial amount of \( H_3N_2Cl_4 \) of 72 milligrams, and a half-life of 30 days. Find the decay constant, \(k\), and determine how long it takes for 95% of the material to decay.
### Solution:
#### 1. Finding the Decay Constant, \(k\):
The relationship between the half-life \(T_{1/2}\) and the decay constant \(k\) is given by the formula:
\[
T_{1/2} = \frac{\ln(2)}{k}
\]
Given:
\[
T_{1/2} = 30 \ \text{days}
\]
Rearrange the formula to solve for \(k\):
\[
k = \frac{\ln(2)}{T_{1/2}}
\]
Substitute the given half-life:
\[
k = \frac{\ln(2)}{30} \approx \frac{0.693}{30} \approx 0.0231 \ \text{days}^{-1}
\]
#### 2. Determining the Time for 95% Decay:
We need to find the time \(t\) when 95% of the material has decayed, which means 5% of the original material remains.
The decay formula is:
\[
N(t) = N_0 e^{-kt}
\]
Where:
- \(N(t)\) is the amount of substance remaining at time \(t\),
- \(N_0\) is the initial amount of the substance,
- \(k\) is the decay constant,
- \(t\) is the time.
Since 5% of the original amount remains:
\[
0.05 N_0 = N_0 e^{-kt}
\]
Simplifying the equation by dividing both sides by \(N_0\):
\[
0.05 = e^{-kt}
\]
Take the natural logarithm of both sides:
\[
\ln(0.05) = -kt
\]
Solve for \(t\):
\[
t = \frac{\ln(0.05)}{-k}
\]
Substitute the value of \(k\):
\[
t = \frac{\ln(0.05)}{-0.0231}
\]
Calculate the value of \(\ln(0.05)\):
\[
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