. A reaction that is second-order in one reactant has a rate constant of 2.2 x 10-2 L/(mol • ). If the initial concentration of the reactant is 0.360 mol/L, how long (in seconds) will it take for the concentration to become 0.180 mol/L?
. A reaction that is second-order in one reactant has a rate constant of 2.2 x 10-2 L/(mol • ). If the initial concentration of the reactant is 0.360 mol/L, how long (in seconds) will it take for the concentration to become 0.180 mol/L?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![### Question on Second-Order Reaction Kinetics
A reaction that is second-order in one reactant has a rate constant of \( 2.2 \times 10^{-2} \) L/(mol·s). If the initial concentration of the reactant is 0.360 mol/L, how long (in seconds) will it take for the concentration to become 0.180 mol/L?
### Explanation
In a second-order reaction where the rate law is given by:
\[ \text{Rate} = k[A]^2 \]
the integrated rate equation for a second-order reaction is:
\[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \]
where:
- \([A]_t\) is the concentration of the reactant at time \( t \)
- \( [A]_0 \) is the initial concentration of the reactant
- \( k \) is the rate constant
- \( t \) is the time
Given:
- The rate constant, \( k = 2.2 \times 10^{-2} \) L/(mol·s)
- The initial concentration, \([A]_0 = 0.360 \) mol/L
- The final concentration, \([A]_t = 0.180 \) mol/L
Plugging in the values, we get:
\[ \frac{1}{0.180} = (2.2 \times 10^{-2}) \times t + \frac{1}{0.360} \]
Solve for \( t \):
\[ \frac{1}{0.180} = 5.556 \]
\[ \frac{1}{0.360} = 2.778 \]
\[ 5.556 = (2.2 \times 10^{-2}) t + 2.778 \]
\[ 5.556 - 2.778 = (2.2 \times 10^{-2}) t \]
\[ 2.778 = (2.2 \times 10^{-2}) t \]
\[ t = \frac{2.778}{2.2 \times 10^{-2}} \]
Finally:
\[ t \approx 126.27 \text{ seconds} \]
Therefore, it will take approximately \( 126.27 \) seconds for the concentration to decrease from 0.360 mol/L to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1243504b-5612-4510-ba6c-7b5ec7c70d6e%2F8802da57-be7e-4392-8caf-15acbc7f8ab5%2Fny6pkm5_processed.png&w=3840&q=75)
Transcribed Image Text:### Question on Second-Order Reaction Kinetics
A reaction that is second-order in one reactant has a rate constant of \( 2.2 \times 10^{-2} \) L/(mol·s). If the initial concentration of the reactant is 0.360 mol/L, how long (in seconds) will it take for the concentration to become 0.180 mol/L?
### Explanation
In a second-order reaction where the rate law is given by:
\[ \text{Rate} = k[A]^2 \]
the integrated rate equation for a second-order reaction is:
\[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \]
where:
- \([A]_t\) is the concentration of the reactant at time \( t \)
- \( [A]_0 \) is the initial concentration of the reactant
- \( k \) is the rate constant
- \( t \) is the time
Given:
- The rate constant, \( k = 2.2 \times 10^{-2} \) L/(mol·s)
- The initial concentration, \([A]_0 = 0.360 \) mol/L
- The final concentration, \([A]_t = 0.180 \) mol/L
Plugging in the values, we get:
\[ \frac{1}{0.180} = (2.2 \times 10^{-2}) \times t + \frac{1}{0.360} \]
Solve for \( t \):
\[ \frac{1}{0.180} = 5.556 \]
\[ \frac{1}{0.360} = 2.778 \]
\[ 5.556 = (2.2 \times 10^{-2}) t + 2.778 \]
\[ 5.556 - 2.778 = (2.2 \times 10^{-2}) t \]
\[ 2.778 = (2.2 \times 10^{-2}) t \]
\[ t = \frac{2.778}{2.2 \times 10^{-2}} \]
Finally:
\[ t \approx 126.27 \text{ seconds} \]
Therefore, it will take approximately \( 126.27 \) seconds for the concentration to decrease from 0.360 mol/L to
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