At a certain temperature the rate of this reaction is second order in H₂CO3 with a rate constant of 2.75M H₂CO3(aq) → H₂O (aq) + CO₂(aq) Suppose a vessel contains H₂CO3 at a concentration of 0.130M. Calculate how long it takes for the concentration of H₂CO3 to decrease to 0.0286 M. You may assume no other reaction is important. Round your answer to 2 significant digits.
At a certain temperature the rate of this reaction is second order in H₂CO3 with a rate constant of 2.75M H₂CO3(aq) → H₂O (aq) + CO₂(aq) Suppose a vessel contains H₂CO3 at a concentration of 0.130M. Calculate how long it takes for the concentration of H₂CO3 to decrease to 0.0286 M. You may assume no other reaction is important. Round your answer to 2 significant digits.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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![**Second-Order Reaction Rate Example**
For an educational context on chemical kinetics, consider the following problem:
At a certain temperature, the rate of a reaction is second order in H\(_{2}\)CO\(_{3}\), with a rate constant of 2.75 M\(^{-1}\)s\(^{-1}\). The reaction is as follows:
\[ \text{H}_{2}\text{CO}_{3} \text{(aq)} \rightarrow \text{H}_{2}\text{O} \text{(aq)} + \text{CO}_{2} \text{(aq)} \]
### Problem:
Suppose a vessel contains H\(_{2}\)CO\(_{3}\) at a concentration of 0.130 M. Calculate how long it takes for the concentration of H\(_{2}\)CO\(_{3}\) to decrease to 0.0286 M. You may assume no other reaction is important.
Round your answer to 2 significant digits.
### Solution:
To find the time required for the concentration to decrease to 0.0286 M, we can use the integrated rate law for second-order reactions:
\[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \]
Where:
- [A]\(_{t}\) = concentration at time \( t \)
- \( [A]_{0} \) = initial concentration
- \( k \) = rate constant
- \( t \) = time
Given:
- \( [A]_{0} = 0.130 \) M
- \( [A]_{t} = 0.0286 \) M
- \( k = 2.75 \) M\(^{-1}\)s\(^{-1}\)
Plugging in the known values:
\[ \frac{1}{0.0286 \text{ M}} = (2.75 \text{ M}^{-1}\text{s}^{-1}) t + \frac{1}{0.130 \text{ M}} \]
Simplify the equation to solve for \( t \):
\[ \frac{1}{0.0286} - \frac{1}{0.130} = 2.75 t \]
\[ 34.965 \text{ M}^{-1} - 7.692 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe361a5c2-e3a6-4cd8-bee8-5dcb720cb10b%2Fe9437f59-9c27-48a6-b391-5bfc87d324ca%2Fikwhy2c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Second-Order Reaction Rate Example**
For an educational context on chemical kinetics, consider the following problem:
At a certain temperature, the rate of a reaction is second order in H\(_{2}\)CO\(_{3}\), with a rate constant of 2.75 M\(^{-1}\)s\(^{-1}\). The reaction is as follows:
\[ \text{H}_{2}\text{CO}_{3} \text{(aq)} \rightarrow \text{H}_{2}\text{O} \text{(aq)} + \text{CO}_{2} \text{(aq)} \]
### Problem:
Suppose a vessel contains H\(_{2}\)CO\(_{3}\) at a concentration of 0.130 M. Calculate how long it takes for the concentration of H\(_{2}\)CO\(_{3}\) to decrease to 0.0286 M. You may assume no other reaction is important.
Round your answer to 2 significant digits.
### Solution:
To find the time required for the concentration to decrease to 0.0286 M, we can use the integrated rate law for second-order reactions:
\[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \]
Where:
- [A]\(_{t}\) = concentration at time \( t \)
- \( [A]_{0} \) = initial concentration
- \( k \) = rate constant
- \( t \) = time
Given:
- \( [A]_{0} = 0.130 \) M
- \( [A]_{t} = 0.0286 \) M
- \( k = 2.75 \) M\(^{-1}\)s\(^{-1}\)
Plugging in the known values:
\[ \frac{1}{0.0286 \text{ M}} = (2.75 \text{ M}^{-1}\text{s}^{-1}) t + \frac{1}{0.130 \text{ M}} \]
Simplify the equation to solve for \( t \):
\[ \frac{1}{0.0286} - \frac{1}{0.130} = 2.75 t \]
\[ 34.965 \text{ M}^{-1} - 7.692 \
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