At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.0159M¹¹; H,CO, (aq) →H,O(aq)+CO,(aq) Suppose a vessel contains H₂CO3 at a concentration of 1.04M. Calculate the concentration of H₂CO3 in the vessel 610. seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. M
At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.0159M¹¹; H,CO, (aq) →H,O(aq)+CO,(aq) Suppose a vessel contains H₂CO3 at a concentration of 1.04M. Calculate the concentration of H₂CO3 in the vessel 610. seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. M
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![### Second-Order Kinetics Example
At a certain temperature, this reaction follows second-order kinetics with a rate constant of \( 0.0159 \, M^{-1} \cdot s^{-1} \):
\[ \text{H}_2\text{CO}_3 (aq) \rightarrow \text{H}_2\text{O} (aq) + \text{CO}_2 (aq) \]
Suppose a vessel contains \(\text{H}_2\text{CO}_3\) at a concentration of \(1.04 \, M\). Calculate the concentration of \(\text{H}_2\text{CO}_3\) in the vessel 610 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
### Given Data
1. Initial concentration of \(\text{H}_2\text{CO}_3\) (\( [\text{H}_2\text{CO}_3]_0 \)) = 1.04 M
2. Rate constant (k) = \( 0.0159 \, M^{-1} \cdot s^{-1} \)
3. Time (t) = 610 seconds
### Second-Order Kinetics Formula
For a second-order reaction, the concentration at time \( t \) can be calculated using the formula:
\[ \frac{1}{[\text{A}]_t} = kt + \frac{1}{[\text{A}]_0} \]
Where:
- \([A]_t\) is the concentration at time \( t \)
- \( k \) is the rate constant
- \( [A]_0 \) is the initial concentration
- \( t \) is the time
### Calculation
1. Substitute the given values into the second-order kinetics formula:
\[ \frac{1}{[\text{H}_2\text{CO}_3]_t} = 0.0159 \times 610 + \frac{1}{1.04} \]
2. Perform the calculation:
\[ \frac{1}{[\text{H}_2\text{CO}_3]_t} = 0.0159 \times 610 + 0.9615 \]
\[ \frac{1}{[\text{H}_2\text{CO}_3]_t} = 9.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12c0a0e9-4d98-45bb-8c42-bb6fd08d9097%2Fbd7bd1b6-40a5-4c71-952d-34aabbfb3a78%2Fs83je6q_processed.png&w=3840&q=75)
Transcribed Image Text:### Second-Order Kinetics Example
At a certain temperature, this reaction follows second-order kinetics with a rate constant of \( 0.0159 \, M^{-1} \cdot s^{-1} \):
\[ \text{H}_2\text{CO}_3 (aq) \rightarrow \text{H}_2\text{O} (aq) + \text{CO}_2 (aq) \]
Suppose a vessel contains \(\text{H}_2\text{CO}_3\) at a concentration of \(1.04 \, M\). Calculate the concentration of \(\text{H}_2\text{CO}_3\) in the vessel 610 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
### Given Data
1. Initial concentration of \(\text{H}_2\text{CO}_3\) (\( [\text{H}_2\text{CO}_3]_0 \)) = 1.04 M
2. Rate constant (k) = \( 0.0159 \, M^{-1} \cdot s^{-1} \)
3. Time (t) = 610 seconds
### Second-Order Kinetics Formula
For a second-order reaction, the concentration at time \( t \) can be calculated using the formula:
\[ \frac{1}{[\text{A}]_t} = kt + \frac{1}{[\text{A}]_0} \]
Where:
- \([A]_t\) is the concentration at time \( t \)
- \( k \) is the rate constant
- \( [A]_0 \) is the initial concentration
- \( t \) is the time
### Calculation
1. Substitute the given values into the second-order kinetics formula:
\[ \frac{1}{[\text{H}_2\text{CO}_3]_t} = 0.0159 \times 610 + \frac{1}{1.04} \]
2. Perform the calculation:
\[ \frac{1}{[\text{H}_2\text{CO}_3]_t} = 0.0159 \times 610 + 0.9615 \]
\[ \frac{1}{[\text{H}_2\text{CO}_3]_t} = 9.
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