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.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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 \
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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