Consider this reaction: 2H₂PO4 (aq) → P₂O, (aq) + 3H₂O (aq) - At a certain temperature it obeys this rate law. rate = (0.048 M¹.s¹) [H₂PO₂] S Suppose a vessel contains H₂PO at a concentration of 1.06 M. Calculate how long it takes for the concentration of H₂PO to decrease to 0.053 M. You may assume no other reaction is important. Round your answer to 2 significant digits. ☐x10
Consider this reaction: 2H₂PO4 (aq) → P₂O, (aq) + 3H₂O (aq) - At a certain temperature it obeys this rate law. rate = (0.048 M¹.s¹) [H₂PO₂] S Suppose a vessel contains H₂PO at a concentration of 1.06 M. Calculate how long it takes for the concentration of H₂PO to decrease to 0.053 M. You may assume no other reaction is important. Round your answer to 2 significant digits. ☐x10
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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![### Chemical Kinetics: Reaction Rate Problem
Consider the following reaction:
\[ 2H_3PO_4(aq) \rightarrow P_2O_5(aq) + 3H_2O(aq) \]
At a certain temperature, it follows this rate law:
\[ \text{rate} = (0.048 \, \text{M}^{-1} \cdot \text{s}^{-1}) [H_3PO_4]^2 \]
#### Problem Statement:
Suppose a vessel contains \( H_3PO_4 \) at an initial concentration of \( 1.06 \, \text{M} \). Calculate the time required for the concentration of \( H_3PO_4 \) to decrease to \( 0.053 \, \text{M} \), assuming no other reactions are significant.
Please round your answer to two significant digits.
#### Input Box:
- **Answer box placeholder:** _s (seconds)
---
#### Explanation of Calculation:
To solve for the required time (\( t \)), we need to integrate the rate law for a second-order reaction, which is in the form:
\[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \]
where:
- \([A]_t\) is the concentration of \( H_3PO_4 \) at time \( t \)
- \([A]_0\) is the initial concentration of \( H_3PO_4 \)
- \( k \) is the rate constant
- \( t \) is time
Given data:
- \([H_3PO_4]_0 = 1.06 \, \text{M}\)
- \([H_3PO_4]_t = 0.053 \, \text{M}\)
- \( k = 0.048 \, \text{M}^{-1} \cdot \text{s}^{-1} \)
By substituting the given concentrations and rate constant into the integrated rate law, we can solve for \( t \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F550116fe-4392-44b3-9444-167bfb5405dc%2F73787c2c-5fbc-4c62-a09c-720783badb09%2Fbsmp2o7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemical Kinetics: Reaction Rate Problem
Consider the following reaction:
\[ 2H_3PO_4(aq) \rightarrow P_2O_5(aq) + 3H_2O(aq) \]
At a certain temperature, it follows this rate law:
\[ \text{rate} = (0.048 \, \text{M}^{-1} \cdot \text{s}^{-1}) [H_3PO_4]^2 \]
#### Problem Statement:
Suppose a vessel contains \( H_3PO_4 \) at an initial concentration of \( 1.06 \, \text{M} \). Calculate the time required for the concentration of \( H_3PO_4 \) to decrease to \( 0.053 \, \text{M} \), assuming no other reactions are significant.
Please round your answer to two significant digits.
#### Input Box:
- **Answer box placeholder:** _s (seconds)
---
#### Explanation of Calculation:
To solve for the required time (\( t \)), we need to integrate the rate law for a second-order reaction, which is in the form:
\[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \]
where:
- \([A]_t\) is the concentration of \( H_3PO_4 \) at time \( t \)
- \([A]_0\) is the initial concentration of \( H_3PO_4 \)
- \( k \) is the rate constant
- \( t \) is time
Given data:
- \([H_3PO_4]_0 = 1.06 \, \text{M}\)
- \([H_3PO_4]_t = 0.053 \, \text{M}\)
- \( k = 0.048 \, \text{M}^{-1} \cdot \text{s}^{-1} \)
By substituting the given concentrations and rate constant into the integrated rate law, we can solve for \( t \).
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