Consider this reaction: 2C1,0; (g) → 2C1, (g) +50, (3) At a certain temperature it obeys this rate law. rate = (0.00575 M-!.5)[c1,0, Suppose a vessel contains Cl,O; at a concentration of 0.900M. Calculate the concentration of Cl,0, in the vessel 580. seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
Consider this reaction: 2C1,0; (g) → 2C1, (g) +50, (3) At a certain temperature it obeys this rate law. rate = (0.00575 M-!.5)[c1,0, Suppose a vessel contains Cl,O; at a concentration of 0.900M. Calculate the concentration of Cl,0, in the vessel 580. seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
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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![### Reaction Kinetics Example
Consider this reaction:
\[ 2Cl_2O_5 (g) \rightarrow 2Cl_2 (g) + 5O_2 (g) \]
At a certain temperature, it obeys this rate law:
\[ \text{rate} = \left(0.00575 \text{ M}^{-1} \cdot \text{s}^{-1} \right) [Cl_2O_5]^2 \]
**Problem:**
Suppose a vessel contains \( Cl_2O_5 \) at a concentration of \( 0.900\,M \). Calculate the concentration of \( Cl_2O_5 \) in the vessel 580 seconds later. You may assume no other reaction is important.
*Round your answer to 2 significant digits.*
**Answer Box:**
\[ \boxed{\phantom{M}} M \]
You should use the integrated rate law for a second-order reaction to find the concentration after a given time. The general form of the integrated rate law for a second-order reaction is:
\[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \]
where:
- \( [A]_t \) is the concentration of \( A \) at time \( t \)
- \( [A]_0 \) is the initial concentration of \( A \)
- \( k \) is the rate constant
- \( t \) is the time
Given:
- \( [Cl_2O_5]_0 = 0.900 \, M \)
- \( k = 0.00575 \, \text{M}^{-1} \cdot \text{s}^{-1} \)
- \( t = 580 \, \text{s} \)
Calculation:
\[ \frac{1}{[Cl_2O_5]_t} = \frac{1}{0.900 \, M} + (0.00575 \, \text{M}^{-1} \cdot \text{s}^{-1})(580 \, \text{s}) \]
\[ \frac{1}{[Cl_2O_5]_t} = 1.111 + 3.335 = 4.446 \]
\[ [Cl_2O_5]_t = \frac{1}{4.446} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F10caddcf-72cd-48c5-b185-0b83dd19834e%2F4b736764-9ec4-40a3-8c2c-fb18f956ff73%2Fzro2zi_processed.png&w=3840&q=75)
Transcribed Image Text:### Reaction Kinetics Example
Consider this reaction:
\[ 2Cl_2O_5 (g) \rightarrow 2Cl_2 (g) + 5O_2 (g) \]
At a certain temperature, it obeys this rate law:
\[ \text{rate} = \left(0.00575 \text{ M}^{-1} \cdot \text{s}^{-1} \right) [Cl_2O_5]^2 \]
**Problem:**
Suppose a vessel contains \( Cl_2O_5 \) at a concentration of \( 0.900\,M \). Calculate the concentration of \( Cl_2O_5 \) in the vessel 580 seconds later. You may assume no other reaction is important.
*Round your answer to 2 significant digits.*
**Answer Box:**
\[ \boxed{\phantom{M}} M \]
You should use the integrated rate law for a second-order reaction to find the concentration after a given time. The general form of the integrated rate law for a second-order reaction is:
\[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \]
where:
- \( [A]_t \) is the concentration of \( A \) at time \( t \)
- \( [A]_0 \) is the initial concentration of \( A \)
- \( k \) is the rate constant
- \( t \) is the time
Given:
- \( [Cl_2O_5]_0 = 0.900 \, M \)
- \( k = 0.00575 \, \text{M}^{-1} \cdot \text{s}^{-1} \)
- \( t = 580 \, \text{s} \)
Calculation:
\[ \frac{1}{[Cl_2O_5]_t} = \frac{1}{0.900 \, M} + (0.00575 \, \text{M}^{-1} \cdot \text{s}^{-1})(580 \, \text{s}) \]
\[ \frac{1}{[Cl_2O_5]_t} = 1.111 + 3.335 = 4.446 \]
\[ [Cl_2O_5]_t = \frac{1}{4.446} \
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