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.

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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} \
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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