Consider this reaction: 280, (g) → 2S0, (g) + O, (g) At a certain temperature it obeys this rate law. rate = (0.0149 s1)[so,] Suppose a vessel contains SO, at a concentration of 1.30M. Calculate the concentration of SO, in the vessel 62.0 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
Consider this reaction: 280, (g) → 2S0, (g) + O, (g) At a certain temperature it obeys this rate law. rate = (0.0149 s1)[so,] Suppose a vessel contains SO, at a concentration of 1.30M. Calculate the concentration of SO, in the vessel 62.0 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
Chemistry
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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
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![### Reaction Rate and Concentration Calculation
Consider the following reaction:
\[ 2SO_3 (g) \rightarrow 2SO_2 (g) + O_2 (g) \]
At a certain temperature, this reaction follows the rate law given by:
\[ \text{rate} = (0.0149 \, s^{-1})[SO_3] \]
#### Problem Statement
Suppose a vessel contains \( SO_3 \) at a concentration of \( 1.30 \, M \). Calculate the concentration of \( SO_3 \) in the vessel after 62.0 seconds. You may assume no other reaction is important. Round your answer to 2 significant digits.
#### Solution
To calculate the concentration of \( SO_3 \) after a given time, we use the first-order integrated rate law formula:
\[ [SO_3]_t = [SO_3]_0 e^{-kt} \]
where:
- \([SO_3]_t\) is the concentration of \( SO_3 \) at time \( t \),
- \([SO_3]_0\) is the initial concentration of \( SO_3 \),
- \( k \) is the rate constant (\( 0.0149 \, s^{-1} \)),
- \( t \) is the time (62.0 seconds),
- \( e \) is the base of the natural logarithm.
Now, substitute the given values into the formula.
1. Initial concentration, \( [SO_3]_0 = 1.30 \, M \)
2. Rate constant, \( k = 0.0149 \, s^{-1} \)
3. Time, \( t = 62.0 \, s \)
\[ [SO_3]_t = 1.30 \, M \cdot e^{-(0.0149 \, s^{-1})(62.0 \, s)} \]
Using a calculator to compute the exponent:
\[ [SO_3]_t = 1.30 \, M \cdot e^{-0.9238} \]
\[ [SO_3]_t \approx 1.30 \, M \cdot 0.3966 \]
\[ [SO_3]_t \approx 0.5156 \, M \]
Rounding to 2 significant digits:
\[ [SO_3]_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ff5d2d4-0b71-46f6-9878-a59ffd2a7f15%2F2fd7211a-468a-4b6b-ae32-8aa7f096291c%2Fffgsdod_processed.png&w=3840&q=75)
Transcribed Image Text:### Reaction Rate and Concentration Calculation
Consider the following reaction:
\[ 2SO_3 (g) \rightarrow 2SO_2 (g) + O_2 (g) \]
At a certain temperature, this reaction follows the rate law given by:
\[ \text{rate} = (0.0149 \, s^{-1})[SO_3] \]
#### Problem Statement
Suppose a vessel contains \( SO_3 \) at a concentration of \( 1.30 \, M \). Calculate the concentration of \( SO_3 \) in the vessel after 62.0 seconds. You may assume no other reaction is important. Round your answer to 2 significant digits.
#### Solution
To calculate the concentration of \( SO_3 \) after a given time, we use the first-order integrated rate law formula:
\[ [SO_3]_t = [SO_3]_0 e^{-kt} \]
where:
- \([SO_3]_t\) is the concentration of \( SO_3 \) at time \( t \),
- \([SO_3]_0\) is the initial concentration of \( SO_3 \),
- \( k \) is the rate constant (\( 0.0149 \, s^{-1} \)),
- \( t \) is the time (62.0 seconds),
- \( e \) is the base of the natural logarithm.
Now, substitute the given values into the formula.
1. Initial concentration, \( [SO_3]_0 = 1.30 \, M \)
2. Rate constant, \( k = 0.0149 \, s^{-1} \)
3. Time, \( t = 62.0 \, s \)
\[ [SO_3]_t = 1.30 \, M \cdot e^{-(0.0149 \, s^{-1})(62.0 \, s)} \]
Using a calculator to compute the exponent:
\[ [SO_3]_t = 1.30 \, M \cdot e^{-0.9238} \]
\[ [SO_3]_t \approx 1.30 \, M \cdot 0.3966 \]
\[ [SO_3]_t \approx 0.5156 \, M \]
Rounding to 2 significant digits:
\[ [SO_3]_
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