21.The decomposition of N20(g) to nitrogen and oxygen has the rate law rate = k[N2O]² second order in N2O. where k = 1.1 x 10-3 /M s at 565 °C. If the initial concentration of a sample of N20 is 0.15 M, what is the concentration after 30 minutes (1800 seconds) at 565°C? а. 9.5 х 10 м b. 2.2 x 103 M c. 0.029 M d. 0.12 M е. 0.46 М
21.The decomposition of N20(g) to nitrogen and oxygen has the rate law rate = k[N2O]² second order in N2O. where k = 1.1 x 10-3 /M s at 565 °C. If the initial concentration of a sample of N20 is 0.15 M, what is the concentration after 30 minutes (1800 seconds) at 565°C? а. 9.5 х 10 м b. 2.2 x 103 M c. 0.029 M d. 0.12 M е. 0.46 М
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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21) please see attached
![**Decomposition of N₂O - Second Order Reaction**
**Question:**
The decomposition of N₂O(g) to nitrogen and oxygen has the rate law \( \text{rate} = k[\text{N}_2\text{O}]^2 \).
**Constants:**
- \( k = 1.1 \times 10^{-3} \text{ /M s at 565°C} \).
**Given:**
- Initial concentration of N₂O: \( 0.15 \text{ M} \).
- Time: \( 30 \text{ minutes (1800 seconds)} \).
**Question:**
What is the concentration after 30 minutes (1800 seconds) at 565°C?
**Options:**
a. \( 9.5 \times 10^{-3} \text{ M} \)
b. \( 2.2 \times 10^{-3} \text{ M} \)
c. \( 0.029 \text{ M} \)
d. \( 0.12 \text{ M} \)
e. \( 0.46 \text{ M} \)
---
To approach this question, you'll need to use the second-order rate law integrated form:
\[
\frac{1}{[\text{N}_2\text{O}]_t} = \frac{1}{[\text{N}_2\text{O}]_0} + kt
\]
Where:
- \([\text{N}_2\text{O}]_t\) is the concentration at time \( t \).
- \([\text{N}_2\text{O}]_0\) is the initial concentration.
- \( k \) is the rate constant.
- \( t \) is the time.
Plug in the known values to solve for \([\text{N}_2\text{O}]_t\):
\[
\frac{1}{[\text{N}_2\text{O}]_t} = \frac{1}{0.15 \text{ M}} + (1.1 \times 10^{-3} \text{ /M s}) (1800 \text{ s})
\]
Now calculate the right-hand side:
\[
\frac{1}{0.15} + (1.1 \times 10^{-3} \times 1800)
\]
\[
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48deb947-3d41-43de-ac43-bef03e01cc59%2Fc0106aa4-9344-4392-a844-5220a5a6224a%2F03we4dj.png&w=3840&q=75)
Transcribed Image Text:**Decomposition of N₂O - Second Order Reaction**
**Question:**
The decomposition of N₂O(g) to nitrogen and oxygen has the rate law \( \text{rate} = k[\text{N}_2\text{O}]^2 \).
**Constants:**
- \( k = 1.1 \times 10^{-3} \text{ /M s at 565°C} \).
**Given:**
- Initial concentration of N₂O: \( 0.15 \text{ M} \).
- Time: \( 30 \text{ minutes (1800 seconds)} \).
**Question:**
What is the concentration after 30 minutes (1800 seconds) at 565°C?
**Options:**
a. \( 9.5 \times 10^{-3} \text{ M} \)
b. \( 2.2 \times 10^{-3} \text{ M} \)
c. \( 0.029 \text{ M} \)
d. \( 0.12 \text{ M} \)
e. \( 0.46 \text{ M} \)
---
To approach this question, you'll need to use the second-order rate law integrated form:
\[
\frac{1}{[\text{N}_2\text{O}]_t} = \frac{1}{[\text{N}_2\text{O}]_0} + kt
\]
Where:
- \([\text{N}_2\text{O}]_t\) is the concentration at time \( t \).
- \([\text{N}_2\text{O}]_0\) is the initial concentration.
- \( k \) is the rate constant.
- \( t \) is the time.
Plug in the known values to solve for \([\text{N}_2\text{O}]_t\):
\[
\frac{1}{[\text{N}_2\text{O}]_t} = \frac{1}{0.15 \text{ M}} + (1.1 \times 10^{-3} \text{ /M s}) (1800 \text{ s})
\]
Now calculate the right-hand side:
\[
\frac{1}{0.15} + (1.1 \times 10^{-3} \times 1800)
\]
\[
=
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