At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.0618 M 2NH3 (g) → N₂ (g) + 3H₂(g) Suppose a vessel contains NH3 at a concentration of 1.25 M. Calculate the concentration of NH3 in the vessel 68.0 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. OM S ? 0.9 X
At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.0618 M 2NH3 (g) → N₂ (g) + 3H₂(g) Suppose a vessel contains NH3 at a concentration of 1.25 M. Calculate the concentration of NH3 in the vessel 68.0 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. OM S ? 0.9 X
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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![### Second-Order Reaction Kinetics
At a certain temperature, the reaction described follows second-order kinetics with a rate constant of \(0.0618 \, M^{-1} \, s^{-1}\):
\[2NH_3(g) \rightarrow N_2(g) + 3H_2(g)\]
Suppose a vessel contains \(NH_3\) at a concentration of \(1.25 \, M\). Calculate the concentration of \(NH_3\) in the vessel 68.0 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
**Input:**
\[ \boxed{\,\,\,\,\,\,} \, M \]
For more assistance:
\[ \square \times 10 \, \boxed{x} \, \boxed{\Rightarrow} \, \boxed{?} \]
**Explanation:**
This problem presents a second-order reaction where the rate law is given by:
\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]
- \([A]\) is the concentration of \(NH_3\) at time \(t\).
- \([A_0]\) is the initial concentration of \(NH_3\) (1.25 M).
- \(k\) is the rate constant (0.0618 \(M^{-1} s^{-1}\)).
- \(t\) is the time (68.0 seconds).
### Calculation Steps
1. Plug in the values into the second-order rate equation:
\[
\frac{1}{[A]} - \frac{1}{1.25} = (0.0618)(68.0)
\]
2. Solve for \( [A] \):
\[
\frac{1}{[A]} = \frac{1}{1.25} + (0.0618 \times 68.0)
\]
3. Simplify:
\[
\frac{1}{[A]} = 0.8 + 4.2024 = 5.0024
\]
4. Take the reciprocal to find the concentration at \( t = 68.0 \) seconds:
\[
[A] = \frac{1}{5.0024} \approx 0.2 \, M](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F550116fe-4392-44b3-9444-167bfb5405dc%2Fd634c23f-0374-4fbc-8e97-8010e4f62ba8%2Fln0q58_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Second-Order Reaction Kinetics
At a certain temperature, the reaction described follows second-order kinetics with a rate constant of \(0.0618 \, M^{-1} \, s^{-1}\):
\[2NH_3(g) \rightarrow N_2(g) + 3H_2(g)\]
Suppose a vessel contains \(NH_3\) at a concentration of \(1.25 \, M\). Calculate the concentration of \(NH_3\) in the vessel 68.0 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
**Input:**
\[ \boxed{\,\,\,\,\,\,} \, M \]
For more assistance:
\[ \square \times 10 \, \boxed{x} \, \boxed{\Rightarrow} \, \boxed{?} \]
**Explanation:**
This problem presents a second-order reaction where the rate law is given by:
\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]
- \([A]\) is the concentration of \(NH_3\) at time \(t\).
- \([A_0]\) is the initial concentration of \(NH_3\) (1.25 M).
- \(k\) is the rate constant (0.0618 \(M^{-1} s^{-1}\)).
- \(t\) is the time (68.0 seconds).
### Calculation Steps
1. Plug in the values into the second-order rate equation:
\[
\frac{1}{[A]} - \frac{1}{1.25} = (0.0618)(68.0)
\]
2. Solve for \( [A] \):
\[
\frac{1}{[A]} = \frac{1}{1.25} + (0.0618 \times 68.0)
\]
3. Simplify:
\[
\frac{1}{[A]} = 0.8 + 4.2024 = 5.0024
\]
4. Take the reciprocal to find the concentration at \( t = 68.0 \) seconds:
\[
[A] = \frac{1}{5.0024} \approx 0.2 \, M
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